Answer:
The balanced chemical equation is:
2As2O3 + 3C → 3CO2 + 4As
To find out how many grams of carbon dioxide can be produced, we need to use stoichiometry.
First, we need to determine which reactant is limiting. We can do this by calculating the amount of carbon that reacts with As2O3:
1.00 g C × (1 mol C / 12.01 g) × (2 mol As2O3 / 3 mol C) × (197.84 g As2O3 / 1 mol As2O3) = 2.60 g As2O3
This means that only 2.60 g of the As2O3 will react, and the rest will be in excess.
Now we can use the balanced equation to calculate the amount of CO2 that will be produced:
2 mol As2O3 : 3 mol CO2
2.60 g As2O3 × (1 mol As2O3 / 197.84 g) × (3 mol CO2 / 2 mol As2O3) × (44.01 g CO2 / 1 mol CO2) = 3.56 g CO2
Therefore, 3.56 grams of carbon dioxide can be produced.
If 44. 0 grams of sodium reacts with 10. 0 grams of chlorine gas, how many grams of sodium chloride could potentially be formed?
2 na(s) + cl2(g) ⟶2 nacl(s)
If 44.0 grams of sodium reacts with 10.0 grams of chlorine gas, 54.0 grams of sodium chloride could potentially be formed in the reaction: 2 Na(s) + Cl₂(g) ⟶ 2 NaCl(s).
1. Calculate the moles of sodium and chlorine:
- moles of Na = mass (g) / molar mass = 44.0 g / 22.99 g/mol = 1.91 mol
- moles of Cl₂ = mass (g) / molar mass = 10.0 g / 70.90 g/mol = 0.141 mol
2. Determine the limiting reactant by dividing the moles of each reactant by their stoichiometric coefficients:
- Na: 1.91 mol / 2 = 0.955
- Cl₂: 0.141 mol / 1 = 0.141
3. Since the value for Cl₂ is lower, chlorine gas is the limiting reactant.
4. Calculate the moles of NaCl produced using the stoichiometry of the reaction:
- moles of NaCl = moles of Cl₂ × (2 moles of NaCl / 1 mole of Cl₂) = 0.141 × 2 = 0.282 mol
5. Calculate the mass of NaCl produced:
- mass of NaCl = moles × molar mass = 0.282 mol × 58.44 g/mol = 54.0 g
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What is the molality of a solution of naphthalene dissolved in chloroform if the solution has a boiling point of 63. 2 °C?
The molality of the naphthalene solution in chloroform is approximately 0.551 mol/kg.
To calculate the molality of a solution of naphthalene dissolved in chloroform, we need to use the boiling point elevation formula: ΔTb = Kb * molality, where ΔTb is the change in boiling point, Kb is the boiling point elevation constant, and molality is the moles of solute per kilogram of solvent.
First, we need to find the change in boiling point (ΔTb) by subtracting the normal boiling point of chloroform from the given boiling point. The normal boiling point of chloroform is 61.2°C. Therefore, ΔTb = 63.2°C - 61.2°C = 2.0°C.
Next, we need to find the boiling point elevation constant (Kb) for chloroform. The Kb value for chloroform is 3.63 °C/kg/mol.
Now, we can use the boiling point elevation formula to solve for molality:
2.0°C = 3.63 °C/kg/mol * molality
Rearranging the equation and solving for molality, we get:
molality = 2.0°C / 3.63 °C/kg/mol = 0.551 mol/kg
Therefore, the molality of the naphthalene solution in chloroform is approximately 0.551 mol/kg.
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Translate the following balanced chemical equation into words.
PCl5(s) + 4H2O(l) → H3PO4(aq) + 5HCl(aq)
A. Phosphorus pentachloride and water yield phosphoric acid and hydrochloric acid.
B. Phosphorus pentachloride and phosphoric acid yield water and hydrochloric acid.
C. Phosphorus pentachloride and water yield phosphorous acid and chloric acid.
D. Phosphorus hexachloride and water yield phosphoric acid and hydrochloric acid.
Translating the given balanced chemical equation into words :A.)Phosphorus pentachloride and water yield phosphoric acid and hydrochloric acid.
What is Phosphorus pentachloride?Phosphorus pentachloride and water react to yield phosphoric acid and hydrochloric acid. Balanced chemical equation shows that for every one mole of PCl₅ and four moles of H₂O that react, one mole of H₃PO₄ and five moles of HCl are produced.
Phosphorus pentachloride (PCl₅) is a chemical compound composed of one phosphorus atom and five chlorine atoms. It is yellowish-white crystalline solid that is highly reactive and can decompose violently when exposed to water or moist air.
PCl₅ is primarily used as a chlorinating agent in organic chemistry, where it is used to convert alcohols, carboxylic acids, and other functional groups into the corresponding chlorides.
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Compound t is a white crystalline solid.
naoh ca)
a neat
when a sample of t was mixed with aqueous sodium hydroxide and heated, a pungent smelling
gas was produced which turned damp red litmus paper blue. this same gas produced dense
white smoke with hydrogen chloride gas. hcl cg)
further testing of a solution of t with barium chloride solution produced a dense white precipitate
which did not dissolve when dilute hydrochloric acid was added to the mixture.
wh3
nh₃ + ball → nh y cut ba
what is the identity of compound t?
a
ammonium carbonate
b ammonium sulfate
csodium carbonate
d
sodium sulfate
The compounds A, B and C are ammonium chloride, ammonia gas and silver chloride respectively.
Ammonium chloride, is a white crystalline solid which is soluble in water. On heating with sodium hydroxide it will produce ammonia gas, which is a colorless gas and has a odor or pungent smell.
So, Ammonia gas will turn red litmus into blue as it's pH is 11.6.
When Ammonium chloride reacts with silver nitrate in presence of dilute Nitric acid , it produces Silver chloride and Ammonium nitrate
AgCl is soluble in Ammonia because it can form complexes which
makes it behave like an ion, making it soluble.
Therefore, Compound A is Ammonium chloride,
B is ammonia gas
and C is Silver chloride.
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The complete question is
A white solid A when heated with sodium hydroxide solution gives a pungent gas B which turns red litmus paper blue. The solid when dissolved in dilute nitric acid is treated with Silver nitrate solution to give white precipitate C, which is soluble in ammonia.
(A) What are the substance A, B, and C?
the acid-dissociation constant for benzoic acid (c6h5cooh) is 6.3 x 10-5. calculate the equilibrium concentrations of h3o c6h5coo-, and c6h5cooh in the solution if the initial concentration of c6h5cooh is 0.050 m.
At equilibrium, the concentrations of [tex]H_{3} O+, C_{6} H_{5}COO[/tex]-, and [tex]C_{6} H_{5}COO[/tex] in the solution will be 0.038 M, 0.038 M, and 0.012 M, respectively.
First, we can write the chemical equation for the dissociation of benzoic acid in water as follows: [tex]C_{6}H5COOH + H_{2}O[/tex] ⇌[tex]C_{6}H_{5}COO- + H_{3}O[/tex]
The acid dissociation constant, Ka, is given as 6.3 × 10^-5.
[tex]Ka = [C_{6}H_{5}COO-][H_{3}O+] / [C_{6}H_{5}COOH][/tex]
We can assume that the initial concentration of [tex]C_{6}H_{5}COOH[/tex] is equal to its concentration at equilibrium, x. Thus, at equilibrium:
[tex][C_{6}H_{5}COOH] = x M \\[/tex]
[tex][C_{6}H_{5}COO-] = y M \\[/tex]
[tex][H_{3}O+] = y M\\[/tex]
Using the equilibrium expression and the given value of Ka, we can solve for the values of x and y:
[tex]Ka = [C_{6}H_{5}COO-][H_{3}O+] / [C_{6}H_{5}COOH]\\6.3 * 10^-5 = y^2 / x[/tex]
Since we know that the initial concentration of benzoic acid is 0.050 M, we can write: [tex]x + y = 0.050 M[/tex]
Now we have two equations and two unknowns. Solving for x and y:
[tex]x = 0.012 M\\y = 0.038 M[/tex]
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An aqueous solution of sodium acetate, nach3coo, is made by dissolving 13.8 grams of sodium acetate in sufficient water in a 300. ml volumetric flask, and then adding enough water to fill the flask to the mark. what is the weight/volume percentage of sodium acetate in the solution?
The weight/volume percentage of sodium acetate in the solution is 4.6%.
The weight/volume percentage of sodium acetate in the solution can be calculated using the formula:
Weight/volume percentage = (Weight of solute ÷ Volume of solution) x 100%
In this case, the weight of sodium acetate is 13.8 grams and the volume of solution is 300 ml.
Therefore,
Weight/volume percentage = (13.8 g ÷ 300 ml) x 100%
Weight/volume percentage = 0.046 x 100%
Weight/volume percentage = 4.6%
Therefore, the weight/volume percentage of sodium acetate in the solution is 4.6%.
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A sample of gas initially at 1. 4 atm and occupies 720 ml whats the final pressure in atm when the volume changes to 820 mL?
The final pressure of the gas when the volume changes from 720 mL to 820 mL is approximately 1.22 atm.
To solve this problem, we can use Boyle's Law, which states that the product of the initial pressure and volume (P1V1) is equal to the product of the final pressure and volume (P2V2) for a gas at a constant temperature:
P1V1 = P2V2
Given the initial pressure (P1) is 1.4 atm and the initial volume (V1) is 720 mL, we need to find the final pressure (P2) when the volume (V2) changes to 820 mL. Rearrange the formula to solve for P2:
P2 = P1V1 / V2
Substitute the given values:
P2 = (1.4 atm × 720 mL) / 820 mL
P2 ≈ 1.22 atm
Therefore, the final pressure of the gas is approximately 1.22 atm.
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What mass in grams of sucrose must be dissolved in 2000 grams of water to make a 0. 1m solution?
We need to dissolve 6.85 grams of sucrose in 2000 grams of water to make a 0.1 M solution.
To calculate the mass of sucrose needed to make a 0.1 molar solution in 2000 grams of water, we need to use the formula:
[tex]m = n *M * MW[/tex]
Step 1: Calculate the number of moles of sucrose needed
Molarity (M) = 0.1 mol/L
volume of solution = 2000 grams of water ÷ density of water = 2000 mL
We need to calculate the number of moles of sucrose that would be present in 2000 mL of a 0.1 M solution:
moles of solute (n) = [tex]M * V = 0.1 mol/L *2.0 L = 0.2 moles[/tex]
Step 2: Calculate the mass of sucrose needed
Molecular weight of sucrose is 342.3 g/mol.
We can use the formula:
[tex]m = n * M * MW \\m = 0.2 moles *0.1 mol/L * 342.3 g/mol = 6.85 g[/tex]
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Balance the Following Equations:
Instruction: While balancing equation write the physical state of
reactants and products as well as any reaction conditions.
1) CuSO4 + KI →Cu2I2 + K2SO4 + I2
2) NH3 + O2 →NO + H2O
3)Fe2O3 + CO → Fe + CO2
4) Cu + AgNO3 → Cu(NO3)2 + Ag
5) Pb(NO3)2 + H2SO4 → PbSO4 + HNO3
6) CaCO3 + HCl → CaCl2 + H2O(l) + CO2
7)MnO2 + HCl → MnCl2 + H2O + Cl2
I will report any comments that are not appropriate for the question asked or simply typed something for the points. Only answer if u know
While balancing equation write the physical state of reactants and products as well as any reaction conditions.
What is reactants ?Reactants are the substances that are present at the start of a chemical reaction. They are typically the substances that are used up during the reaction and are converted into different products. Reactants are usually written on the left side of a chemical equation, while the products are written on the right side. Reactants are essential components of any chemical reaction and are essential in order for the reaction to take place. Reactants are also known as substrates or starting materials.
Balancing the Following Equations:
1) CuSO4(s) + 2KI(aq) → Cu2I2(s) + K2SO4(aq) + I2(g)
2) 2NH3(g) + O2(g) → 2NO(g) + 2H2O(g)
3) 3Fe2O3(s) + 4CO(g) → 6Fe(s) + 3CO2(g)
4) Cu(s) + 2AgNO3(aq) → Cu(NO3)2(aq) + 2Ag(s)
5) 2Pb(NO3)2(aq) + H2SO4(aq) → PbSO4(s) + 2HNO3(aq)
6) CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
7) 2MnO2(s) + 4HCl(aq) → 2MnCl2(aq) + 2H2O(l) + Cl2(g)
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Calculate E0, E, and ΔG for the following cell reactions (a) Mg(s) + Sn2+(aq) ⇌ Mg2+(aq) + Sn(s) where [Mg2+] = 0. 055 M and [Sn2+] = 0. 055 M
The E° for cell reaction is - 2.37 V and 2.23 V and E for cell reaction = 2.22V and ΔG = - 428.39kJ/mol.
The formula for solving the equation for given cell is as follows :
E°cell , Ecell and Δ[tex]G_{rnx}[/tex]
The standard cell potential is the potential of cell at standard condition of 1MConcentration and pressure 1 atm E°cell
calculation :
E°cell = E° cathode - E° anode it is calculated using the Nernst equation which is discussed below :
Ecell = E°cell -- [tex]\frac{RT}{nF}[/tex] 1n K = E°cell -- [tex]\frac{0.0591}{n}[/tex]log [tex]\frac{Products}{Reactants}[/tex]
Here, F is the Faraday's constant, R is the gas constant, T is the temperature, and n is the number of transferred electrons. K is the equilibrium constant.
The Gibbs free energy is the greatest work that is finished by a framework . The standard cell potential is without like energy by the recipe as follows;and F is Faraday's steady.
A system's maximum amount of work is referred to as its Gibbs free energy. The standard cell potential is connected with the free energy by the recipe as follows: Δ G = -n F Ecell
Here, E cell is cell potential
Δ G is the free energy n is the quantity of electrons moved and F is Faraday's steady.
The given net cell equation is as follows: Mg + Sn²⁺⇒ Mg²⁺ + SnOxidation :
Mg ⇒ Mg ²⁺ + 2e⁻ E⁰anode = - 2.37 V
Reduction:Sn²⁺ + 2e⁻⇒ Sn E⁰
So, cathode = - 0.14V
The standard cell potential is calculated as follows:E⁰ cell = - 0.14 V- (- 2.37 V ) = 2.23 V
The half reaction potentials for the oxidation and reduction are determined. They are subbed in the equation and the standard cell potential is determined.
Number of electrons transferred , n = 2 ,[Mg²⁺] = 0.055M , [ Sn²⁺ ] = 0.030 M The Nernst equation for reaction :
Ecell = E °cell = [tex]\frac{0.0591}{n}[/tex]log Mg ²⁺ / Sn²⁺
The cell potential for reaction is :
Ecell = 2.23V - [tex]\frac{0.0591}{2}[/tex]log[tex]\frac{0.055M}{0.030M}[/tex]= 2.22V
The values are substituted for the reaction calculated here in the Nernst equation and cell potential.
Calculation for the free energy for reaction ,
] Δ[tex]G_{rxn}[/tex] = -nFE cell
= - 2 × 96485 C/ mol ×2.22 V
= --428393J/mol × [tex]\frac{1KJ}{1000J}[/tex] = - 428.39kJ/mol
The cell potential for the response is subbed in the recipe and free energy for the response is determined
Nernst equation :
The standard electrode potential, absolute temperature, the number of electrons involved in the redox reaction, and activities (often approximated by concentrations) of the chemical species undergoing reduction and oxidation, respectively, can all be used to calculate the reduction potential of a half-cell or full cell reaction using the Nernst equation, a chemical thermodynamic relationship.
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If a 2000g block of metal lost 3120 j of heat energy is cooled from 212 c to 200 c, what is the specific heat of the metal
Explanation:
3120 j / (2000 g * (212-200 C) ) = .13 j /( g C)
A quantity of gas is at a temperature of 20°C, a pressure of 760 torr and occupies a volume of 2. 00 L. If the pressure is changed to 730 torr, what will be the new volume? Assume that there is no temperature change
The new volume of the gas, assuming constant temperature and a change in pressure from 760 torr to 730 torr, is 2.09 L.
Using the Boyle's Law equation,
P₁V₁ = P₂V₂,
where P is pressure and V is volume, we can solve for V₂ by plugging in the given values in the equation:
(760 torr)(2.00 L) = (730 torr)(V₂)
Solving for V₂, we get:
V₂ = (760 torr)(2.00 L) / (730 torr) = 2.09 L
Therefore, the new volume of the gas is 2.09 L.
This result makes sense because according to Boyle's Law, as pressure decreases, volume increases proportionally, assuming a constant temperature.
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Consider the chemical equation for the combustion of ammonia: 4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(g) Which statement provides the correct and standard interpretation of the chemical equation in terms of the volume of gases at STP? A. 4 L of NH3(g) react with 7 L of O2(g) to produce 4 L of NO2(g) and 6 L of H2O(g). B. 12 L of NH3(g) react with 14 L of O2(g) to produce 8 L of NO2(g) and 6 L of H2O(g). C. 22.4 L of NH3(g) react with 22.4 L of O2(g) to produce 22.4 L of NO2(g) and 22.4 L of H2O(g). D. 89.6 L of NH3(g) react with 156.8 L of O2(g) to produce 89.6 L of NO2(g) and 134.4 L of H2O(g).
The correct interpretation is 89.6 L of [tex]NH_{3}[/tex](g) react with 156.8 L of [tex]O_{2}[/tex](g) to produce 89.6 L of [tex]NO_{2}[/tex](g) and 134.4 L of [tex]H_{2} O[/tex](g).
What is the correct interpretation?We know that one mole of a gas does occupy 22.4 L. We can now use this to obtain the number of volumes of the gas based on the stoichiometric coefficient that has been given in the problem.
Molar volume of a gas refers to the volume occupied by one mole of a gas at a specific temperature and pressure. This value is useful in many applications of chemistry, such as in stoichiometric calculations and the determination of gas densities.
Using the stoichiometric coefficients we can see that the volume of the gases are;
Ammonia - 89.6 L
Oxygen - 156.8 L
Nitrogen dioxide - 89.6 L
Water - 134.4 L
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A sample of bromine gas occupies 2. 65 L at 1. 20 atm. What pressure (in kPa) would this sample of gas exert in 1. 50L container at the same temperature? show work
ASAP PLEASE
We can use the ideal gas law to calculate the pressure of the bromine gas in the 1.5 L container. The ideal gas law is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
Since we know the temperature and the volume, we can rearrange the ideal gas law to solve for P, the pressure. We can use the pressure and volume from the first container to calculate the number of moles. Plugging in all of the known values, we get:
P1V1 = nRT
n = P1V1/RT
P2 = (P1V1/RT) * (V2/V1)
Using the values from the question, we get:
P2 = (1.20 atm * 2.65 L)/(0.08206 L·atm·mol-1·K-1 * 298 K) * (1.50 L/2.65 L)
This gives us a pressure of 1.04 atm in the 1.5 L container, which is equal to 1040 kPa.
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What mass in grams of hydrogen gas is produced if 20. 0 mol of zn are added to excess hydrochloric acid according to the equation
zn(s) +2hcl(aq) --> zncl₂(aq) + h₂(g)?
First, we need to use stoichiometry to find out how many moles of hydrogen gas are produced. From the balanced chemical equation, we can see that for every 1 mole of zinc (Zn), 1 mole of hydrogen gas (H2) is produced. Therefore, if we have 20.0 mol of Zn, we will also produce 20.0 mol of H2.
Next, we can use the formula for the mass of a gas:
mass = molar mass x number of moles
The molar mass of hydrogen gas is approximately 2.02 g/mol. Therefore, the mass of 20.0 mol of hydrogen gas would be:
mass = 2.02 g/mol x 20.0 mol
mass = 40.4 g
So, 40.4 grams of hydrogen gas are produced when 20.0 mol of Zn are added to excess hydrochloric acid.
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Fe(NO3)2 + Al = Fe + Al(NO3)3 identify what's being oxidized and reduced
In the given chemical equation:
Fe(NO3)2 + Al → Fe + Al(NO3)3
Iron (Fe) is being reduced because it is gaining electrons and its oxidation state is decreasing from +2 to 0 (elemental state).
Aluminum (Al) is being oxidized because it is losing electrons and its oxidation state is increasing from 0 (elemental state) to +3.
Therefore, Fe(NO3)2 is the oxidizing agent, and Al is the reducing agent in this reaction.
How dose the angle of a light beam affect the intensity and the amount of light reflected or transmitted?
The angle of a light beam affects the intensity and the amount of light reflected or transmitted through a process known as the "angle of incidence." When a light beam strikes a surface, the angle between the incoming light beam and the surface is called the angle of incidence. This angle plays a crucial role in determining the amount of light reflected or transmitted.
When the angle of incidence is small (light beam nearly perpendicular to the surface), more light is transmitted through the surface, and less is reflected. As the angle of incidence increases (light beam more parallel to the surface), the amount of light reflected also increases, while the intensity of the transmitted light decreases.
This phenomenon occurs due to the interaction of light with the surface material, which can either absorb, transmit, or reflect the incoming light, depending on the angle of incidence and the material's properties. The angle at which the light beam is incident on the surface also affects the intensity of the reflected light.
At a specific angle, called the "critical angle," the light beam is no longer transmitted but is entirely reflected, a phenomenon called "total internal reflection." The critical angle depends on the refractive indices of the two materials at the interface. When the angle of incidence is greater than the critical angle, all the light is reflected, and none is transmitted.
In summary, the angle of a light beam significantly influences the intensity and the amount of light reflected or transmitted by a surface. The angle of incidence determines the amount of light reflection, with a smaller angle leading to more transmission and a larger angle leading to increased reflection. The critical angle, in particular, plays a crucial role in determining the behavior of the light beam at the surface.
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Nitrogen oxide (NO) has been found to be a key component in many biological processes. It also can react with oxygen to give the brown gas NO2. When one mole of NO reacts with oxygen, 57. 0 kJ of heat are evolved. What is ΔH when 8. 00 g of nitrogen oxide react?
NO(g) + ½O2(g) → NO2(g) ΔH = –57. 0 kJ
The enthalpy change when 8.00 g of nitrogen oxide react is -15.162 kJ for the given chemical reaction.
The molar mass of NO = 30.01 g/mol
8.00 g of NO = 8.00 g / 30.01 g/mol
8.00 g of NO = 0.266 mol of NO
Heat rejection = 57. 0 kJ
Here, 1 mole of NO reacts with 1/2 mole of Oxygen to produce 1 mole of [tex]NO_{2}[/tex]
The amount of Oxygen required for 0.266 mol of NO is calculated as:
The amount of Oxygen = 0.266 mol NO x (1/2) mol [tex]O_{2}[/tex] / 1 mol NO
The amount of Oxygen required = 0.133 mol [tex]O_{2}[/tex]
The heat reaction will be:
-57.0 kJ/mol x 0.266 mol NO = -15.162 kJ
Therefore, we can conclude that the enthalpy change is -15.162 kJ.
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What is the concentration of a solution that contains 25. 0 g NaOH in 500 cm3
of water?
The concentration of the solution containing 25.0 g NaOH in 500 cm³ of water is approximately 1.25 M (moles per liter).
To find the concentration of a solution containing 25.0 g NaOH in 500 cm³ of water, follow these steps:
1. Convert grams of NaOH to moles. The molar mass of NaOH is approximately 40 g/mol (Na = 23 g/mol, O = 16 g/mol, H = 1 g/mol).
25.0 g NaOH × (1 mol NaOH / 40 g NaOH) ≈ 0.625 mol NaOH
2. Convert the volume of water from cm³ to liters (L).
500 cm³ × (1 L / 1000 cm³) = 0.5 L
3. Calculate the concentration of the solution in moles per liter (M).
Concentration = moles of solute/volume of solvent (in liters)
Concentration = 0.625 mol NaOH / 0.5 L ≈ 1.25 M
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The pressure at the bottom of a lake is 2. 35 atm. If water saturated with oxygen
(concentration 0,34 g/L) is carried by a current to to a depth where the solubility
of oxygen is 0. 21 g/L, what is the pressure of the water.
The pressure at the bottom of a lake is given as 2.35 atm, and we are asked to find the pressure of the water. Since water is the fluid in question, we can assume that it is incompressible and that its density is constant. To find the pressure of the water, we can use the following formula:
Pressure = Density x Acceleration due to gravity x Height
Here, the height refers to the depth of the lake, which we can assume to be the same as the height of the water column. The acceleration due to gravity is a constant, and the density of water is given as 0.21 g/L.
Substituting these values in the formula, we get:
Pressure = 0.21 g/L x 9.8 m/s^2 x Depth
Since the pressure at the bottom of the lake is given as 2.35 atm, we can convert this to SI units using the conversion factor:
1 atm = 101325 Pa
Therefore, 2.35 atm = 2.35 x 101325 Pa = 2.38 x 10^5 Pa
Substituting this value in the formula, we can solve for the depth:
2.38 x 10^5 Pa = 0.21 g/L x 9.8 m/s^2 x Depth
Depth = 114.7 m
Therefore, the pressure of the water at this depth is:
Pressure = 0.21 g/L x 9.8 m/s^2 x 114.7 m = 240.3 kPa
In conclusion, the pressure of the water at the bottom of the lake is 240.3 kPa. This is the pressure exerted by the water column due to its weight, and it is in addition to the atmospheric pressure. Understanding the pressure of fluids is important in many fields, such as hydrology, engineering, and physics.
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What is the normal boiling point of a 3.45mol solution of kbr that has density of 1.10gml?(ka for h2o is 0.512°c kg/mole)
The normal boiling point of the 3.45 mol solution of KBr is 104.7384°C.
The normal boiling point of a 3.45 mol solution of KBr with a density of 1.10 g/mL can be calculated using the formula:
ΔT = Kb * molality
where ΔT is the boiling point elevation, Kb is the molal boiling point elevation constant for water (0.512°C kg/mol), and molality is the number of moles of solute per kilogram of solvent.
First, we need to calculate the mass of the solvent (water) required to dissolve 3.45 mol of KBr. The molar mass of KBr is 119 g/mol, so 3.45 mol of KBr would weigh 409.55 g.
Since the density of the solution is given as 1.10 g/mL, the volume of the solution is:
V = m / ρ = 409.55 g / 1.10 g/mL = 372.32 mL
So, the mass of the water is:
mH2O = V * ρH2O = 372.32 mL * 1 g/mL = 372.32 g
The molality of the solution can be calculated as follows:
molality = moles of solute / mass of solvent (in kg) = 3.45 mol / 0.37232 kg = 9.27 mol/kg
Substituting the values in the formula for boiling point elevation:
ΔT = 0.512°C kg/mol * 9.27 mol/kg = 4.7384°C
The normal boiling point of pure water is 100°C, so the boiling point of the KBr solution would be:
Boiling point = 100°C + ΔT = 100°C + 4.7384°C = 104.7384°C
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8.
What is true concerning the concentrations of H3O+ and OH- ions in the acidic solutions?
There are more H3O+
There are fewer H3O+
They are in equal amounts
There are more hydroxide ions
9.
What is true concerning the concentrations of H3O+ and OH- ions in the basic solutions?
There are more H3O+
There are fewer H3O+
They are in equal amounts
There are more hydronium ions
calculate the molality of a solition with 85 g of KOH added to 590. g of water
A Gas Thermometer Measures Temperature By Measuring The Pressure Of A Gas Inside The Fixed Volume Container. A Thermometer Reads A Pressure Of 780 mmHg at 0C. What Is The Temperature When The Thermometer Reads A Pressure Of 800 mmHg?
The temperature when the thermometer reads a pressure of 800 mmHg is approximately 282.2 K (or 9.1 °C).
To solve this problem, we can use the ideal gas law:
PV = nRT
We can use this equation to calculate the temperature of the gas when the pressure is 800 mmHg.
First, we need to convert the pressures from mmHg to atm, since R is in units of L·atm/K·mol.
1 atm = 760 mmHg
780 mmHg = 1.026 atm
800 mmHg = 1.053 atm
Next, we can set up a ratio of the two pressures and temperatures:
P1/T1 = P2/T2
[tex](1.026 atm) / (273.15 K) = (1.053 atm) / T2[/tex]
Solving for T2, we get:
[tex]T2 = (1.053 atm) / (1.026 atm/273.15 K) \\T2 = 282.2 K[/tex]
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Draw all possible lewis structures (including resonance structures for methyl azide (ch3n3 using lewis structure rules. One or more of your structures may seem unstable or unlikely; include them in your answer as long as they do not violate lewis structure rules. For each resonance structure, assign formal charges to all atoms that have formal charge. Draw all possible lewis dot structure for methyl azide. Include all hydrogen atoms and nonbonding electrons. Show the formal charges of all atoms
2) If a solid line represents a covalent bond and a dotted line represents intermolecular attraction, which of the choices shows a hydrogen bond?
H−H
H3N⋅⋅⋅⋅⋅⋅H−O−H
H
3
N
⋅
⋅
⋅
⋅
⋅
⋅
H
−
O
−
H
H4C⋅⋅⋅⋅⋅⋅H−F
H
4
C
⋅
⋅
⋅
⋅
⋅
⋅
H
−
F
H2O⋅⋅⋅⋅⋅⋅H−CH3
H2O⋅⋅⋅⋅⋅⋅H−CH3 shows intermolecular attraction between water molecules and methane molecules, but not a hydrogen bond specifically.
A hydrogen bond is a type of intermolecular attraction that occurs when a hydrogen atom is bonded to an electronegative atom such as nitrogen, oxygen, or fluorine, and it is attracted to another electronegative atom in a nearby molecule. This attraction is represented by a dotted line.
Looking at the choices provided, the only option that shows a hydrogen bond is H3N⋅⋅⋅⋅⋅⋅H−O−H. In this molecule, the hydrogen atom in the H−O−H group is bonded to the highly electronegative oxygen atom, and it forms a hydrogen bond with the lone pair of electrons on the nitrogen atom in the H3N group.
The dotted line between the H and N represents the hydrogen bond.
In contrast, the other options do not show a hydrogen bond. H−H represents a simple covalent bond between two hydrogen atoms, while H4C⋅⋅⋅⋅⋅⋅H−F represents a covalent bond between a carbon atom and a fluorine atom, with no electronegative atoms capable of forming a hydrogen bond. H2O⋅⋅⋅⋅⋅⋅H−CH3 shows intermolecular attraction between water molecules and methane molecules, but not a hydrogen bond specifically.
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How many valence electrons does carbon have available for bonding to other atoms?
a. 2
b. 4
c. 6
d. 8
Answer:
4 valence electrons.
Explanation:
Carbon has 4 valence electrons because it is in the 14th group on the Periodic Table.
explain how electrical conductivity can be used to distinguish between magnesium oxide and silicon oxide
Magnesium metal will conduct electricity via mobile electrons whether it is in the solid or liquid state.
Magnesium oxide will not conduct electricity in the solid state as they are no mobile charge carriers.
Molten (liquid) magnesium oxide has mobile ions and these can transfer electrons via mobile ions. This is electrolysis and the compound is turned back into its elements (magnesium and oxygen).
A balloon containing 2. 6 mol hydrogen has a volume of 3. 9 l. More hydrogen is added to the balloon, giving it a volume of 17. 1 l. How many moles of hydrogen were added? show your work
The moles of hydrogen added to the balloon to give it a volume of 17.1 L were 8.8 mol.
Determine the initial ratio of moles to volume:
Initial moles of hydrogen = 2.6 mol
Initial volume of the balloon (V₁) = 3.9 L
Ratio of moles to volume: 2.6 mol / 3.9 L = 0.6667 mol/L
Final volume of the balloon (V₂) = 17.1 L
Calculate the final moles of hydrogen in the balloon using the initial ratio of moles to volume.
Final moles of hydrogen (H₂) = Ratio of moles to volume * V₂
Final moles of hydrogen (H₂) = 0.6667 mol/L * 17.1 L = 11.4 mol
The number of moles of hydrogen added thus are:
Moles of hydrogen added = Final moles of hydrogen - Initial moles of hydrogen
Moles of hydrogen added = 11.4 mol - 2.6 mol = 8.8 mol
So, 8.8 moles of hydrogen were added to the balloon to give it a volume of 17.1 L.
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What volume of dichloromethane (ch2cl2) is produced when 149 liters of methane (ch4) react according to the following reaction? (all gases are at the same temperature and pressure. ) methane (ch4)(g) carbon tetrachloride(g) dichloromethane (ch2cl2)(g)'
The volume of dichloromethane [tex](CH_2Cl_2)[/tex] produced when 149 liters of methane [tex](CH_4)[/tex] react according to the given reaction is approximately 6.224 x [tex]10^5 J/K*m^3[/tex].
The volume of dichloromethane [tex](CH_2Cl_2)[/tex] produced when 149 liters of methane [tex](CH_4)[/tex] react according to the given reaction is not immediately apparent from the reaction stoichiometry.
The balanced equation for the reaction between methane [tex](CH_4)[/tex] and carbon tetrachloride (CCl4) to form dichloromethane [tex](CH_2Cl_2)[/tex] and carbon dioxide (CO2) is:
[tex](CH_4)[/tex] + [tex]CO_2[/tex] → [tex](CH_2Cl_2)[/tex] + [tex]CO_2[/tex]
The balanced equation shows that 1 mole reacts with 1 mole of CCl4 to produce 1 mole of [tex](CH_2Cl_2)[/tex] and 1 mole of [tex]CO_2[/tex].
The volume of the gas can be calculated using the ideal gas law:
PV = nRT
To find the number of moles of gas, we can use the molecular masses of the reactants and products:
Molar mass of [tex](CH_4)[/tex] = 16.04 g/mol
Molar mass of [tex]CCl_4[/tex] = 89.9 g/mol
Molar mass of [tex](CH_2Cl_2)[/tex] = 70.1 g/mol
Molar mass of [tex]CO_2[/tex] = 44.01 g/mol
The number of moles of [tex](CH_4)[/tex] can be calculated from the initial amount of gas:
149 L of CH4 = 149 x 16.04 g/mol = 2432 g
The number of moles of CCl4 can be calculated from the given volume:
149 L of [tex](CH_4)[/tex] + [tex]CCl_4[/tex] → [tex](CH_2Cl_2)[/tex] + [tex]CO_2[/tex]
The volume of the gas is given as 149 L, so the number of moles of [tex]CCl_4[/tex] can be calculated as:
149 L = 149 x 89.9 g/mol = 13,277 g
The number of moles can be calculated from the given volume and the desired amount of product
149 L of [tex](CH_4)[/tex] + [tex]CCl_4[/tex] → [tex](CH_2Cl_2)[/tex] + [tex]CO_2[/tex]
149 L of [tex](CH_4)[/tex] + [tex]CCl_4[/tex] → 149 x 70.1 g/mol + 13,277 g x 1 mol/13.277 g = 43,691 g
V = nRT
V = 43,691 g x 8.314 J/mol·K = 364,617.5 J/K
1 J/K = 1/1000 L·K
Therefore, the volume of the gas is:
V = 364,617.5 J/K x (1/1000 L·K) = 3.646 x 10^4 L
substitute this value for V in the equation for the volume of [tex](CH_2Cl_2)[/tex] :
PV = nRT
PV = 149 x 8.314 J/mol·K x (3.646 x [tex]10^4[/tex] L)
PV = 6.224 x [tex]10^5 J/K*m^3[/tex].
Therefore, The volume of dichloromethane [tex](CH_2Cl_2)[/tex] produced when 149 liters of methane [tex](CH_4)[/tex] react according to the given reaction is approximately 6.224 x [tex]10^5 J/K*m^3[/tex].
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