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Explanation:

That the mass can neither be created nor destroyed in a chemical reaction.

Adding MnO₄⁻² ions will result in making the solution more green in this reaction.

What is Temperature?

Temperature is a measure of the average kinetic energy of the particles in a substance or system. It is typically measured using a scale such as Celsius, Fahrenheit, or Kelvin, and is commonly used in physics, chemistry, and engineering to describe the behavior of materials and systems.

In this reaction, the MnO₂ react with I⁻ ions to form MnO₄⁻¹ ions and MnO₂. The MnO₄⁻¹ ions are green in color, while the MnO₂ is purple in color. Therefore, the overall color of the solution is a mixture of green and purple.

By adding more MnO₄⁻² ions to the solution, the concentration of green MnO₄⁻¹ ions will increase, and this will result in a more green color of the solution. Hence, option A is the correct answer.

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Nuclear fission is a process where a heavy atomic nucleus is split into two or more lighter nuclei, releasing a large amount of energy.

1.Uranium-235 fissioned by neutron:

U-235 + neutron → Kr-92 + Ba-141 + 3 neutrons + energy

2.Plutonium-239 fissioned by neutron:

Pu-239 + neutron → Sr-95 + Zr-139 + 2 neutrons + energy

3.Californium-252 fissioned by neutron:

Cf-252 + neutron → 2 Sm-126 + 3 neutrons + energy.

Nuclear fission is a process in which a heavy atomic nucleus (such as uranium-235, plutonium-239, or californium-252) is split into two or more lighter nuclei (such as krypton-92 and barium-141), along with the release of a large amount of energy in the form of radiation and kinetic energy of the resulting particles. This process is typically initiated by the absorption of a neutron by the heavy nucleus, which causes it to become unstable and split apart.

In the fission reaction, the mass of the products is slightly less than the mass of the reactant nucleus, due to the conversion of some mass into energy according to Einstein's famous equation [tex]E=mc^2[/tex], where E is the energy released, m is the mass lost, and c is the speed of light.

The energy released in nuclear fission reactions is much greater than that released in chemical reactions, making nuclear fission an attractive source of energy for power generation. However, fission reactions can also be dangerous if not properly controlled, as the released energy can cause explosions or release dangerous radiation.

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Answer:

First, we need to convert the toxicity value from mg/kg to mg for a 70 kg adult:

Toxic dose for a 70 kg adult = 12,000 mg/kg x 70 kg = 840,000 mg

Next, we need to calculate the number of moles of 7-UP in one can:

Concentration of 7-UP in one can = 0.012 M

Volume of one can = 330 mL = 0.33 L

Number of moles of 7-UP in one can = concentration x volume = 0.012 mol/L x 0.33 L = 0.00396 mol

Finally, we can calculate the number of cans of 7-UP a 70 kg adult would have to drink to reach a toxic dose:

Number of cans of 7-UP = toxic dose / (number of moles in one can x molecular weight of 7-UP)

Molecular weight of 7-UP = 338.14 g/mol

Number of cans of 7-UP = 840,000 mg / (0.00396 mol x 338.14 g/mol) ≈ 619 cans

Therefore, a 70 kg adult would have to drink approximately 619 cans of 7-UP to reach a toxic dose.

Explanation:

The range of paracetamol dosages expressed in mg per kg body weight for children who weigh between 20 and 30 pounds and are recommended a dosage of 1.5 teaspoons of Carfemol is between 165.5 mg/kg and 248.1 mg/kg.

What is  Paracetamol?

Paracetamol, also known as acetaminophen, is a common over-the-counter pain reliever and fever reducer. It is used to treat mild to moderate pain, such as headaches, menstrual cramps, toothaches, and backaches. Paracetamol is also used to reduce fever, such as in the case of flu or common cold.

To calculate the range of paracetamol dosages expressed in mg per kg body weight for children who weigh between 20 and 30 pounds and are recommended a dosage of 1.5 teaspoons of Carfemol, we need to convert the weight of the child to kilograms and then use the dosage and concentration of paracetamol in Carfemol to calculate the dosage in mg/kg.

Converting 20 pounds to kilograms:

20 pounds = 20 / 2.205 = 9.07 kg

Converting 30 pounds to kilograms:

30 pounds = 30 / 2.205 = 13.61 kg

Using the dosage recommended for a child of 1.5 teaspoons of Carfemol, we can calculate the dosage of paracetamol for a child weighing between 20 and 30 pounds:

Dosage of paracetamol for a child weighing 9.07 kg = 1.5 teaspoons * 75 mg/0.50 teaspoons = 2,250 mg

Dosage of paracetamol for a child weighing 13.61 kg = 1.5 teaspoons * 75 mg/0.50 teaspoons = 2,250 mg

Now, we can calculate the range of paracetamol dosages expressed in mg/kg body weight:

For a child weighing 9.07 kg: 2,250 mg / 9.07 kg = 248.1 mg/kg

For a child weighing 13.61 kg: 2,250 mg / 13.61 kg = 165.5 mg/kg

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Answer: The correct answer is 120.3g

Explanation:

Atomic mass of calcium, Ca = 40.1

Mass of 1 mole of Ca is 40.1g

Mass of 3.00 moles of Ca = 3 x 40.1 = 120.3g

Hope this helps

The energy (in joules) of 1 mole of photons with this frequency is 1.99 x 10⁻¹⁴ J per photon.

Should a substance happen to have a lot of electrons in a higher level, and a lower level is mostly empty  then a photon can cause an electron to transfer from a higher state to a lower one. This change releases energy and creates a new photon, in addition to the one which caused the transfer. This photon can in turn induce more electrons to fall to a lower state.

use formula

The energy of 1 mole of photons with the given frequency can be calculated using the following equation:

Energy (J) = Avogadro's number x Plank's Constant x Frequency

Therefore,

Energy (J) = 6.02 x 10²³ x 6.626 x 10⁻³⁴ x 5.40 x 10⁴

Energy (J) = 1.99 x 10⁻¹⁴ J

Therefore, the energy of 1 mole of photons with a frequency of 5.40 x 10⁴ Hz is 1.99 x 10⁻¹⁴ J, expressed in scientific notation.

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The chemical equation for the decomposition of water is:

[tex]2 H_2O -- > 2 H_2 + O_2[/tex]

To balance this equation, we need to count the number of atoms of each element on both sides of the equation.

On the left side of the equation, we have:

2 hydrogen atoms (2 H₂O)

2 oxygen atoms (2 H₂O)

On the right side of the equation, we have:

2 hydrogen atoms (2 H₂)

2 oxygen atoms (1 O₂)

We can see that the number of hydrogen atoms is already balanced, but the number of oxygen atoms is not. To balance the equation, we need to add a coefficient in front of O2 so that we have the same number of oxygen atoms on both sides.

The balanced equation is:

[tex]2 H_2O -- > 2 H_2 + 1 O_2[/tex]

A compound is broken down into simpler compounds during a decomposition reaction. Different techniques, such as heating, exposure to light, or the inclusion of a catalyst, can be used to produce this reaction.

The reactant component splits into two or more products, which may be elements or compounds, during decomposition. A synthesis reaction, in which less complex substances join to create a more complex compound, is the antithesis of this reaction.

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The given reaction sequence suggests the use of PPh3 followed by BuLi to form a nucleophilic species that can react with 3-pentanone and Br.

The first step involves the formation of a phosphonium ylide intermediate by the reaction of PPh3 with BuLi. In the second step, the ylide acts as a nucleophile and attacks the carbonyl group of 3-pentanone to form a betaine intermediate. The final step involves the reaction of the betaine with Br to give the desired product.

The major organic product formed in this reaction sequence is (E)-1-bromo-3-(phenyl(phenyl)methylidene)butan-1-one. This is because the reaction proceeds via an SN2 mechanism, which leads to the formation of an alkene with an E configuration. The systematic IUPAC name of the product is (E)-1-bromo-3-(phenyl(phenyl)methylidene)butan-1-one.

In this name, "E" indicates the configuration of the double bond, "1-bromo" indicates the presence of a bromine atom on the first carbon, "3-(phenyl(phenyl)methylidene)" indicates the presence of a substituted phenyl group attached to the third carbon, and "butan-1-one" indicates the presence of a ketone group on the first carbon of a four-carbon chain.

What is the major organic product in the following reaction sequence? Type its systematic IUPAC name in the box below.

1) PPh3

2) BuLi

3) 3-pentanone

Br?

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Answer:

To calculate the amount of solute (NaNO3) that will be left after the saturated solution is evaporated to dryness, we need to use the solubility of NaNO3 in water at 30 degrees Celsius. The solubility of NaNO3 in water at 30 degrees Celsius is 88.0 g/100 mL. First, we need to convert the 50 grams of water to milliliters. The density of water at 30 degrees Celsius is 0.996 g/mL. Therefore: 50 g ÷ 0.996 g/mL = 50.2 mL So, we have 50.2 mL of water in which the solubility of NaNO3 is 88.0 g/100 mL. To calculate the amount of NaNO3 that will be left after the solution is evaporated to dryness, we can

Answer:

aluminium reacts with sulfuric acid to produce aluminium sulfate and hydrogen gas according to the following equation: H2SO4 (aq) + Al (s) = Al2 (SO4)3 (aq) + H2 (g).

Answer:Adding heat to a solid will definitely melt it.

Example: let's say we have a pack of Hershey's chocolate we take it and leave it outside in the heat.what will happen to it? it will melt 100%

the majority of solids melt. It depends on how solid it is.

Example 2 scenario: lets say we have a penny (a solid) it will melt but with a much higher temperature

the melting temperature for the penny is 1984.32 °F which is extremely high.

lets go back to the previous chocolate example--->

before the chocolate melted it wasnt as hard as the penny. the melting temperature would be much lower compared to the penny. the melting temperature for chocolate is 85°F-93°F.

Why do harder solids have a higher melting point than less-hard solids?

the reason for this is density. The penny is much dense than the chocolate.

How is a penny more dense than chocolate?

lets say we have chocolate (before it melted it was a solid) we are able to actually bite and chew the chocolate; but you are unable to bite and chew a copper coin. the reason for this is that copper is much more dense than chocolate.

it will NOT evaporate because the temperature for liquid evaporation is 212° F.

I hope this helps!!!

Answer:

0.61 V

Explanation:

a. The two metals in this voltaic cell are cadmium (Cd) and copper (Cu). Cadmium is oxidized (loses electrons) and copper is reduced (gains electrons).

b. Half-reactions:

Cathode: Cu2+(aq) + 2e- → Cu(s)

Anode: Cd(s) → Cd2+(aq) + 2e-

Overall reaction: Cd(s) + Cu2+(aq) → Cd2+(aq) + Cu(s)

c. The anode is where oxidation occurs, so the anode is the cadmium electrode. The cathode is where reduction occurs, so the cathode is the copper electrode.

d. Standard cell potential (Eᵒcell) can be calculated using the standard reduction potentials (Eᵒred) for each half-reaction.

Eᵒcell = Eᵒred(cathode) - Eᵒred(anode)

Eᵒred(Cu2+(aq) + 2e- → Cu(s)) = +0.34 V (from tables)

Eᵒred(Cd(s) → Cd2+(aq) + 2e-) = -0.40 V (from tables)

Eᵒcell = +0.34 V - (-0.40 V) = +0.74 V

e. The standard free energy change (ΔG°) can be calculated from the standard cell potential (Eᵒcell) using the following equation:

ΔG° = -nFEᵒcell

where n is the number of electrons transferred in the balanced equation (2), and F is the Faraday constant (96,485 C/mol).

ΔG° = -2 x 96,485 C/mol x 0.74 V = -141,800 J/mol = -141.8 kJ/mol

f. The cell diagram for this voltaic cell is:

Cd(s) | Cd(NO3)2 (0.10 M) || Cu(NO3)2 (0.20 M) | Cu(s)

g. The Nernst equation can be used to calculate the cell potential (E) under non-standard conditions, where the concentrations of the species involved are not at their standard state. The Nernst equation is:

E = Eᵒcell - (RT/nF) x ln(Q)

where R is the gas constant (8.314 J/mol-K), T is the temperature in Kelvin, n is the number of electrons transferred (2), F is the Faraday constant (96,485 C/mol), and Q is the reaction quotient, which is the ratio of the concentrations of the products over the concentrations of the reactants raised to their stoichiometric coefficients.

At non-standard conditions, the concentrations of the species involved are:

[Cd2+] = 0.10 M

[Cu2+] = 0.20 M

The reaction quotient (Q) is:

Q = ([Cd2+]/[Cu2+])^2 = (0.10 M / 0.20 M)^2 = 0.25

Plugging in the values:

E = 0.74 V - (8.314 J/mol-K x 298 K / (2 x 96,485 C/mol)) x ln(0.25) = 0.61 V

Therefore, the cell potential at non-standard conditions is 0.61 V.

The percent yield of the reaction is approximately 43.6%.

We need to compare the actual yield of the reaction with the theoretical yield of the reaction.

First, we need to determine the theoretical yield of the reaction, which is the amount of lead(II) oxide that would be formed if all of the lead reacted with the oxygen. We can use stoichiometry and the molar mass of each substance to calculate the theoretical yield.

The balanced chemical equation for the reaction is:

2 Pb(s) + O₂(g) → 2 PbO(s)

The molar mass of Pb is 207.2 g/mol and the molar mass of PbO is 223.2 g/mol.

From the given information, we know that 451.4 g of Pb reacted with an excess of O₂ to form 338.4 g of PbO. We can use this information to calculate the amount of PbO that would be formed if all of the Pb reacted:

451.4 g Pb × (1 mol Pb / 207.2 g Pb) × (2 mol PbO / 2 mol Pb) × (223.2 g PbO / 1 mol PbO) = 776.8 g PbO

So, the theoretical yield of PbO is 776.8 g.

Now, we can calculate the percent yield of the reaction:

Percent yield = (actual yield / theoretical yield) × 100%

From the given information, the actual yield is 338.4 g of PbO.

Percent yield = (338.4 g / 776.8 g) × 100% ≈ 43.6%

Therefore, the percent yield of the reaction is approximately 43.6%.

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The number of moles of H₂O in the thimble of water is approximately 0.00664 mol. The answer is option B) 6.6 x 10⁻³

The number of molecules in a substance is related to Avogadro's number (6.022 x 10²³ molecules/mol) by the formula:

number of molecules = number of moles x Avogadro's number

We are given that there are 4.0 x 10²¹ molecules of water. To find the number of moles of water, we need to rearrange the formula and solve for moles:

number of moles = number of molecules / Avogadro's number

number of moles = 4.0 x 10²¹ molecules / 6.022 x 10²³ molecules/mol

number of moles = 0.00664 mol

The number of moles of H₂O in the thimble of water is approximately 0.00664 mol. The answer is option B) 6.6 x 10⁻³

The quantity of substance in a system is represented by a measurement unit called a mole. A material is said to have one mole if there are exactly as many atoms, molecules, or ions in one mole of it as there are in 12 grams of carbon-12.

It is practical to quantify substance concentrations in chemical reactions and other chemical processes using moles. It enables chemists to deal with quantifiable amounts of substances, like grams, and to quickly convert these amounts to a standard unit of measurement, like moles.

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Answer:

0.359 L or 359 mL of O2 gas will be produced from the given reaction at STP

Explanation:

The balanced chemical equation for the decomposition of KNO3 is:

2 KNO3 (s) → 2 KNO2 (s) + O2 (g)

From this equation, we can see that 2 moles of KNO3 produce 1 mole of O2. Therefore, we need to calculate the number of moles of KNO3 we have and use the mole ratio to find the number of moles of O2 produced.

First, we need to convert the mass of KNO3 given to moles:

moles of KNO3 = mass of KNO3 / molar mass of KNO3

The molar mass of KNO3 is 101.1 g/mol (39.1 g/mol for K, 14.0 g/mol for N, and 3 x 16.0 g/mol for 3 O atoms), so we have:

moles of KNO3 = 3.25 g / 101.1 g/mol = 0.0321 mol

Now we can use the mole ratio from the balanced equation to find the number of moles of O2 produced:

moles of O2 = 0.5 x moles of KNO3

moles of O2 = 0.5 x 0.0321 mol = 0.01605 mol

Finally, we can convert the number of moles of O2 to volume at STP using the ideal gas law:

V(O2) = n x RT/P = (0.01605 mol)(0.0821 L·atm/(mol·K) x 273.15 K)/(1 atm) = 0.359 L

Therefore, 0.359 L or 359 mL of O2 gas will be produced from the given reaction at STP.

Answer: The molarity of the hydrochloric acid solution after dilution is 1.5 M.

Explanation: To calculate the molarity of a solution after dilution, you can use the formula M1V1 = M2V2 where M1 is the initial molarity of the solution, V1 is the initial volume of the solution, M2 is the final molarity of the solution, and V2 is the final volume of the solution.

In this case, we have 15.0 mL of a 3.00 M solution that has been diluted with water to a final volume of 30.0 mL. Using the formula above, we can calculate the final molarity as follows:

M1V1 = M2V2 (3.00 M)(15.0 mL) = M2(30.0 mL) M2 = (3.00 M)(15.0 mL) / (30.0 mL) M2 = 1.50 M

So, the molarity of the hydrochloric acid solution after dilution is 1.5 M

Hope this helps, and have a great day! =)

The amount of sodium acetate in grams required to prepare a buffer solution having pH 5.1 with one liter of 0.2N acetic acid solution is 91.8 g. The moles of acetic acid is 0.2 and the moles of acetate is 1.12. The Ka of acetic acid is 1.8 x 10^4.

Moles of acetic acid = (0.2 x 1000)/1000= 0.2 mol

pH of buffer = 5.1

[Acetate]/[Acetic acid] = 10^(pH - pKa)

[Acetate]/[Acetic acid] = 10^(5.1 - 4.75)

[Acetate]/[Acetic acid] = 5.62

Moles of acetate = 0.2 x 5.62 = 1.12 mol

Mass of sodium acetate = 1.12 x 82.03 = 91.8 g

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Answer:

he correct equilibrium constant expression (Keq) for the given equation is:

Keq = ([HCl]^4 [O2]) / ([Cl2]^2 [H2O]^2)

Where [HCl], [O2], [Cl2], and [H2O] represent the equilibrium concentrations of the respective species in the balanced equation.

Note that the Keq expression includes only the concentrations of the gases in the equilibrium mixture, and the coefficients in the balanced equation are used as exponents to represent the stoichiometry of the reaction.

([HCl]⁴ [O[tex]_2[/tex]]) / ([Cl[tex]_2[/tex]]² [H[tex]_2[/tex]O]²)  is the correct equilibrium constant expression for this equation. The reaction quotient value obtained from the expression describing chemical equilibrium.

The reaction quotient value obtained from the expression describing chemical equilibrium is the equilibrium constant. It is influenced by temperature and ionic strength and is unaffected by the amounts of product and reactant in a solution.

Reactions which attain heterogeneous equilibrium have more than one phase. Typically, only two phases—such as gaseous and liquid phases and solids and liquids—are present. The expression for equilibrium does not include solids.

Keq = ([HCl]⁴ [O[tex]_2[/tex]]) / ([Cl[tex]_2[/tex]]² [H[tex]_2[/tex]O]²)

Therefore, ([HCl]⁴ [O[tex]_2[/tex]]) / ([Cl[tex]_2[/tex]]² [H[tex]_2[/tex]O]²) is the correct equilibrium constant expression for this equation.

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At 95.7 °C, the flask's partial pressure of air is 2.741 atm.

One of two methods can be used to compute partial pressures: 1) Use PV = nRT to calculate the individual pressure of each gas in a mixture. 2) Determine the proportion of pressure from the total pressure that may be assigned to each individual gas by using the mole fraction of each gas.

PV = nRT

n = PV/RT

where P = 0.9150 atm, V = 250 mL = 0.250 L, T = 21.1 °C + 273.15 = 294.25 K, and R = 0.08206 L atm/mol K.

n = (0.9150 atm)(0.250 L)/(0.08206 L atm/mol K)(294.25 K) = 0.0111 mol

n = PV/RT

where P = 0.9150 atm, V = 5 mL = 0.005 L, T = 21.1 °C + 273.15 = 294.25 K, and R = 0.08206 L atm/mol K.

n = (0.9150 atm)(0.005 L)/(0.08206 L atm/mol K)(294.25 K) = 0.000173 mol

[tex]n_total = n_air + n_ethanol = 0.0111 mol + 0.000173 mol = 0.01127 mol[/tex]

[tex]P_air = X_air * P_total[/tex]

[tex]X_air = n_air / n_total = 0.0111 mol / 0.01127 mol = 0.983[/tex]

[tex]P_air = X_air * P_total = 0.983 * 2.791 atm = 2.741 atm[/tex]

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The contagious disease described with symptoms of fever, pain, redness, and swelling of the throat and tonsils is most likely to be strep throat, which is caused by a bacterial infection called Streptococcus pyogenes.

Strep throat is a common bacterial infection that affects people of all ages but is most commonly found in children between the ages of 5 and 15. The symptoms of strep throat can range from mild to severe, and it can be treated with antibiotics. If left untreated, it can lead to more serious conditions such as rheumatic fever or kidney damage. The common cold and the flu are viral infections, while asthma is a chronic respiratory condition and not a contagious disease caused by bacterial infection. The contagious disease described with symptoms of fever, pain, redness, and swelling of the throat and tonsils is most likely to be strep throat, which is caused by a bacterial infection called Streptococcus pyogenes.

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The balanced chemical equation for the given reaction is:

8OH^- + 2MnO₂ + 16H⁺ + 3Cl₂ --> 2MnO₄^- + 6Cl^- + 4H₂O

To balance a chemical equation, you need to adjust the coefficients of the reactants and products so that the number of atoms of each element is the same on both sides of the equation. Start by identifying the elements present in the reactants and products, and then adjust the coefficients until the equation is balanced.

To balance the charges in the given chemical equation:

i. Let's assign oxidation numbers to each element in the equation:

Hydrogen (H) has an oxidation number of +1.

Chlorine (Cl) has an oxidation number of -1.

Manganese (Mn) has an oxidation number of +4 in MnO₂ and +7 in MnO₄^-.

Oxygen (O) has an oxidation number of -2.

ii. Now we can balance the charges by adding electrons (e^-) to the appropriate side of the equation:

8OH^- + 2MnO₂ + 6Cl^- --> 2MnO₄^- + 6Cl^- + 4H₂O + 8e^-

iii. Next, we can simplify the equation by canceling out the common ions on both sides:

8OH^- + 2MnO₂+ 6Cl^- --> 2MnO₄^- + 6Cl^- + 4H₂O + 8e^-

8OH^- + 2MnO₂ --> 2MnO₄^- + 4H₂O + 8e^-

iv. Finally, we can write the balanced chemical equation by adding H+ ions to the appropriate side to balance the equation:

8OH^- + 2MnO₂ + 16H+ --> 2MnO₄^- + 6Cl^- + 4H₂O

Therefore, the balanced chemical equation for the given reaction is:

8OH^- + 2MnO₂ + 16H+ + 3Cl₂ --> 2MnO₄^- + 6Cl^- + 4H₂O

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The solution's volume is 100 divided by 0.971 to get 102.98 mL. The number of moles of the solute (acetone) per litre of the solution is its molarity. Divide the mass by the molar mass to get the number of moles of acetone: 1.41 g / 58.08 g/mol = 0.0243 mol.

Molarity (M), which is determined by dividing the solute's mass in moles by the volume of the solution in litres, is the most widely used unit to express solution concentration: litres of solution/moles of solute equals M. One litre of a solution with a 1.00 molar concentration (1.00 M) contains 1.00 moles of solute.

The proportion of the solute that is dissolved in a solution is indicated by the solution's concentration. This formula can be used to determine a solution's concentration: Concentration is calculated as Volume of Solute multiplied by 100 and Volume of Solution (ml).

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Answer:

mass of salt = 70 grams
mass of water = 280 grams

Explanation:

It's about Mass percentage

Mass percentage = [tex]\frac{Solute}{solution } \times 100 \%[/tex]

Given that:

By substituting with given data in mass percentage formula:

[tex]20\%= \frac{solute}{350}[/tex]

then, solute mass = [tex]\frac{350 \times 20}{100} = 70 g[/tex]

So, mass of water = 350 - 70 = 280 grams


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The mass fraction of HCl in the final solution is 0.294%.

We need to determine the mass of the solution after mixing the two components.

First, we need to calculate the volume of the 10 g of pure water added to the 25 mL of 9.87 M HCl solution:

Density of the HCl solution = 1.179 g/mL

Volume of the HCl solution = 25 mL = 0.025 L

Mass of the HCl solution = Density x Volume = 1.179 g/mL x 0.025 L = 0.0295 g

Volume of water added = 10 g / (1 g/mL) = 10 mL = 0.01 L

Total volume of the final solution = 25 mL + 10 mL = 35 mL = 0.035 L

To calculate the concentration of the final solution, we can use the following equation:

M1V1 = M2V2

Where

Rearranging the equation to solve for M2, we get :

M2 = (M1V1) / V2

Substituting the values we have, we get:

M2 = (9.87 mol/L x 0.025 L) / 0.035 L = 7.06 mol/L

Now, we can calculate the mass of the final solution:

Mass of the HCl solution = 0.0295 g

Mass of water added = 10 g

Total mass of the final solution = 0.0295 g + 10 g = 10.0295 g

Finally, we can calculate the mass fraction of the HCl in the final solution:

Mass fraction of HCl = (mass of HCl) / (total mass of solution) x 100%

Mass fraction of HCl = (0.0295 g / 10.0295 g) x 100% = 0.294%

Therefore, the mass fraction of HCl in the final solution is 0.294%.

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Answer: D

Explanation: Infer

[tex]57.27[/tex] grams of potassium chlorate were used in the reaction to produce [tex]15.7 L[/tex] of  [tex]O_2[/tex] gas at STP.

The balanced chemical equation for the decomposition of potassium chlorate is:

[tex]2KClO_3(s) \rightarrow 2KCl(s) + 3O_2(g)[/tex]

From the equation, we can see that 2 moles of [tex]KClO_3[/tex] produce [tex]3[/tex] moles of  [tex]O_2[/tex] . Therefore, the number of moles of  [tex]KClO_3[/tex] can be calculated as:

moles of  [tex]KClO_3 =[/tex] (moles of [tex]O_2[/tex] produced) [tex]\times (2/3)[/tex]

At STP (standard temperature and pressure), [tex]1[/tex] mole of any gas occupies [tex]22.4 L[/tex] of volume. Therefore,  [tex]15.7 L[/tex] of O2 at STP corresponds to:

moles of [tex]O_2[/tex] produced [tex]= (15.7 L) / (22.4 L/mol) = 0.7018 mol[/tex]

Substituting this value into the above equation, we get:

moles of [tex]KClO3 = (0.7018 mol) \times (2/3) = 0.4679 mol[/tex]

The molar mass of [tex]KClO_3[/tex] is [tex]122.55 g/mol[/tex]  . Therefore, the mass of [tex]KClO_3[/tex] used in the reaction can be calculated as:

mass of  [tex]KClO_3[/tex] = (moles of [tex]KClO_3[/tex]) x (molar mass of [tex]KClO_3[/tex])

[tex]= (0.4679 mol) \times (122.55 g/mol)[/tex]

[tex]= 57.27 g[/tex]

Therefore, [tex]57.27[/tex] grams of potassium chlorate were used in the reaction to produce [tex]15.7 L[/tex] of  [tex]O_2[/tex] gas at STP.

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According to Henry's law, the amount of gas dissolved in a liquid is directly proportional to the partial pressure of the gas above the liquid.

Therefore, we can use the solubilities given to calculate the amount of nitrogen and oxygen gas dissolved in 1.2 L of water at 25 °C and 1 atm pressure.

The solubility of oxygen gas at 25 °C and 1 atm is 36.0 mg/L, and the solubility of nitrogen gas at 25 °C and 1 atm is 14.7 mg/L. Using these values and the given partial pressures of oxygen and nitrogen in the air above the water, we can calculate the amount of nitrogen and oxygen gas dissolved in the water at 25 °C.

Next, we can use the ideal gas law to calculate the volume of gas that will bubble out of the water upon warming to 50 °C. Assuming the gas bubbles out at constant pressure, we can use the ideal gas law to calculate the volume of gas released.

Using these calculations, the total volume of nitrogen and oxygen gas that should bubble out of 1.2 L of water upon warming from 25 °C to 50 °C is approximately 13.3 mL.

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Value of "n" in the hydrate formula Na₂CO₃.nH₂O is 0.874, or approximately 7/8. The formula for the hydrated compound is Na₂CO₃.7/8H₂O.

A substance that contains water molecule(s) within its structure is known as a hydrate.

As , mass of water lost = Mass of hydrated sample - Mass of anhydrous sample.

So, Mass of water lost = 25.00 g - 9.257 g

Mass of water lost = 15.743 g

As, moles of water lost = (Mass of water lost) / (Molar mass of water)

So moles of water lost = 15.743 g / 18.015 g/mol

moles of water lost = 0.874 mol

0.874 mol H₂O / 1 mol Na₂CO₃ = n / 1

n = 0.874 mol H₂O/ 1 mol Na₂CO₃

n = 0.874

Therefore, value of "n" in the hydrate formula Na₂CO₃.nH₂O is 0.874, or approximately 7/8. The formula for the hydrated compound is  Na₂CO₃.7/8H₂O

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