We need to carry at least 187 CCCs on board the space shuttle to absorb all the CO2 produced by the crew during the 18-day mission.
What is the amount of CO2 absorbed?To solve this problem, we need to use the following information:
Each crew member expels 42.0 g of CO2 per hour.The mission is scheduled to last 18 days.There are 6 crew members on board.Each CCC contains 750 g of LIOH.First, we need to calculate the total amount of CO2 that will be expelled during the mission:
Total CO2 = 6 crew members x 42.0 g CO2/hour x 24 hours/day x 18 days = 136,080 g CO2
Next, we need to calculate the amount of LIOH needed to absorb this CO2. The balanced chemical equation for the reaction between CO2 and LIOH is:
CO2 + 2 LIOH → Li2CO3 + H2O
The molar mass of CO2 is 44.01 g/mol, and the molar mass of LIOH is 23.95 g/mol.
This means that 2 moles of LIOH are needed to absorb 1 mole of CO2.
So, to absorb 136,080 g of CO2, we need:
136,080 g CO2 x (1 mol CO2/44.01 g) x (2 mol LIOH/1 mol CO2) x (23.95 g LIOH/1 mol) = 139,648 g LIOH
Since each CCC contains 750 g of LIOH, we need:
139,648 g LIOH / 750 g CCC = 186.2 CCCs
Therefore, we need to carry at least 187 CCCs on board the space shuttle to absorb all the CO2 produced by the crew during the 18-day mission.
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5 moles of a monoatomic ideal gas is compressed reversibly and adiabatically. The initial volume is 6 dm3 and the final volume is 2 dm3. The initial temperature is 27°C.
(i) What would be the final temperature in this process?
(ii) Calculate w, q and ΔE for the process. Given Cv = 20.91 J K−1 mol−1, γ = 1.4
Final temperature: 677.4K. Work done: -7026J.
Heat exchanged: 0J. Change in internal energy: -7026J.
How to solve(i) For an adiabatic process, T1(V1)^γ-1 = T2(V2)^γ-1.
When we substitute the values (γ=1.4, T1=300K, V1=6dm³, V2=2dm³), we get T2 = 677.4K.
(ii) w = -(P1V1 - P2V2)/(γ-1) = -(nRT1 - nRT2)/(γ-1) = -5 * 8.314 * (677.4 - 300) / 0.4 = -7026J.
For adiabatic, q = 0. ΔE = q + w = -7026J (since q=0).
Final temperature: 677.4K. Work done: -7026J.
Heat exchanged: 0J. Change in internal energy: -7026J.
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Solutions of Pb(NO3)2 and NaCl are combined, resulting in concentration of 0.0050 M Pb(NO3)2 and 0.0025 M NaCl immediately upon mixing. Select the correct description of the final solution, given that the Ksp of PbCl2 is 1.70×10^-5.
A. All solutes remain soluble
B. NaNO3 precipitates
C. Pb(NO3)2 precipitates
D. PbCl2 precipitates
Solutions of [tex]Pb(NO_3)^2[/tex] and [tex]NaCl[/tex] are combined, resulting in concentration of 0.0050 M [tex]Pb(NO_3)^2[/tex] and 0.0025 M [tex]NaCl[/tex] immediately upon mixing. The correct description of the final solution, given that the Ksp of [tex]PbCl_2[/tex] is 1.70×10^-5 is All solutes remain soluble. The correct answer is option A
Upon mixing [tex]Pb(NO_3)^2[/tex] and [tex]NaCl[/tex] , the following reaction occurs:
[tex]Pb(NO_3)^2[/tex] + [tex]2NaCl[/tex] → [tex]PbCl_2[/tex] +[tex]2NaNO_3[/tex]
Using the given concentrations of the reactants, the reaction quotient Qc can be calculated as:
Qc =[tex][Pb^2^+][Cl^-]^2[/tex] = [tex](0.0050 M)(0.0025 M)^2[/tex]
Qc [tex]= 3.13[/tex] × [tex]10^{-3}[/tex]
Comparing Qc to the solubility product constant (Ksp) of [tex]PbCl_2[/tex] , we see that Qc < Ksp. This indicates that the system is not at equilibrium and more [tex]PbCl_2[/tex] can dissolve before the product reaches saturation.
Therefore, no precipitation of [tex]PbCl_2[/tex] will occur, and option A is the correct answer: all solutes remain soluble.
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Quantifying chemical reactions
Quantifying chemical reactions is essential in understanding the stoichiometry of a reaction, predicting product formation, and optimizing product yield in industrial applications. Stoichiometric coefficients and limiting reactants are two important tools used in this process.
Quantifying chemical reactions involves measuring the amount of reactants and products involved in a chemical reaction. This is important in determining the stoichiometry of the reaction, which refers to the relative amounts of reactants and products involved. Stoichiometry is a crucial concept in chemistry because it allows scientists to predict the amount of product that will be formed from a given amount of reactant, or vice versa.
One way to quantify chemical reactions is through the use of stoichiometric coefficients. These coefficients represent the number of moles of each reactant and product involved in the reaction. For example, the balanced chemical equation for the reaction between hydrogen gas and oxygen gas to form water is:
[tex]2H2 + O2 → 2H2O[/tex]
This equation tells us that two moles of hydrogen gas react with one mole of oxygen gas to form two moles of water. The stoichiometric coefficients can be used to determine the mass of each reactant and product involved in the reaction, using the molar masses of each substance.
Another way to quantify chemical reactions is through the use of limiting reactants. A limiting reactant is the reactant that is completely consumed in a reaction, limiting the amount of product that can be formed. The amount of product formed will be determined by the amount of limiting reactant present. This concept is important in industrial chemistry, where maximizing product yield is often the goal.
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How much heat, in joules, would be required to raise the temperature of 450 g of
Aluminum (c Al = 0.21 cal/g o C) from 19.5 o C to 31.2 o C?
Answer:
[tex]\huge\boxed{\sf Q = 1105.65\ cal}[/tex]
Explanation:
Given data:Mass = m = 450 g
T₁ = 19.5 °C
T₂ = 31.2 °C
Change in Temperature = ΔT = 31.2 - 19.5 = 11.7 °C
c = 0.21 cal/g °C
Required:Heat = Q = ?
Formula:Q = mcΔT
Solution:Put the given data in the above formula.
Q = (450)(0.21)(11.7)
Q = 1105.65 cal
[tex]\rule[225]{225}{2}[/tex]
arrange the following electrons, represented by their quantum numbers, in increasing order of energy (lowest written first)
(1,0,0,-1/2); (3,1,1,1/2); (2,1,0,-1/2); (2,1,0,-1/2); (3,2,0,-1/2)
The electrons can be arranged in increasing order of energy as follows: (1,0,0,-1/2) < (2,1,0,-1/2) < (2,1,0,-1/2) < (3,1,1,1/2) < (3,2,0,-1/2).
The energy of an electron is determined by its principal quantum number (n), azimuthal quantum number (l), magnetic quantum number (m), and spin quantum number (s). The electrons can be arranged in increasing order of energy by comparing their quantum numbers.
Starting with the lowest energy electron, we have the electron with quantum numbers (1,0,0,-1/2). This electron has the lowest principal quantum number, indicating that it occupies the lowest energy level.
It also has an azimuthal quantum number of zero, which corresponds to the s subshell, and a negative spin quantum number, indicating that its spin is aligned opposite to the magnetic field.
Next, we have the two electrons with quantum numbers (2,1,0,-1/2). These electrons have the same principal quantum number, indicating that they occupy the same energy level.
They both have an azimuthal quantum number of one, which corresponds to the p subshell, and a negative spin quantum number.
Following these electrons, we have the electron with quantum numbers (3,1,1,1/2). This electron has a higher principal quantum number than the previous electrons, indicating that it occupies a higher energy level.
It has an azimuthal quantum number of one, which corresponds to the p subshell, and a positive spin quantum number.
Finally, we have the electron with quantum numbers (3,2,0,-1/2). This electron has the highest azimuthal quantum number of all the electrons, indicating that it occupies the d subshell. It also has a negative spin quantum number.
Therefore, the electrons can be arranged in increasing order of energy as follows: (1,0,0,-1/2) < (2,1,0,-1/2) < (2,1,0,-1/2) < (3,1,1,1/2) < (3,2,0,-1/2).
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What was the effect of the addition of FeCl3 to the sample solution in the dichromate titration? Explain
[tex]FeCl_3[/tex] is added to ensure all reducing agent is oxidized, indicating endpoint in dichromate titration.
In a dichromate titration, [tex]FeCl_3[/tex] is frequently added to the example arrangement as a sign of the endpoint. The expansion of [tex]FeCl_3[/tex] to the example arrangement assists with guaranteeing that the lessening specialist has been all oxidized by the potassium dichromate ([tex]K_2Cr_2O_7[/tex] ) arrangement.
[tex]FeCl_3[/tex] responds with any overabundance[tex]K_2Cr_2O_7[/tex] in the answer for structure a red-earthy colored encourage of[tex]Fe(OH)_3[/tex] , demonstrating that the lessening specialist has been all oxidized. This response is known as a "back-titration" since overabundance[tex]K_2Cr_2O_7[/tex] is added to the arrangement, trailed by the option of [tex]FeCl_3[/tex] to decide how much unreacted [tex]K_2Cr_2O_7[/tex] .
The response somewhere in the range of [tex]FeCl_3[/tex] and [tex]K_2Cr_2O_7[/tex] can be addressed as:
[tex]6FeCl_3 + K_2Cr_2O_7 + 7H_2SO_4 → 3Fe_2(SO_4)_3 + Cr_2(SO_4)_3 + K_2SO_4 + 7H_2O + 3Cl_2[/tex]
The [tex]FeCl_3[/tex] goes about as a pointer in this response since it responds with the overabundance [tex]K_2Cr_2O_7[/tex] until the decreasing specialist has been all oxidized. As of now, the red-earthy colored encourage of [tex]Fe(OH)_3[/tex] structures, showing that the endpoint has been reached.
Without the expansion of [tex]FeCl_3[/tex], it would be challenging to precisely decide the endpoint of the titration. The expansion of [tex]FeCl_3[/tex] is important to guarantee that the lessening specialist has been all oxidized and that the endpoint has been reached, considering a more exact assurance of the grouping of the diminishing specialist in the example arrangement.
In outline, the expansion of [tex]FeCl_3[/tex] to the example arrangement in a dichromate titration is significant in light of the fact that it assists with guaranteeing that the decreasing specialist has been all oxidized, considering a more precise assurance of its focus.
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The complete question is -
What is the reason for a dichromate titration, and how does [tex]FeCl_3[/tex]support deciding the endpoint of the titration? Might you at any point give the compound condition to the response somewhere in the range of [tex]FeCl_3[/tex] and [tex]K_2Cr_2O_7[/tex] and make sense of why [tex]FeCl_3[/tex] goes about as a marker in this response?
What mass (grams) of magnesium chloride would be formed by the compete reaction of 72.8 grams of magnesium?
Mg +FeCl2 --> Fe + MgCl2
Answer: 285.63g of MgCl2.
Explanation:
Very easy stiochemistry question. Use the dimensional analysis. For example 1 m x 100 cm / 1m and meters get canceled out and 1 m is 100 cm.
For the question, start with the given things. You know that it was started with 72.8 grams of magnesium. Convert it to molar mass (to use moles for comparison), and then find the mass of mg.
How many grams of O are in 615g of N2O?
There are approximately 223.2 grams of oxygen in 615 grams of N2O.
To find the number of grams of O in 615g of N2O, we first need to understand the chemical formula of N2O. N2O is a compound made up of two nitrogen atoms (N) and one oxygen atom (O). Therefore, the molecular weight of N2O would be:
(2 x atomic weight of N) + (1 x atomic weight of O)
= (2 x 14.01 g/mol) + (1 x 16.00 g/mol)
= 44.01 g/mol
Now, to calculate the number of grams of O in 615g of N2O, we need to know the proportion of O in the compound. Since there is only one oxygen atom in each molecule of N2O, we can find the proportion of O by dividing the atomic weight of O by the molecular weight of N2O:
Atomic weight of O / Molecular weight of N2O
= 16.00 g/mol / 44.01 g/mol
= 0.363
This means that oxygen makes up 36.3% of the total weight of N2O. To find the number of grams of O in 615g of N2O, we can multiply the total weight by the proportion of O:
615g x 0.363
= 223.2g
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What mass of oxygen would be released by the thermal decomposition of 918.7 grams of Mercury (II) Oxide?
HgO --> Hg + O2
Answer:
Explanation:
[tex]\frac{918.7 g}{1} *\frac{1}{216.59m } = 4.241 mol[/tex] To start off the mol of HgO must be found.
After that the molar ratio between HgO and O must be found but in this case its 1:1
[tex]4.241 mol HgO*\frac{1 molO}{1molHgO} = 4.241 mol O[/tex] the mols of HgO is put on the bottom to cancel out with the other one leaving just mols of oxygen. Finally to find g of oxygen it must be multiplied by its molar mass.
[tex]\frac{4.241 molO}{1} * \frac{15.999 g}{mol} = 67.85 g[/tex] Oxygen
in an experiment, 1 mol A, 2 mol B and 1 mol D were mixed and allowed to come to equilibrium at 25C. The resulting mixture was found to contain 0.9 mol of C at a total pressure of 1.00 bar. Find the mole fractions of each species at equilibrium
The mole fractions of each species at equilibrium are 0.25 for A and D, 0.5 for B, and 0.225 for C.
we can use the principles of chemical equilibrium and the mole fraction formula.
First, we need to write the balanced chemical equation for the reaction involving A, B, C, and D. Let's assume that the reaction is:
A + 2B <=> C + D
where A, B, C, and D are the chemical species, and the coefficients indicate their stoichiometric ratios.
Next, we need to write the expression for the equilibrium constant, Kc, for this reaction:
Kc = [C][D] / [A][B]²
where [X] denotes the molar concentration of species X at equilibrium.
Since we know the initial moles of A, B, and D, we can calculate their total moles in the mixture:
Total moles = 1 mol A + 2 mol B + 1 mol D = 4 mol
We also know that the final mixture contains 0.9 mol of C. Therefore, the molar concentration of C at equilibrium is:
[C] = 0.9 mol / 4 L = 0.225 M
Since we have only one unknown, we can use the equilibrium constant expression to calculate the molar concentration of D:
Kc = [C][D] / [A][B]²
0.9 = (0.225)(D) / (1)(2²)
D = 1.8
Therefore, the molar concentration of D at equilibrium is 1.8 M.
Using the law of conservation of mass, we can also calculate the molar concentration of A and B at equilibrium:
[A] = 1 mol / 4 L = 0.25 M
[B] = 2 mol / 4 L = 0.5 M
Mole fraction of X = moles of X / total moles
Mole fraction of A = 1 mol / 4 mol = 0.25
Mole fraction of B = 2 mol / 4 mol = 0.5
Mole fraction of C = 0.9 mol / 4 mol = 0.225
Mole fraction of D = 1 mol / 4 mol = 0.25
Therefore, the mole fractions of each species at equilibrium are 0.25 for A and D, 0.5 for B, and 0.225 for C.
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Question 28
(1 mark)
Two other minerals can be seen in the photo:
galena, a dark grey mineral with the formula PbS
iron pyrite, a gold-coloured mineral with the formula FeS2
Compare their chemical formulas, by writing down one similarity and one difference between these two minerals.
Note: Pb = lead, Fe = iron, S = sulfur.
Question 29
(1 mark)
Wanting to create the beautiful golden colour of iron pyrite, FeS2, in the lab, a student mixes together black powdered iron (Fe) and yellow powdered sulfur (S). The result is a dull, yellowish grey powder. Propose why this attempt failed?
Question 30
(1 mark)
The student can vary the proportions of iron and sulfur by adding more of each powder to the mixture. Clarify why the same thing isn't true for the compound iron pyrite.
PbS, sometimes referred to as galena, is made up of lead (Pb) and sulfur (S) atoms and has a straightforward structure. Each lead atom forms a tetrahedral link with four sulfur atoms, while each sulfur atom forms a covalent bond with two lead atoms.
Each sulfur atom forms an octahedral link with six iron atoms, while each iron atom forms a covalent bond with two sulfur atoms. FeS2's crystal structure is a cubic, tightly packed lattice.
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Dimensional analysis with shapes
The surface area of the rectangular prism is 0.034 square meters.
For a rectangular prism with length l, width w, and height h, the surface area is:
Surface area = 2lw + 2lh + 2wh
Substituting the given values, we get:
Surface area = 2(10 cm x 5 cm) + 2(10 cm x 8 cm) + 2(5 cm x 8 cm)
Surface area = 100 cm² + 160 cm² + 80 cm² = 340 cm²
We can use dimensional analysis. So the conversion factor is:
1 m² / 10,000 cm²
Multiplying the surface area by this conversion factor, we get:
Surface area = 340 cm² x (1 m² / 10,000 cm²)
Surface area = 0.034 m²
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--The complete Question is, What is the surface area of a rectangular prism that has a length of 10 cm, a width of 5 cm, and a height of 8 cm? Use dimensional analysis to convert the answer to square meters--
The iodine monobromide molecule, IBr, has a bond length of 249 pm and a dipole moment of 1.21 D. (a) Which atom of the molecule is expected to have a negative charge? (b) Calculate the effective charges on the I and Br atoms in IBr in units of the electronic charge, e.
a. Br will have the negative charge
b. The effective charges on the I and Br atoms are approximately +1.012e and -1.012e, respectively.
How to determine the negative chargea. To identify which atom within an IBr molecule will have a partial negative charge, we must consider each atom's electronegativity.
On the periodic table, iodine (I) has an electronegativity value of 2.66 while bromine (Br) boasts 2.96; since Br has higher electronegativity it will attract electrons more strongly and hence have an even stronger partial negative charge.
B. To calculate the effective charges on the I and Br atoms in IBr, we can use the dipole moment equation:
μ = Q * d
where μ is the dipole moment, Q is the effective charge, and d is the bond length.
We are given the dipole moment (μ) as 1.21 D, and the bond length (d) as 249 pm. However, we need to convert the units to the SI system before proceeding with the calculation.
[tex]1 D (Debye) = 3.336 * 10^-^3^0 cm,\\\\1 pm = 10^-^1^2 m.[/tex]
Now we can solve for the effective charge (Q):[tex]u = 1.21 D * (3.336 × 10^-^3^0 Cm/D)\\ \\= 4.03656 * 10^-^3^0 cm\\d = 249 pm * (10^-12 m/pm) = 2.49 * 10^-^1^0 m[/tex]
Q = μ / d
[tex]Q = (4.03656 * 10^-^3^0 cm) / (2.49 *10^-^1^0 m) \\\\\\=1.62151 * 10^-^2^0 C[/tex]
This effective charge represents the charge difference between the I and Br atoms. To express the charges in units of the elementary charge (e), we need to divide the effective charge by the elementary charge value (e = 1.602 × 10^-19 C):
Q_e =[tex]\frac{(1.62151 * 10^-^2^0 C)}{(1.602 * 10^-^1^9 C)} = 1.012[/tex]
The effective charges on the I and Br atoms are approximately +1.012e and -1.012e, respectively.
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A pH of 5 is considered to be neutral
Answer:
No,a pH of 5 is slightly acidic,not neutral. A pH of 7 is considered neutral
Identity the number of bonding pairs and lone pairs of electrons n2
There are 3 bonding pairs and 7 lone pairs of electrons in N2.
What is an electronAn atom's nucleus is orbited by an electron, a subatomic particle with a negative charge. Along with protons and neutrons, it is one of the elementary particles that make up matter. The mass of an electron is exceedingly small, it is roughly 1/1836 that of a proton.
To determine the number of lone pairs of electrons in N2, we need to subtract the number of bonding pairs from the total number of valence electrons:
Number of lone pairs = Total number of valence electrons - Number of bonding pairs
Number of lone pairs = 10 - 3
Number of lone pairs = 7
Therefore, there are 3 bonding pairs and 7 lone pairs of electrons in N2.
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Question
How many moles of Na₂S₂O3 are needed to dissolve 0.65 mol of AgBr in a solution volume of
1.0 L, if Ksp for AgBris 3.3 x 10-13 and K for the complex ion [Ag(S₂03)2] is 4.7 × 10¹3?
Remember to use correct significant figures in your answer (round your answer to the nearest
tenth). Do not include units in your response.
The precipitation of an ionic substance from solution occurs when the ionic product exceeds the value of its solubility product at that temperature. Here the moles of Na₂S₂O₃ needed is
The solubility product of a sparingly soluble salt is defined as the product of the molar concentrations of its ions in a saturated solution of it at a given temperature.
Here the concentration of Ag⁺ ions = √Ksp = √3.3 × 10⁻¹³ = 1.81 × 10⁻¹³.
Moles of Ag⁺ ions: (1.82 x 10⁻¹³ M) x 1.0 L = 1.82 x 10⁻¹³ mol Ag⁺
Use the stoichiometry of the reaction to find the moles of Na₂S₂O₃ needed: 1 mol Na₂S₂O₃ / 2 mol Ag⁺ = 0.5 mol Na₂S₂O₃/mol Ag⁺
Moles of Na₂S₂O₃ required: 0.5 mol Na₂S₂O₃/mol Ag⁺ x 1.82 x 10⁻¹³ mol Ag⁺ = 9.1 x 10⁻¹⁴ mol Na₂S₂O₃
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A bag of potato chips is sealed in a factory near seal level. The atmospheric pressure is 99.82 kPa. What is the difference in Pa between the pressure in the bag and the atmospheric pressure?
The difference in Pa between the pressure in the bag and the atmospheric pressure is 1.505 kPa.
How to obtain the difference in pressureTo obtain the difference in pressure, we first need to know the atmospheric pressure near sea level. This is 760 mm Hg. When we convert this to pascals, we will have, 101.32472 kPa.
Now, the difference in pressure will be obtained by subtracting the atmospheric pressure in the bag from the atmospheric pressure near sea level and this is:
101.32472 kPa - 99.82 kPa
= 1.505 kPa.
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Draw a model of the four types of nuclear decay and explain each. Pick the same element (Si-32) to start with.
Sure, I can explain the four types of nuclear decay and provide a model for each using Si-32 as an example.
Si-32 is a radioactive isotope of Silicon with 14 protons and 18 neutrons.
1. Alpha Decay:
In alpha decay, an unstable nucleus emits an alpha particle, which consists of two protons and two neutrons, reducing the atomic number by two and the mass number by four. This makes the resulting nucleus a different element.
Model: Si-32 → alpha particle + Mg-28
Explanation: Si-32 decays into an alpha particle (two protons and two neutrons) and becomes Mg-28.
2. Beta Decay:
In beta decay, a neutron is converted into a proton and an electron. The proton stays in the nucleus, and the electron is emitted as a beta particle. This increases the atomic number by one while keeping the mass number the same.
Model: Si-32 → beta particle + P-32
Explanation: Si-32 decays into a beta particle (an electron) and becomes P-32.
3. Gamma Decay:
Gamma decay occurs when an unstable nucleus emits high-energy photons called gamma rays. Unlike alpha and beta decay, gamma decay does not change the atomic number or mass number of the nucleus.
Model: Si-32 → Si-32 + gamma ray
Explanation: Si-32 emits a gamma ray but remains Si-32.
4. Electron Capture:
In electron capture, an unstable nucleus absorbs an electron from an inner shell, converting a proton into a neutron. This reduces the atomic number by one while keeping the mass number the same.
Model: Si-32 + electron → Al-32
Explanation: Si-32 captures an electron and becomes Al-32.
These four types of nuclear decay can occur in radioactive isotopes, and they result in a change in the atomic number and/or mass number of the nucleus.
Ethane burns in oxygen according to the following equation: 2 C2H6 + 7 O2 → 4 CO2 + 6 H2O
(a) How many liters of O2 at 41 °C and 0.307 atm will be needed to burn 8.57 L of C2H6 at 41 °C and 0.307 atm?
(b) How many liters of CO2 at 41 °C and 0.307 atm will be produced? Report your answers to parts (a) and (b) to 3 significant figures.
a) We need 32.6 liters of [tex]O_2[/tex] at 41 °C and 0.307 atm to burn 8.57 L of [tex]C_2H_6[/tex] at 41 °C and 0.307 atm
b) 18.5 liters of [tex]CO_2[/tex] will be produced at 41 °C and 0.307 atm.
To answer this question, we will use the ideal gas law, which relates pressure, volume, temperature, and number of moles of a gas. We will also use stoichiometry to relate the amount of ethane and oxygen consumed and the amount of carbon dioxide and water produced.
(a) To determine how many liters of [tex]O_2[/tex] are needed to burn 8.57 L of [tex]C_2H_6[/tex] , we first need to convert the volume of ethane to moles using the ideal gas law:
n([tex]C_2H_6[/tex] ) = PV/RT = (0.307 atm)(8.57 L)/(0.0821 L·atm/mol·K)(314 K) = 0.342 mol
From the balanced equation, we see that 2 moles of [tex]C_2H_6[/tex] react with 7 moles of [tex]O_2[/tex] . Therefore, the amount of [tex]O_2[/tex] needed is:
n([tex]O_2[/tex]) = (7/2) n([tex]C_2H_6[/tex]) = (7/2)(0.342 mol) = 1.20 mol
Now we can use the ideal gas law again to calculate the volume of [tex]O_2[/tex] needed:
V([tex]O_2[/tex] ) = n([tex]O_2[/tex])RT/P = (1.20 mol)(0.0821 L·atm/mol·K)(314 K)/(0.307 atm) = 32.6 L
Therefore, 32.6 liters of [tex]O_2[/tex] are needed to burn 8.57 L of [tex]C_2H_6[/tex] at at 41 °C and 0.307 atm
(b) From the balanced equation, we see that 2 moles of [tex]C_2H_6[/tex] produce 4 moles of [tex]CO_2[/tex] . Therefore, the amount of [tex]CO_2[/tex] produced is:
n([tex]CO_2[/tex]) = 2 n([tex]C_2H_6[/tex]) = 2(0.342 mol) = 0.684 mol
V([tex]CO_2[/tex]) = n([tex]CO_2[/tex])RT/P = (0.684 mol)(0.0821 L·atm/mol·K)(314 K)/(0.307 atm) = 18.5 L
Therefore, 18.5 liters of [tex]CO_2[/tex] at 41 °C and 0.307 atm will be produced.
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Need help matching pairs of structures to diastereomers, enantiomers, constitutional isomers, not isomers, diff representations of the same?
A pair of molecules which exist in two forms that are mirror images of each other but cannot be superimposed one upon the other are called the enantiomers. They are present in pairs and have similar molecular shape.
The compounds with the same molecular formula but are non-superimposable non-mirror images are called diastereomers. They have distinct physical properties and molecular shape.
The constitutional isomers have the same molecular formula but have different bonding atomic organization and bonding patterns.
So here:
1st structure is constitutional isomers (c), 2nd structures are enantiomers (b) and the 3rd are completely different not isomers (d).
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Based on the solubility rules, which one of these phosphate compounds is insoluble in water?
A) Li2CO3
B) Na3PO4
C) Ba(OH)2
D) (NH4)3PO4
(NH4)3PO4 is insoluble in water. The correct option is D
What is solubility rules ?According to their chemical formula and ionic charges, ionic compounds generally follow a set of solubility laws that define their solubility patterns in water. These guidelines aid in determining whether an ionic compound will dissolve in water or not as well as if it will precipitate when combined with other ionic compounds.
Therefore, (NH4)3PO4 is the compound that is expected to be insoluble in water based on the solubility rules.
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In a few sentences, briefly summarize how you were able to determine the amount of Vitamin C in fruit juice using DCPIP.
50cm³ of 1.0M hydrochloric acid reacted with excess zinc. i) Write the equation for the reaction. ii) How many mole of aqueous hydrogen ions were present in the acid solution? iii) Calculate the volume of gas evolved at s.t.p. [Molar volume = 22.4 dm³ at s.t.p. of gas].
i) The equation for the reaction between hydrochloric acid and zinc is:
[tex]Zn + 2HCl → ZnCl2 + H2[/tex]
ii) n(HCl) = C × V = 1.0M × 0.05 L = 0.05 moles
iii) The volume of gas evolved at STP is 0.544 L or 544 mL.
The concentration of hydrochloric acid is 1.0M, which means that there is 1 mole of hydrochloric acid in 1 liter (1000 cm³) of solution. The volume of the hydrochloric acid used is 50 cm³, which is 0.05 liters.
According to the stoichiometry of the reaction, each mole of hydrochloric acid produces one mole of hydrogen ions, so the number of moles of hydrogen ions in the solution is also 0.05 moles.
The volume of gas evolved can be calculated from the ideal gas law:
PV = nRT
where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of the gas, R is the gas constant (0.0821 L·atm/K·mol), and T is the temperature of the gas in Kelvin. At standard temperature and pressure (STP), the pressure is 1 atm and the temperature is 273 K. The molar volume of a gas at STP is 22.4 L/mol.
From the equation for the reaction, we know that one mole of hydrogen gas is produced for every two moles of hydrochloric acid used. Therefore, the number of moles of hydrogen gas produced is:
n(H2) = 0.5 × n(HCl) = 0.5 × 0.05 moles = 0.025 moles
Using the ideal gas law, we can calculate the volume of hydrogen gas produced at STP:
V(H2) = n(H2) × RT/P = 0.025 mol × 0.0821 L·atm/K·mol × 273 K/1 atm = 0.544 L
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How many grams of magnesium oxide would be formed if 28.2 grams of magnesium was burned?
Mg + O2 --> MgO
When 28.2 grams of Mg is burned, 46.7 grams of MgO will be formed.
How to determine the amount of MgO formed when 28.2 grams of Mg is burnedThe balanced chemical equation for the combustion of magnesium is:
2 Mg + O2 --> 2 MgO
This equation shows that 2 moles of Mg react with 1 mole of O2 to produce 2 moles of MgO.
To determine the amount of MgO formed when 28.2 grams of Mg is burned, we first need to convert the given mass of Mg to moles:
molar mass of Mg = 24.31 g/mol
moles of Mg = mass of Mg / molar mass of Mg
moles of Mg = 28.2 g / 24.31 g/mol
moles of Mg = 1.16 mol
According to the balanced chemical equation, 2 moles of Mg produce 2 moles of MgO. Therefore, we can use the mole ratio to calculate the moles of MgO formed:
moles of MgO = moles of Mg x (2 moles of MgO / 2 moles of Mg)
moles of MgO = 1.16 mol x 1
moles of MgO = 1.16 mol
Finally, we can convert the moles of MgO to grams using its molar mass:
molar mass of MgO = 40.31 g/mol
mass of MgO = moles of MgO x molar mass of MgO
mass of MgO = 1.16 mol x 40.31 g/mol
mass of MgO = 46.7 g
Therefore, when 28.2 grams of Mg is burned, 46.7 grams of MgO will be formed.
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Н
HOH
14
Н-С-С-С-Н
I
ННН
List the number of each atom in the formulas above:
H
НН Н
Н-С-С-С-О-Н
LI
НН Н
DONE
Н Н
H
Н-С-С-О-С-Н
II
Н Н
H
Answer:
Explanation:
It seems like you’re trying to count the number of atoms in some chemical formulas. Here’s the list of the number of each atom in the formulas you provided:
Formula 1: Н - 1 Formula 2: H - 1, O - 1 Formula 3: Н - 14 Formula 4: Н - 2, C - 3 Formula 5: I - 1 Formula 6: Н - 3 Formula 7: H - 2 Formula 8: Н - 2, C - 3, O - 1 Formula 9: Li - 1 Formula 10: Н - 2 Formula 11: Н - 2 Formula 12: H - 1 Formula 13: Н - 2, C - 2, O - 1 Formula 14: II
How do I find solution concentration
To find the solution concentration, you need to know the amount of solute and the volume of the solution.
The solution concentration is typically expressed in terms of molarity (moles of solute per liter of solution). To calculate the molarity of a solution, divide the moles of solute by the volume of the solution in liters.
Another way to express solution concentration is in terms of percent by mass or volume, which is calculated by dividing the mass or volume of the solute by the mass or volume of the solution and multiplying by 100.
To find the solution concentration, you'll need to calculate the ratio of solute (substance being dissolved) to solvent (substance doing the dissolving) in the mixture.
Concentration is commonly expressed in units like molarity (M), mass/volume percent, or parts per million (ppm).
To calculate molarity (M), divide the moles of solute by the volume of the solvent (in liters). The formula is:
Molarity (M) = moles of solute / volume of solvent (L)
For mass/volume percent, divide the mass of the solute by the total volume of the solution and multiply by 100. The formula is:
Mass/volume percent = (mass of solute / total volume of solution) x 100
For parts per million (ppm), divide the mass of the solute by the total mass of the solution and multiply by 1,000,000.
The formula is:
ppm = (mass of solute / total mass of solution) x 1,000,000
Choose the appropriate formula based on the units required for your specific problem.
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You have 20.7 grams of water at -25.34 °C. You want to warm it to 155.0 °C. Use the information below to calculate how much heat this will require.
Csolid = 2.09 J/(g·°C)
ΔHfus = 333 J/g
Cvapor = 2.03 J/(g·°C)
ΔHvap = 2260 J/g
Answer:
Cvapor = 2.03 J/(g·°C)heu
Which of the following is an example of an environmental impact of
agriculture?
O high use of gold, copper, and silver
O high use of rock supplies
O high use of mineral resources
O high use of water
Ne
Answer:
B
self explanatory
Explanation:
A container of hellum has 4.3 moles of gas in a container with a volume of 3.9 liters and a pressure of 201.6kPa at 298K. A container of xenon has a volume of 3.9 liters
and a pressure of 201.6kPa at 298K. How many moles of xenon gas is present?
The ideal gas law states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature in Kelvin. Rearranging the equation, we get:
n = PV/RT
For the container of helium:
n = (201.6 kPa) x (3.9 L) / [(8.31 J/mol*K) x (298 K)] = 0.0688 mol
Now, using the same equation for the container of xenon:
n = (201.6 kPa) x (3.9 L) / [(8.31 J/mol*K) x (298 K)] = 0.0688 mol
Therefore, there are also 0.0688 moles of xenon gas present in the container.
Haw many valance electrons in the following atoms.
O Na Sr
Answer:O has 6, Na has 1, and Sr has 2.
Explanation: