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April 7, 2021 at 1:41 pm #374150Godwin EmmanuelKeymaster
1. At what temperature will 0.654 moles of neon gas occupy 12.30 L at 1.95 atm?
2. A sample of argon gas at STP occupies 56.2 L. Determine the number of moles of argon.
3. A gas is compressed in a 25 L storage tank under a pressure of 150 kPa and a temp of 21°C. How many moles of gas are contained in the tank?
4. What volume will 45 grams of hydrogen gas occupy at 1.05 atm and 25°C?
5. At what temperature will 5.00 g of Cl2 exert a pressure of 900 mm Hg at a volume of 750.0 mL?
6. A 143.03 g sample of a gas is at 1789.7 torr, 256.8 K and occupies 29.1 L. What is the chemical formula for this gas? (hint: find the molar mass first)
7. A 28.8 g piece of dry ice (solid carbon dioxide) is allowed to sublime (convert directly from a solid to a gas) into a large balloon. Assuming that all of the carbon dioxide ends up in the balloon, what will the volume of the balloon be at a temperature of 22°C and a pressure of 742 mm Hg?
8. Calculate the density of NO2 gas at 0.97 atm and 35°C.
9. Ammonium nitrate decomposes explosively upon heating according to the following equation:April 7, 2021 at 1:45 pm #374151Godwin EmmanuelKeymasterDont worry, Here are the solutions
1) At what temperature will 0.654 moles of neon gas occupy 12.30 L at 1.95 atm?
Answer
T = unknown, n = 0.654 mol, V = 12.30 L, P = 1.95 atm.
We can use the ideal gas equation, PV = nRT, and find the temperature of the gas.
PV=nRT; PV=nRT; where R = 0.08206 Latm/mol.K
Substitute all the given information in the ideal gas equation we have
1.95 atm.12.30 L = 0.654mol * 0.08206L.atm/mol.K*T
(23.985L.atm)/(0.05367L.atm/k )=(0.05367L.atm/K*T)/(0.05367L.atm/K)
Divide both the side by 0.05367 L.atm/K we have
T = 446.9K. = 446.9K – 273.15 = 173.8 °C
Hence the neon gas will be at 447 Kelvin temperature. Which in degree celsius will be 174 °C?2) A sample of argon gas at STP occupies 56.2 L. Determine the number of moles of argon.
Answer
At STP, T = 273.15K and P = 1atm.
We are given V = 56.2 L.
Gas constant, R = 0.08206 L.atm/mol.K.
Substitute all this information in the ideal gas equation, PV = nRT we have
PV = nRTPV=nRT
1atm * 56.2L = n * 0.08206 L.atm/mol.K* 273.15K
56.2L.atm = n * 22.415 L.atm/mol
divide both the side by 22.415L.atm/mol, we have
(56L.atm)/(22.415L.atm/mol )=(n*22.415L.atm/mol)/(22.415L.atm/mol)
n = 2.5073 moles of Argon.
In the question, we are given 56.2 L in 3 significant figure, so our final answer must also be in 3 significant figures.
Hence there would be 2.51 moles of Argon present in the given sample.3) A gas is compressed in a 25 L storage tank under a pressure of 150 kPa and a temp of 21°C. How many moles of gas are contained in the tank?
Answer
We are given Volume, V = 25L, Pressure, P = 150 kPa, Temperature T = 21°C.
In the ideal gas equation, pressure must be used in the ‘átm’ unit and temperature must be in the Kelvin unit.
Pressure P in atm = 150kPa*1atm/101.325kPa =1.4804atm.
Temperature in Kelvin=21°C=21+273.15=294.15K
Gas constant, R = 0.08206 L.atm/mol.K
Substitute all this information in the ideal gas equation we have
PV = nRTPV=nRT
1.480atm * 25L = n * 0.08206 L.atm/mol.K * 294.15K
37L.atm = n * 22.415 L.atm/mol
Divide both the side by 22.415L.atm/mol, we have
(37L.atm)/(22.415L.atm/mol )=(n*22.415L.atm/mol)/(22.415L.atm/mol)
n = (37L.atm)/(22.415L.atm/mol)
n = 1.6507 moles.
in the question, we are given pressure and volume both in 2 significant figure, so our final answer must also, be in 2 significant figures.
Hence there would be 1.7 moles of gas in the tank.4) What volume will 45 grams of hydrogen gas occupy at 1.05 atm and 25°C?
Answer
Here instead of ‘moles of hydrogen gas’ we are given the grams of hydrogen gas.
So we will first convert 45 grams of H2 gas to moles by using the molar mass of H2 gas.
Molar mass of H2 = 2 * atomic mass of H = 2 * 1.00794 g/mol = 2.016 g/mol
Moles of H2 = 45g of H2 * 〖1mole H〗_2/〖2.016g of H〗_2 = 22.32 mol of H2
So now we have
n = 22.32 moles, P = 1.05 atm .
T = 25°C = 25 + 273.15 = 298.15K.
Gas constant, R = 0.08206 L.atm/mol.K
Substitute this information in the ideal gas equation we have
PV = nRTPV=nRT
1.05 atm * V = 22.32 mol * 0.08206 L.atm/mol.K * 298.15K
1.05 atm * V = 546.09 L.atm.
divide both the side by 1.05 atm, we have
(1.05atm*V)/(1.05atm )=(546.09 L.atm)/1.05atm
V = 520.08 L.
In the given question the quantity with the least number of the significant figure is 45 grams.
It is in 2 significant figures. So our final answer must also be in 2 significant figures.
Hence the volume occupied by the hydrogen gas will be 520 L.5) At what temperature will 5.00 g of Cl2 exert a pressure of 900 mm Hg at a volume of 750.0 mL?
Answer
Here too we are given a mass of Cl2 gas. So first we will have to convert it to grams by using the molar mass of Cl2.
Molar mass of Cl2 = 2 * atomic mass of Cl = 2 * 35.453 g/mol = 70.906 g/mol
Moles of Cl2 = 5.00 g of Cl2 * 〖1mole Cl〗_2/〖70.906g of Cl〗_2 = 0.07052 moles of Cl2
Convert 750.0mL to ‘L’ and 900 mm Hg to ‘atm’
Pressure in ‘atm’ = 900mmHg * 1atm/760mmHg = 1.184atm (since 1atm – 760mmHg)
So now we have
Moles of Cl2 = 0.07052 mol
Volume occupied by Cl2 = 0.750 L
Pressure = 1.184 atm.
Gas constant, R = 0.08206 L.atm/mol.K
Plug all this information in the idea gas equation we have
PV = nRTPV=nRT
1.184 atm * 0.750L = 0.07052 mol * 0.08206 L.atm/mol.K * T
0.8882 L.atm = 0.005787 Latm/K * T
Divide both the side by 0.005784 L.atm/mol, we have
(0.8882L.atm)/(0.005787L.atm/K )=(0.005787L.atm/K*T)/(0.005787L.atm/mol)
T= 153.48K
Which in 3 significant figure is 153 K or in
Degree Celsius is = 153.45 K – 273.15 = 119.66 °C = 120 °C
Hence the temperature of the Cl2 gas will be 153 K or 120 °C.6) A 143.03 g sample of a gas is at 1789.7 torr, 256.8 K and occupies 29.1 L. What is the chemical formula for this gas? (hint: find the molar mass first)
Answer
The best we can do with this information is find the molar mass, and take a guess as to its identity from that
This formula is derived from PV = nRT by substitutions
m/VRT = M P
Using R compatible with L, Torr, mol and K
143.03/29.1(62.4)(256.8) = M (1789.7)
M = 44.01
This agrees very closely with the molar mass of the common gas is Carbon dioxide (CO2)7) A 28.8 g piece of dry ice (solid carbon dioxide) is allowed to sublime (convert directly from a solid to a gas) into a large balloon. Assuming that all of the carbon dioxide ends up in the balloon, what will the volume of the balloon be at a temperature of 22°C and a pressure of 742 mm Hg?
Answer
n = 28.8g CO2 * mol/44g=0.655mol
where T = 22+273 = 295K
P= 743mmHg * (1 atm)/(760 mmhg) = 0.978atm
v=nRT/p=(0.655mol)(0.0821)(295K)/0.978atm=16.2L8) Calculate the density of NO2 gas at 0.97 atm and 35°C.
Answer
Density = mass/volume
The molar volume, (one mole NO2 is 46grams)
Gas constant: 0.082057338 Latm/Kmol
So volume of one mole is
V=nRT/P =(1*.083*(273+35))/0.970
V= 1.77g/L
9) Ammonium nitrate decomposes explosively upon heating according to the following equation:
2NH4NO3(s) —–> 2NH3 (g) + CO2 (g) + H2O (g)
Calculate the total volume of gas produced at 125°C and 748 mm Hg when 1.55 kg of ammonium nitrate completely decomposes.Answer
Our first step will be to balance the given reaction.
NH4NO3(s) —–> N2 (g) + O2 (g) + H2O (g)
We will first balance the element which is not present in a single elemental form. So N and O will be balanced afterward. First, we will balance hydrogen.
There is 4 H on the left side and 2 on the right side, so to balance the H write 2 in front of H2O.
This will balance H on both the side.
NH4NO3(s) —–> N2 (g) + O2 (g) + 2 H2O (g)
Nitrogen is already balanced here. There are 3 O on the left side and 4 on the right side.
To balance the oxygen we will have to write a proper coefficient in front of O2.
Placing a fraction 1/2 in front of O2 will make oxygen equal on both the side.
NH4NO3(s) —–> N2 (g) + 1/2 O2 (g) + 2 H2O (g)
We can’t keep fractions in our balanced equation. To remove fraction multiply the whole reaction by 2.
2 * ( NH4NO3(s) —–> N2 (g) + 1/2 O2 (g) + 2 H2O (g) )
2 NH4NO3(s) —–> 2 N2 (g) + 1 O2 (g) + 2 H2O (g)
So now this reaction is balanced.
Step 1: Convert 1.55 kg of NH4 NO3 to moles by using the molar mass of NH4NO3
1.55 kg NH4NO3 in grams = 1.55 kg * 1000g/1kg = 1550 grams of NH4NO3
molar mass of NH4 NO3 = 2 * atomic mass of N + 3 * atomic mass of O + 4 * atomic mass of H
= 2 * 14.007 g/mol + 3 * 15.999 g/mol + 4 * 1.00794 g/mol = 80.0426g/mol
Moles of NH4NO3 = 1550 g of NH4NO3 * 1mol NH4NO3 / 80.0426 g NH4NO3
= 19.365 mol NH4NO3
In the balanced reaction, we can see that
2 NH4NO3(s) —–> 2 N2 (g) + 1 O2 (g) + 2 H2O (g)
2 moles of NH4 NO3 gives total = 2 + 1 + 2 = 5 moles of gas.
So 19.365 moles of NH4NO3 will give
= 19.365 mol NH4NO3 * 5 mol gas / 2 mol NH4 NO3 = 48.412 mol gas
Step 2: To find the volume occupied by the gas.
We have moles of gas = 48.412 moles
T = 125°C = 398.15K
Pressure P = 748 mm Hg, which in átm = 748mmHg * 1atm/(760 mm Hg) = 0.984 atm .
Volume = unknown
Gas constant R = 0.08206 L.atm/mol.K
Moles of gas = 48.412 mol
Plug all this information in the ideal gas equation, PV = nRT
0.984 atm * V = 48.412 mol * 0.08206 L.atm/mol.K * 398.15K
0.984atm * V = 1581.7 Latm
Divide both the side by 0.984 atm we have
V = (1581.73Latm)/(0.984 atm) = 1607.4 L
Hence the volume occupied by the combusted gas after completely decompostion of NH4NO3 is = 1600Liters. 
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