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Vladimir [108]
4 years ago
9

15. How are molecules and moles different from mass?

Chemistry
2 answers:
Ierofanga [76]4 years ago
7 0

Answer:

mass gives the mass of atoms while molecular weight gives the mass of molecules

kumpel [21]4 years ago
3 0
Mass gives the mass of atoms
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To solve the question you use the following formula:

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First, if you want to get the amount of production from the equation, you need to use the units of mol. So, we need to know the molar mass of C which is 15.75/12=1.31 mol. Then we need to use the less part(O2) to get the production amount which is 0.116 mol. Then we can get the answer in units of g: 0.116*(12+16*2)=5.104 g.
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4 years ago
I’ve been stuck on these 5 questions!? Can you guys help?!
Varvara68 [4.7K]

Answer:

1. 0.224 moles of oxygen

3. 143.36 L oxygen gas

5. 0.059 atm

10. 5.14 atm

11. 307 K

Explanation:

1. You have to use the ideal gas law: PV=nRT where P is pressure in atm, V is volume in liters, n is number of moles, R is the constant 0.08206 L atm mol^-1 K^-1, and T is the temperature in Kelvins where K=degrees celsius+273.15 . By rearranging the equation, you solve for n, which is n=(PV)/(RT)

P= 28.3 atm

V=0.193 L

R= 0.08206 L atm mol^-1 K^-1

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Plugging the values in,

n=(28.3 atm x 0.193 L)/(0.08206 L atm mol^-1 K^-1 x 297.65 K)

n= 0.224 moles of oxygen

3. At STP, there are 22.4 L of gas for every mole of gas present. So 6.4 moles of oxygen would mean that there are:

6.4 mol x 22.4 L= 143.36 L oxygen gas

5. You have to use the ideal gas law: PV=nRT where P is pressure in atm, V is volume in liters, n is number of moles, R is the constant 0.08206 L atm mol^-1 K^-1, and T is the temperature in Kelvins where K=degrees celsius+273.15 . By rearranging the equation, you solve for P, which is       P=(nRT)/V

n= 0.72 g converting to moles, divide by molar mass of oxygen gas:        0.72 g/32g= 0.0225 moles

V=9.3 L

R= 0.08206 L atm mol^-1 K^-1

T= 23.0+273.15= 296.15 K

Plugging the values in,

P=(0.0225 moles x 0.08206 L atm mol^-1 K^-1 x 296.15 K)/ 9.3 L

P= 0.059 atm

10. Ideal gas law again using the same equation as 5 above: You have to use the ideal gas law: PV=nRT where P is pressure in atm, V is volume in liters, n is number of moles, R is the constant 0.08206 L atm mol^-1 K^-1, and T is the temperature in Kelvins where K=degrees celsius+273.15 . By rearranging the equation, you solve for P, which is P=(nRT)/V

n= 0.108 mol

R=0.08206 L atm mol^-1 K^-1

T=20.0+273.15= 293.15 K

V= 0.505 L

Plugging the values in,

P=(0.108 mol x 0.08206 L atm mol^-1 K^-1 x 293.15 K)/0.505 L

P= 5.14 atm

11. You have to use the ideal gas law: PV=nRT where P is pressure in atm, V is volume in liters, n is number of moles, R is the constant 0.08206 L atm mol^-1 K^-1, and T is the temperature in Kelvins where K=degrees celsius+273.15 . By rearranging the equation, you solve for Y, which is T=(PV)/(nR)

P= 0.988 atm

V= 1.20 L

n= 0.0470 mol

R=0.08206 L atm mol^-1 K^-1

Plugging the numbers in,

T=(0.988 atm x 1.20 L)/(0.0470 mol x 0.08206 L atm mol^-1 K^-1)

T= 307 K

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