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3 years ago

The density (g/L) of CO2 gas at STP ( 273 K, 1.00 atm) is

Chemistry

1 answer:

Triss [41]3 years ago

3 0

The density : 1.96 g/L

<h3>Further explanation</h3>

Ideal gas Law

Pv=nRT

where

P = pressure, atm , N/m²

V = volume, liter

n = number of moles

R = gas constant = 0.082 l.atm / mol K (P= atm, v= liter),or 8,314 J/mol K (P=Pa or N/m², v= m³)

T = temperature, Kelvin

so for density,

MW CO₂ = 44 g/mol

$\tt \rho=\dfrac{P\times MW}{RT}\\\\\rho=\dfrac{1\times 44}{0.082\times 273}\\\\\rho=1.96~g/L$

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3 years ago

A company produces a colorless vinegar that is 5.0% HC2H3O2 in water. Using thymol blue as an indicator, a student titrates a 15

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<h3>What is a neutralization reaction?</h3>

A neutralization reaction is a reaction between an acid and a base to form salt and water only.

The reaction between vinegar and NaOH is a neutralization reaction which produces a salt and water only.

NaOH is composed of a positive sodium ion, Na+ and a negative hydroxide ion, OH-

Therefore, the negative ion in the NaOH(aq) used in this titration is hydroxide ion, OH-.

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2 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

T= 24.5+273.15= 297.65 K

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

4 0

3 years ago

If 3.85 mL of 10.0 M NaOH is used for a reaction, calculate the number of moles of NaOH that were used.

Lena [83]

The number of mole of NaOH that was used, given that 3.85 mL of 10.0 M NaOH is used in the reaction is 0.0385 mole

<h3>How do I determine the number of mole of NaOH?</h3>

We know that molarity is related to number of mole and volume according to the following formula:

Molarity = number of mole / Volume

With the above formula, we can determine the number of mole of NaOH. Details below

The following data were obtained from the question:

Volume = 3.85 mL = 3.85 / 1000 = 0.00385 L
Molarity = 10.0 M
Number of mole of NaOH =?

Molarity = number of mole / Volume

Cross multiply

Number of mole = molarity × volume

Number of mole of NaOH = 10 × 0.00385

Number of mole of NaOH = 0.0385 mole

Thus, the number of mole of NaOH is 0.0385 mole

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2 years ago