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Using the correct answer from Part B, calculate the volume of a rectangular prism with a length of 5.6 cm, a width of 2.1 cm, an

Alenkinab [10]

5.6 x 2.1 x 6.6 = 77.616 cm3

5 0

2 years ago

Suppose the gas resulting from the sublimation of 1.00 g carbon dioxide is collected over water at 25.0◦c into a 1.00 l containe

jeyben [28]

Answer:

0.55 atm

Explanation:

First of all, we need to calculate the number of moles corresponding to 1.00 g of carbon dioxide. This is given by

$n=\frac{m}{M_m}$

where

m = 1.00 g is the mass of the gas

Mm = 44.0 g/mol is the molar mass of the gas

Substituting,

$n=\frac{1.00 g}{44.0 g/mol}=0.0227 mol$

Now we can find the pressure of the gas by using the ideal gas law:

$pV=nRT$

where

p is the gas pressure

V = 1.00 L is the volume

n = 0.0227 mol is the number of moles

R = 0.082 L/(atm K mol) is the gas constant

T = 25.0 C + 273 = 298 K is the temperature of the gas

Solving the formula for p, we find

$p=\frac{nRT}{V}=\frac{(0.0227 mol)(0.082 L/(atm K mol))(298 K)}{1.00 L}=0.55 atm$

8 0

3 years ago

A solid powder is known to be a mixture of NaCl and Na2CO3, but the relative amounts of each compound in the sample are unknown.

AveGali [126]

Answer:

Concentration of sodium carbonate in the solution before the addition of HCl is 0.004881 mol/L.

Explanation:

$Na_2CO_3(aq) +2 HCl(aq)\rightarrow 2NaCl(aq) + H_2O(l)+CO_2(g)$

Molarity of HCl solution = 0.1174 M

Volume of HCl solution = 83.15 mL = 0.08315 L

Moles of HCl = n

$molarity=\frac{moles}{Volume (L)}$

$0.1174 M=\frac{n}{0.08315 L}$

$n=0.1174 M\times 0.08315 L=0.009762 mol$

According to reaction , 2 moles of HCl reacts with 1 mole of sodium carbonate.

Then 0.009762 mol of HCl will recat with:

$\frac{1}{2}\times 0.009762 mol=0.004881 mol$

Moles of Sodium carbonate = 0.004881 mol

Volume of the sodium carbonate containing solution taken = 1L

Concentration of sodium carbonate in the solution before the addition of HCl:

$[Na_2CO_3]=\frac{0.004881 mol}{1 L}=0.004881 mol/L$

4 0

2 years ago

A container of 56mL is at STP, The pressure is increased to 3.6atm and the volume to 162 mL.

Sindrei [870]

Answer:

T2 = 2843.1 oK. This is a huge temperature. Check it for errors.

Explanation:

Remark

This is the same question as the other one I've answered. Only the numbers have been altered.

Givens

v1 = 56 mL

P1 = 1 atm

T1 = 273o K

v2 = 162

P2 = 3.6 atm

T2 = ?

Formula

Vi * P1 / T1 = V2 * P2/T2

Solution

Rearrange the formula so T2 is on the left

T2 = V2 P2 * T1 / (V1 * P1) Now just put the numbers in.

T2 = 162 * 3.6* 273 / (56 *1)

T2 = 159213.6/56

T2 = 2843.1

3 0

2 years ago

If you breathe normally you have about 0.11 moles of air in your lungs and

Firdavs [7]

Answer:

5.89 Liters

Explanation:

Lung capacity=(2.4/0.11) x 0.27

4 0

2 years ago