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Katarina [22]
3 years ago
10

if a sample of oxygen gas occupies a volume of 2.15 At a pressure of 0.572 atm and a temperature of 25 c what volume would this

sample occupy at STP
Chemistry
1 answer:
MaRussiya [10]3 years ago
8 0
Answer:
            1.126 L

Solution:

Let's assume the gas is behaving ideally, then initial and final states of given gas are given as,

                                      P₁ V₁ / T₁  =  P₂ V₂ / T₂   ---- (1)
Data Given;

                  P₁  =  0.572 atm

                  V₁  =  2.15 L

                  T₁  =  25 °C + 273  =  298 K

                  P₁  =  1.00 atm

                  V₂  =  ?

                  T₂  =  0 °C + 273  =  273 K

Solving equation 1 for V₂,

                 V₂  =  P₁ V₁ T₂ / T₁ P₂

Putting Values,

                 V₂  =  (0.572 atm × 2.15 L × 273 K) ÷ (298 K × 1 atm)

                 V₂  =  1.126 L
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The percent by mass of bicarbonate (HCO3−) in a certain Alka-Seltzer product is 32.5 percent. Calculate the volume of CO2 genera
nevsk [136]

Answer:

The volume of carbon dioxide gas generated 468 mL.

Explanation:

The percent by mass of bicarbonate in a certain Alka-Seltzer = 32.5%

Mass of tablet = 3.45 g

Mass of bicarbonate =3.45 g\times \frac{32.5}{100}=1.121 mol

Moles of bicarbonate ion = \frac{1.121 g/mol}{61 g/mol}=0.01840 mol

HCO_3^{-}(aq)+HCl(aq)\rightarrow H_2O(l)+CO_2(g)+Cl^-(aq)

According to reaction, 1 mole of bicarbonate ion gives with 1 mole of carbon dioxide gas , then 0.01840 mole of bicarbonate ion will give:

\frac{1}{1}\times 0.01840 mol=0.01840 mol of carbon dioxide gas

Moles of carbon dioxide gas  n = 0.01840 mol

Pressure of the carbon dioxide gas = P = 1.00 atm

Temperature of the carbon dioxide gas = T = 37°C = 37+273 K=310 K

Volume of the carbon dioxide gas = V

PV=nRT (ideal gas equation)

V=\frac{nRT}{P}=\frac{0.01840 mol\times 0.0821 atm L/mol K\times 310 K}{1.00 atm}=0.468 L

1 L = 1000 mL

0.468 L =0.468 × 1000 mL = 468 mL

The volume of carbon dioxide gas generated 468 mL.

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