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Nuetrik [128]
3 years ago
6

B. If the initial pressure of gas inside the balloon was 1.65 atm, what is the new pressure?

Chemistry
1 answer:
Orlov [11]3 years ago
8 0

Question 1 :

V1/T1 = V2/T2

3.0L/273K = V2/373K

To get the value of Z, cross multiply

3.0L x 373K = 273K x V2

1119 = 273V2

Divide both sides by 273

1119/273 = 273V2/273

4.10L = V2

The new volume is 4.10 liters

Question 2 :

P1/T1 = P2 /T2

P1 = 880 kPA= 880 *10^3 Pa

T1 = 250 K

T2 = 303 K

P2 =?

Substituting for P2

P2 = P1 T2/ T1

P2 = 880 kPa * 303 / 250

P2 = 266,640 kPa/ 250

P2 = 1066.56 kPa.

The new pressure of the gas is 1066.56 kPa

Question 3 :

Given that:

Volume of gas V = 4.80L

(since 1 liter = 1dm3

4.80L = 4.80dm3)

Temperature T = 62°C

Convert Celsius to Kelvin

(62°C + 273 = 335K)

Pressure P = 2.9 atm

Number of moles of gas N = ?

Apply ideal gas equation

pV = nRT

2.9atm x 4.8dm3 = n x (0.0082 atm dm3 K-1 mol-1 x 335K)

13.92 atm dm3 = nx 2.747 atm dm3 mol-1

n = 13.92/2.747

n = 5.08 moles

There are 5.08 moles of gas contained in the sample

Question 4 :

Volume of gas V = 3.47L

(since 1 liter = 1dm3

3.47L = 3.47dm3)

Temperature T = 85.0°C

Convert Celsius to Kelvin

(85.0°C + 273 = 358K)

Pressure P = ?

Number of moles of gas N = 0.100 mole

Apply ideal gas equation

pV = nRT

p x 3.47dm3 = 0.10 x (0.0082 atm dm3 K-1 mol-1 x 358K)

p x 3.47dm3 = 0.29 atm dm3

p = (0.29 atm dm3 / 3.47 dm3)

p = 0.085 atm

If 1 atm = 760 mm Hg

0.085atm = 0.085 x 760

= 64.6 mm Hg

The pressure of the gas is 64.6 mm hg

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A mixture contains NaHCO3 together with unreactive components. A 1.75 g sample of the mixture reacts with HA to produce 0.561 g
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Answer:

\%NaHCO_3=61.2\%

Explanation:

Hello.

In this case, since the undergoing chemical reaction is only between the sodium bicarbonate and the acid HA:

NaHCO_3+HA\rightarrow NaA+H_2O+CO_2

For 0.561 g of yielded carbon dioxide (molar mass 44 g/mol), the following mass of sodium bicarbonate (molar mass 84 g/mol) that reacted was:

m_{NaHCO_3}=0.561gCO_2*\frac{1molCO_2}{44gCO_2} *\frac{1molNaHCO_3}{1molCO_2} *\frac{84gNaHCO_3}{1molNaHCO_3} \\\\m_{NaHCO_3}=1.071g

Considering the 1:1 mole ratio between CO2 and NaHCO3. Finally, the percent by mass of NaHCO3 is computed by dividing the mass of reacted NaHCO3 and t the mixture:

\%NaHCO_3=\frac{1.071g}{1.75g}*100\%\\ \\\%NaHCO_3=61.2\%

Best regards.

5 0
3 years ago
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