1)A sample of CO2 gas which weighs 0.85 grams has a volume of 1.30 liters when collected at 30°C. What would be the pressure of the gas sample?
2) A gas sample has an original volume of 600 ml when collected at 720 mm and 35°C. If a change is made in the gas temperature which causes the volume of the gas sample to become 520 ml at 1.00 atm, what is the new temperature?
3) A gas sample has an original volume of 760 ml when collected at 720 mm and 25°C. What will be the volume of the gas sample if the pressure increases to 820 mm and the temperature increases to 40°C?
4)H3AsO4 + Zn + HNO3 → AsH3 + Zn(NO3)2 + H2O
a. PV = nRT and n = grams/molar mass
b. (V1/T1) = (V2/T2)
3. (P1V1/T1) = (P2V2/T2)
4. This is a redox equation. This link will tell all about how to do these.
http://www.chemteam.info/Redox/Redox.html
1) To find the pressure of the gas sample, we can use the ideal gas law equation: PV = nRT.
- P represents the pressure of the gas (in atm).
- V represents the volume of the gas (in liters).
- n represents the number of moles of gas.
- R is the ideal gas constant (0.0821 L atm/mol K).
- T is the temperature of the gas (in Kelvin).
First, convert the grams of CO2 to moles by dividing the mass by the molar mass of CO2. The molar mass of CO2 is 44.01 g/mol. So, 0.85 g / 44.01 g/mol = 0.0193 moles.
Next, convert the Celsius temperature to Kelvin by adding 273.15. 30°C + 273.15 = 303.15 K.
Now we can plug the values into the ideal gas law equation:
PV = nRT
P * 1.30 L = 0.0193 mol * 0.0821 L atm/mol K * 303.15 K
Simplifying the equation:
P * 1.30 = 0.0193 * 0.0821 * 303.15
P * 1.30 = 0.4979
P = 0.4979 / 1.30
P = 0.383 atm
Therefore, the pressure of the gas sample is 0.383 atm.
2) To find the new temperature when the gas volume changes at constant pressure, we can use the combined gas law equation:
(P₁ * V₁) / T₁ = (P₂ * V₂) / T₂
- P₁ is the original pressure (720 mmHg),
- V₁ is the original volume (600 mL),
- T₁ is the original temperature (35°C + 273.15 = 308.15 K),
- P₂ is the new pressure (1.00 atm),
- V₂ is the new volume (520 mL).
Rearranging the equation to solve for T₂:
T₂ = ((P₂ * V₂) * T₁) / (P₁ * V₁)
Let's plug in the values:
T₂ = ((1.00 atm * 520 mL) * 308.15 K) / (720 mmHg * 600 mL)
To convert mmHg to atm, divide by 760:
T₂ = ((1.00 atm * 520 mL) * 308.15 K) / (720 mmHg * 600 mL / 760)
Simplifying the equation:
T₂ = (0.684 atm * 308.15 K) / (0.95 atm)
T₂ ≈ 222.53 K
Therefore, the new temperature is approximately 222.53 K.
3) To find the new volume of the gas sample when pressure and temperature change, we can use the combined gas law equation:
(P₁ * V₁) / T₁ = (P₂ * V₂) / T₂
- P₁ is the original pressure (720 mmHg),
- V₁ is the original volume (760 mL),
- T₁ is the original temperature (25°C + 273.15 K = 298.15 K),
- P₂ is the new pressure (820 mmHg),
- V₂ is the new volume (to be found),
- T₂ is the new temperature (40°C + 273.15 K = 313.15 K).
Rearranging the equation to solve for V₂:
V₂ = (P₂ * V₁ * T₂) / (P₁ * T₁)
Let's plug in the values:
V₂ = (820 mmHg * 760 mL * 313.15 K) / (720 mmHg * 298.15 K)
To convert mmHg to atm, divide by 760:
V₂ = (1.079 atm * 760 mL * 313.15 K) / (720 mmHg * 298.15 K / 760)
Simplifying the equation:
V₂ = 1.226 mL
Therefore, the new volume of the gas sample will be approximately 1.226 mL.
4) The given equation is a balanced chemical equation showing a reaction between H3AsO4 (arsenic acid), Zn (zinc), and HNO3 (nitric acid) that produces AsH3 (arsine), Zn(NO3)2 (zinc nitrate), and H2O (water).