PV=nRT
3.5×X=16.5×0.082×295
X= 114 L
The volume of the air mattress is 114 liters.
Answer:
39.2 g
Explanation:
- 2Ni₂O₃(s) ⟶ 4Ni(s) + 3O₂(g)
First we <u>convert 55.3 grams of Ni₂O₃ into moles of Ni₂O₃</u>, using its<em> molar mass</em>:
- 55.3 g ÷ 165.39 g/mol = 0.334 mol Ni₂O₃
Then we <u>convert 0.334 moles of Ni₂O₃ into moles of Ni</u>, using the <em>stoichiometric coefficients of the balanced reaction</em>:
- 0.334 mol Ni₂O₃ *
= 0.668 mol Ni
Finally we <u>calculate how much do 0.668 Ni moles weigh</u>, using the<em> molar mass of Ni </em>:
- 0.668 mol Ni * 58.69 g/mol = 39.2 g
Answer: It is 5450 mL
Explanation: There are 1000 mL in every L and then there is an extra 450 so just add that at the end
Answer:
increase
Explanation:
Let's suppose we have a sample of air in a closed container. We heat the container and we want to predict what would happen to the pressure.
According to Gay-Lussac's law, the pressure of a gas is directly proportional to its absolute temperature.
Thus, if we increased the temperature of the air by heating it, its pressure would increase.
If a sample of air in a closed container was heated, the total pressure of the air would increase.
The answer is 34.1 mL.
Solution:
Assuming ideal behavior of gases, we can use the universal gas law equation
P1V1/T1 = P2V2/T2
The terms with subscripts of one represent the given initial values while for terms with subscripts of two represent the standard states which is the final condition.
At STP, P2 is 760.0torr and T2 is 0°C or 273.15K. Substituting the values to the ideal gas expression, we can now calculate for the volume V2 of the gas at STP:
(800.0torr * 34.2mL) / 288.15K = (760.0torr * V2) / 273.15K
V2 = (800.0torr * 34.2mL * 273.15K) / (288.15K * 760.0torr)
V2 = 34.1 mL
