Answer:
1) 5.74 L.
2) 67.158 g/mol.
Explanation:
<u><em>Q1:</em></u>
- We suppose that the two gases CH₄ and O₂ behave ideally.
- We can use the general gas law of ideal gases:
<u><em>PV = nRT</em></u>
P is the pressure of the gases (P = 950 torr / 760 = 1.25 atm).
V is the volume of the gases (V = ??? L).
R is the general gas constant (R = 0.082 L.atm/mol.K).
T is the temperature of the gases (T = 400 K).
n is the no. of moles of the gases.
- <em>n = no. of moles of CH₄ + no. of moles of O₂.</em>
no. of moles of CH₄ = mass / molar mass = (2.0 g) / (16.0 g/mol) = 0.125 mol.
no. of moles of O₂ = mass / molar mass = (3.0 g) / (32.0 g/mol) = 0.09375 mol.
∴ n = no. of moles of CH₄ + no. of moles of O₂ = 0.125 mol + 0.09375 mol = 0.21875 mol.
∴ V = nRT/P = (0.21875 mol)(0.082 L.atm/mol.K)(400 K) / (1.25 atm) = 5.74 L.
<em><u>Q2:</u></em>
- We can also use the general gas law: PV = nRT.
- Each term in the law is defined in the answer of the first question.
- n = mass / molar mass, we substitute by this value in the gas law.
<em>PV = (mass/molar mass)RT.</em>
The molar mass = (mass)RT/PV.
The mass of the gas = 0.90 g.
R = 0.082 L.atm/mol.K.
T = 273 K at STP conditions.
P = 1.0 atm at STP conditions.
V = 300.0 mL = 0.30 L.
- The molar mass = (mass)RT/PV = (0.90 g)(0.082 L.atm/mol.K)(273.0 K) / (1.0 atm)(0.30 L) = 67.158 g/mol.
Answer:
the stream contains more water and moves faster
Explanation:
Answer:
3270m
Explanation:
1 minute= 109m
so
30 minutes= 109 × 30 = 3270m
(whatever you do on one side you have to do to the other)
hope this helps :)
Answer:
Atom is a particle that defines a chemical element, and is normally surrounded by 2 electrons.
Also, how do you that font ?
