Answer:
648.5 mL
Explanation:
Here we will assume that the pressure of the gas is constant, since it is not given or specified.
Therefore, we can use Charle's law, which states that:
"For an ideal gas kept at constant pressure, the volume of the gas is proportional to its absolute temperature"
Mathematically:
where
V is the volume of the gas
T is its absolute temperature
The equation can be rewritten as
where in this problem we have:
is the initial volume of the gas
is the initial temperature
is the final temperature
Solving for V2, we find the final volume of the gas:
545mm Hg in Kilopascals is 72.6607
I hope this helps you. Good luck stay safe, healthy and, happy!<3
Answer:
34g
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
H2S + 2AgNO3 —> 2HNO3 + Ag2S
Next, we shall determine the number of mole of H2S required to react with 2 moles of AgNO3.
This is illustrated below:
From the balanced equation above,
We can see that 1 mole of H2S is required to react completely with 2 moles of AgNO3.
Finally, we shall convert 1 mole of H2S to grams. This is shown below:
Number of mole H2S = 1 mole
Molar mass of H2S = (2x1) + 32 = 34g/mol
Mass = number of mole x molar Mass
Mass of H2S = 1 x 34
Mass of H2S = 34g
Therefore, 34g of H2S is needed to react with 2 moles of AgNO3.
Answer:
0.121 moles of aluminum metal are required to produce 4.04 L of hydrogen gas at 1.11 atm and 27 °C by reaction with HCl
Explanation:
This is the reaction:
2 Al(s) + 6 HCl(aq) → 2 AlCl₃ (aq) + 3 H₂(g)
To make 3 moles of H₂, we need 2 moles of Al.
By conditions given, we will find out how many moles of H₂ do we have.
Let's use the Ideal Gas Law
P. V = n . R . T
1.11 atm . 4.04L = n . 0.082 L.atm/mol.K . 300K
(1.11 atm . 4.04L) / (0.082 mol.K/L.atm . 300K) = n
0.182 mol = n
So the rule of three will be:
If 3 moles of H₂ came from 2 moles of Al
0.182 moles of H₂ will come from x
(0.182 .2) / 3 = 0.121 moles
Answer: 26.5 mm Hg
Explanation:
The vapor pressure is determined by Clausius Clapeyron equation:
where,
= initial pressure at 
= ?
= final pressure at 
= 100 mm Hg
= enthalpy of vaporisation = 28.0 kJ/mol =28000 J/mol
R = gas constant = 8.314 J/mole.K
= initial temperature = 
= final temperature =
Now put all the given values in this formula, we get
![\log (\frac{P_1}{100})=\frac{28000}{2.303\times 8.314J/mole.K}[\frac{1}{299.5}-\frac{1}{267.9}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BP_1%7D%7B100%7D%29%3D%5Cfrac%7B28000%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B299.5%7D-%5Cfrac%7B1%7D%7B267.9%7D%5D)
Thus the vapor pressure of 
in mmHg at 26.5 ∘C is 26.5
