1) we calculate the molar mass of He (helium) and Kr (Krypton).
atomic mass (He)=4 u
atomic mas (Kr)=83.8 u
Therefore the molar mass will be:
molar mass(He)=4 g/mol
molar mass(Kr)=83.8 g/mol.
1) We can find the next equation:
mass=molar mass x number of moles.
x=number of moles of helium
y=number of moles of helium.
(4 g/mol) x +(83.8 g/mol)y=103.75 g
Therefore, we have the next equation:
(1)
4x+83.8y=103.75
2) We can find other equation:
We have 30% helium atoms and 70% Kryptum atoms, therefore we have 30% Helium moles and 70% of Krypton moles.
1 mol is always 6.022 * 10²³ atoms or molecules, (in this case atoms).
Then:
x=number of moles of helium
y=number of moles of helium.
(x+y)=number of moles of our sample.
x=30% of (x+y)
Therefore, we have this other equation:
(2)
x=0.3(x+y)
With the equations(1) and (2), we have the next system of equations:
4x+83.8y=103.75
x=0.3(x+y) ⇒ x=0.3x+0.3y ⇒ x-0.3x=0.3y ⇒ 0.7 x=0.3y ⇒ x=0.3y/0.7
⇒x=3y/7
We solve this system of equations by substitution method.
x=3y/7
4(3y/7)+83.8y=103.75
lower common multiple)7
12y+586.6y=726.25
598.6y=726.25
y=1.21
x=3y/7=3(1.21)/7=0.52
We have 0.52 moles of helium and 1.21 moles of Krypton.
1 mol=6.022 * 10²³ atoms
Total number of particles=(6.022 *10²³ atoms /1 mol) (number of moles of He+ number of moles of Kr).
Total number of particles=6.022 * 10²³ (0.52+1.21)=6.022 * 10²³ (1.73)=
=1.044 * 10²⁴ atoms.
Answer: The sample have 1.044 * 10²⁴ atoms.
Answer:
The final pressure of the gas mixture after the addition of the Ar gas is P₂= 2.25 atm
Explanation:
Using the ideal gas law
PV=nRT
if the Volume V = constant (rigid container) and assuming that the Ar added is at the same temperature as the gases that were in the container before the addition, the only way to increase P is by the number of moles n . Therefore
Inicial state ) P₁V=n₁RT
Final state ) P₂V=n₂RT
dividing both equations
P₂/P₁ = n₂/n₁ → P₂= P₁ * n₂/n₁
now we have to determine P₁ and n₂ /n₁.
For P₁ , we use the Dalton`s law , where p ar1 is the partial pressure of the argon initially and x ar1 is the initial molar fraction of argon (=0.5 since is equimolar mixture of 2 components)
p ar₁ = P₁ * x ar₁ → P₁ = p ar₁ / x ar₁ = 0.75 atm / 0.5 = 1.5 atm
n₁ = n ar₁ + n N₁ = n ar₁ + n ar₁ = 2 n ar₁
n₂ = n ar₂ + n N₂ = 2 n ar₁ + n ar₁ = 3 n ar₁
n₂ /n₁ = 3/2
therefore
P₂= P₁ * n₂/n₁ = 1.5 atm * 3/2 = 2.25 atm
P₂= 2.25 atm
Answer:
Maintaining a high starting-material concentration can render this reaction favorable.
Explanation:
A reaction is <em>favorable</em> when <em>ΔG < 0</em> (<em>exergonic</em>). ΔG depends on the temperature and on the reaction of reactants and products as established in the following expression:
ΔG = ΔG° + R.T.lnQ
where,
ΔG° is the standard Gibbs free energy
R is the ideal gas constant
T is the absolute temperature
Q is the reaction quotient
To make ΔG < 0 when ΔG° > 0 we need to make the term R.T.lnQ < 0. Since T is always positive we need lnQ to be negative, what happens when Q < 1. Q < 1 implies the concentration of reactants being greater than the concentration of products, that is, maintaining a high starting-material concentration will make Q < 1.
Answer:

Explanation:
Hello,
In this case, we apply the Gay-Lussac's law which allows us to understand the pressure-temperature behavior as a directly proportional relationship:

Thus, we solve for the final pressure P2 to obtain it as shown below:

Hence, we notice that the temperature doubles as well as the pressure.
Best regards.
A. N₂ (g) + 3 H₂ (g) --> 2 NH₃ (g)
B. The value for standard enthalpy of formation is empirical given that the reactants involved were pure elements. So, you can search this on the internet or in any textbook. The Hf for NH₃ is -46.0 kJ/mol.
C. C (s) + O₂ (g) --> CO₂ (g)
D. The Hf for CO₂ is <span>-393.5 kJ/mol
E. 4 Fe (s) + 3 O</span>₂ (g) --> 2 Fe₂O₃ (s)
F. The Hf for solid Fe₂O₃ is -826.0 kJ/mol.
G. C (s) + 2 H₂ (g) --> CH₄ (g)
H. The Hf for methane gas is -74.9 kJ/mol.