The sample of argon gas that has the same number of atoms as a 100 milliliter sample of helium gas at 1.0 atm and 300 is 100. mL at 1.0 atm and 300. K
The correct option is D.
<h3>What is the number of moles of gases in the given samples?</h3>
The number of moles of gases in each of the given samples of gas is found below using the ideal gas equation.
The ideal gas equation is: PV/RT = n
where;
- P is pressure
- V is volume
- n is number of moles of gas
- T is temperature of gas
- R is molar gas constant = 0.082 atm.L/mol/K
Moles of gas in the given helium gas sample:
P = 1.0 atm, V = 100 mL or 0.1 L, T = 300 K
n = 1 * 0.1 / 0.082 * 300
n = 0.00406 moles
For the argon gas sample:
A. n = 1 * 0.05 / 0.082 * 300
n = 0.00203 moles
B. n = 0.5 * 0.05 / 0.082 * 300
n = 0.00102 moles
C. n = 0.5 * 0.1 / 0.082 * 300
n = 0.00203 moles
D. n = 1 * 0.1 / 0.082 * 300
n = 0.00406 moles
Learn more about ideal gas equation at: brainly.com/question/24236411
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The matter will be consumed by other living organisms and the blood will settle to the bottom of the body
Answer:
Water pressure 0.5 atm
Total Pressure= 2.27 atm
Explanation:
To answer this problem, one has to realize that there are two processes that increase the temperature of the sealed vessel.
First, the dry air in the sealed vessel will be heated which will cause its pressure to increase and it can be determined by the equation:
P₁ x T₂ = P₂ x T₁ ∴ P₂ = P₁ x T₂ / T₁
For the second process, we have an amount of n moles of water which will be released when the copper sulfate is heated. In this case, to determine the value of the the water gas we will use the gas law:
PV = nRT ∴ P = nRT/V
n will we calculated from the quantity of sample.
2.50 g CuSo₄ 5H₂O x 1 mol/ 249.69 g = 0.01 mol CuSo₄ 5H₂O
the amount water of hydration is
= 0.01 mol CuSo₄ 5H₂O * 5 mol H₂O / 1 mol CuSo₄ 5H₂O
= 0.05 mo H₂O
pressure of dry air at the final temperature,
P₂ = 1 atm x 500 K/ 300 K = 1.67 atm
Pressure of water :
P (H₂O) 0.05 mol x 0.08206 Latm/kmol x 500 K/ 4 L = 0.5 atm
∴ Total Pressure = 1.67 atm
H2O Pressure = 0.5 atm
M1V1 = M2V2
.200 (.025) = 1.60 X 10 -2 (V2)
V2 = .315 L
1.60 x 10-2 M in 315 mL
Answer:
-372000 J or -372 KJ
Explanation:
We have the electrochemical reaction as;
Mg(s) + Fe^2+(aq)→ Mg^2+(aq) + Fe(s)
We must first calculate the E∘cell from;
E∘cathode - E∘anode
E∘cathode = -0.44 V
E∘anode = -2.37 V
Hence;
E∘cell = -0.44 V -(-2.37 V)
E∘cell = 1.93 V
n= 2 since two electrons were transferred
F=96,500C/(mol e−)
ΔG∘=−nFE∘
ΔG∘= -( 2 * 96,500 * 1.93)
ΔG∘= -372000 J or -372 KJ
