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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Latent heat, also called the heat of vaporization, is the amount of energy necessary to change a liquid to a vapour at constant temperature and pressure. The energy required to melt a solid to a liquid is called the heat of fusion
<span>S-shadows tells that the earths interior is liquid.</span>
The question is incomplete, the complete question is
Which is NOT correct for when the silver and vanadium half-cells are connected via a salt bridge and a potentiometer? Ag^+ + 1 e^- rightarrow Ag Edegree = 0.7993 V V^2+ + 2e^- right arrow V E degree =-1.125 V Ag+ is reduced V is oxidized 1.924 V V2^+ is reduced Ag is oxidized I and II III, IV, and V I, II, and III III only IV and V
Answer:
only IV and V
Explanation:
If we look at the values of reduction potential for the two species, we will discover that vanadium has a negative reduction potential indicating its tendency towards oxidation.
On the other hand, solve has a positive reduction potential indicating a tendency towards reduction.
This implies that vanadium must be oxidized and silver reduced and not the not her way ground? Hence the answer above.
