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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Any substance that accept a proton by definition is considered to be BRONSTED LOWRY BASE.
Bronsted Lowry defined acid and base on the basis of donating or accepting protons. In the Bronsted Lowry classification of acid and base, an acid is defined as a substance which donate proton while a base is defined as a substance which accept proton.
Answer:
Explanation:
The positive (protons) and negative (electrons) charges balance each other in a neutral atom, which has a net zero charge. Because protons and neutrons each have a mass of 1, the mass of an atom is equal to the number of protons and neutrons of that atom.
Sapphire.
It is a stone not a metal.
Answer:
9.4
Explanation:
The equation for the reaction can be represented as:
+ 
⇄ 
The ICE table can be represented as:
+ 
⇄
Initial 0.27 0.49 0.0
Change -x -2x x
Equilibrium 0.27 - x 0.49 -2x x
We can now say that the concentration of at equilibrium is x;
Let's not forget that at equilibrium = 0.11 M
So:
x = [] = 0.11 M
[] = 0.27 - x
[] = 0.27 - 0.11
[] = 0.16 M
[] = (0.49 - 2x)
[] = (0.49 - 2(0.11))
[] = 0.49 - 0.22
[] = 0.27 M
![K_C = \frac{[CH_3OH]}{[CO][H_2]^2}](https://tex.z-dn.net/?f=K_C%20%3D%20%5Cfrac%7B%5BCH_3OH%5D%7D%7B%5BCO%5D%5BH_2%5D%5E2%7D)
= 9.4
∴ The equilibrium constant at that temperature = 9.4
