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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Answer:
A pH scale reading 13 indicates a strong base.
Explanation:
From my understanding:
1 -4 is a strong acid
4 - 7 is weak acid
7 - 9 is a weak base
9 - 14 is a strong base
Acid Rain is formed when chemicals in the air get into rain and up the acidity levels!!!!!!
Hope this helps guys!
Potassium K
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Answer : The correct option is, (b) +115 J/mol.K
Explanation :
Formula used :
where,
= change in entropy
= change in enthalpy of vaporization = 40.5 kJ/mol
= boiling point temperature = 352 K
Now put all the given values in the above formula, we get:
Therefore, the standard entropy of vaporization of ethanol at its boiling point is +115 J/mol.K
