The mass would still be the same 25.0 g but the volume would be bigger
I would say the answer is A.
Answer:
The correct answer is c) 134L
Explanation:
We use the formula PV =nRT. The normal conditions of temperature and pressure are 273K and 1 atm, we use the gas constant = 0, 082 l atm / K mol.
1 atm x V = 5, 98 mol x 0, 082 l atm / K mol x 273 K
V = 5, 98 mol x 0, 082 l atm / K mol x 273 K / 1 atm
V = 133, 86828 l
Answer:
3.052 × 10^24 particles
Explanation:
To get the number of particles (nA) in a substance, we multiply the number of moles of the substance by Avogadro's number (6.02 × 10^23)
The mass of Li2O given in this question is as follows: 151grams.
To convert this mass value to moles, we use;
moles = mass/molar mass
Molar mass of Li2O = 6.9(2) + 16
= 13.8 + 16
= 29.8g/mol
Mole = 151/29.8g
mole = 5.07moles
number of particles (nA) of Li2O = 5.07 × 6.02 × 10^23
= 30.52 × 10^23
= 3.052 × 10^24 particles.
