Answer:
M = 20.5 g/mol
Explanation:
Given data:
Volume of gas = 1.20 L
Mass of gas = 1.10 g
Temperature and pressure = standard
Solution:
First of all we will calculate the density.
Formula:
d = mass/ volume
d = 1.10 g/ 1.20 L
d = 0.92 g/L
Now we will calculate the molar mass.
d = PM/RT
0.92 g/L = 1 atm × M / 0.0821 atm.L/mol.K ×273.15 K
M = 0.92 g/L × 0.0821 atm.L/mol.K ×273.15 K / 1 atm
M = 20.5 g/mol
Answer:
12.4 × 10∧3 atoms
Explanation:
Given data:
moles of oxygen molecule= 1.0000 x 10-20 mol
atoms =?
Solution:
32 g O2 = 1 mol = 6.02 × 10∧23
1.0000 x 10∧-20 mol × 6.02 × 10∧23 × 2 = 12.4 × 10∧3 atoms
Answer : The mass of reactant 
remain would be, 0.20 grams.
Solution : Given,
Moles of = 0.40 mol
Moles of 
= 0.15 mol
Molar mass of = 2 g/mole
First we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
From the balanced reaction we conclude that
As, 1 mole of react with 2 mole of
So, 0.15 moles of react with 
moles of
From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.
The moles of reactant remain = 0.40 - 0.30 = 0.10 mole
Now we have to calculate the mass of reactant remain.
Therefore, the mass of reactant remain would be, 0.20 grams.
The answer is (3) pH 3 to pH 1. The pH is related to the concentration of H3O+ with the relationship: pH = -lg c(H+). So when concentration of H3O+ increase, the pH will decrease. And decrease 2 when when increase hundredfold. Because 100=10^2.
