First, we need to get n1 (no.of moles of water ): when
mass of water = 0.0203 g and the volume = 1.39 L
∴ n1 = mass / molar mass of water
= 0.0203g / 18 g/mol
= 0.00113 moles
then we need to get n2 (no of moles of water) after the mass has changed:
when the mass of water = 0.146 g
n2 = mass / molar mass
= 0.146g / 18 g/ mol
= 0.008 moles
so by using the ideal gas formula and when the volume is not changed:
So, P1/n1 = P2/n2
when we have P1 = 1.02 atm
and n1= 0.00113 moles
and n2 = 0.008 moles
so we solve for P2 and get the pressure
∴P2 = P1*n2 / n1
=1.02 atm *0.008 moles / 0.00113 moles
= 7.22 atm
∴the new pressure will be 7.22 atm
Elements in the third row can break the octet rule
Answer:
C. NaOH acts as a reactant in the reaction
Explanation:
Because during the saponification process, Na+ replaces the H+ in the fatty acid been used for the saponification process
Answer:
15.75 grams of HNO3 was used and dissolved in 2.5 liters of solvent, to make a 0.10 M solution
Explanation:
Step 1: Data given
Nitric acid = HNO3
Molar mass of H = 1.01 g/mol
Molar mass of N = 14.0 g/mol
Molar mass O = 16.0 g/mol
Number of moles nitric acid (HNO3) = 0.25 moles
Molairty = 0.10 M
Step 2: Calculate molar mass of nitric acid
Molar mass HNO3 = Molar mass H + molar mass N + molar mass (3*O)
Molar mass HNO3 = 1.01 + 14.0 + 3*16.0
Molar mass HNO3 = 63.01 g/mol
Step 3: Calculate mass of solute use
Mass HNO3 = moles HNO3 * molar mass HNO3
Mass HNO3 = 0.25 moles * 63.01 g/mol
Mass HNO3 = 15.75 grams
15.75 grams of HNO3 was used and dissolved in 2.5 liters of solvent, to make a 0.10 M solution
