Answer: 31.8 g
Explanation:
To calculate the moles :
According to stoichiometry :
1 mole of 
require 3 moles of 
Thus 0.59 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent as it is present in more amount than required.
As 1 mole of give = 2 moles of 
Thus 0.59 moles of give =
of
Mass of 
Thus 31.8 g of will be produced from the given masses of both reactants.
Exchange rate is 2 guilders for every 3 florins
The conversion factor will be
2 guilders / 3 florins = 3 florins / 2 guilders = 1
so if we wish to convert 17 guilders to florins we will use following conversion factor
3 florins / 2 guilders = B / C
D = 17guilders X 3 florins / 2 guilders = 25.5 florins
Answer:
Given
mass of H2O (m) =35.6g
molarmass (mr) = H2O ), 1x2+16=18g/mol
moles of H2O (n) =?
sln
n=m/mr
n=35.6g/18g/mol
n=1.978moles
the moles of H2O are 1.978moles
Answer:
9.91 mL
Explanation:
Using the combined gas law equation as follows;
P1V1/T1 = P2V2/T2
Where;
P1 = initial pressure (torr)
P2 = final pressure (torr)
V1 = initial volume (mL)
V2 = final volume (mL)
T1 = initial temperature (K)
T2 = final temperature (K)
According to the information provided in this question;
V1 = 15.0mL
V2 = ?
P1 = 760 torr
P2 = 1252 torr
T1 = 10°C = 10 + 273 = 283K
T2 = 35°C = 35 + 273 = 308K
Using P1V1/T1 = P2V2/T2
760 × 15/283 = 1252 × V2/308
11400/283 = 1252V2/308
Cross multiply
11400 × 308 = 283 × 1252V2
3511200 = 354316V2
V2 = 3511200 ÷ 354316
V2 = 9.91 mL
