The molar mass of CO2 can be calculated as follows;
CO2 — 12 + (16x2) = 12+ 32 = 44 g
Therefore molar mass of CO2 is 44 g/mol
In 44 g of CO2 there’s 1 mol of CO2
Then 1 g of CO2 there’s 1/44 mol of CO2
Therefore in 78.3 g of CO2 there’s — 1/44 x 78.3 =1.78 mol of CO2
Answer: 75%
Explanation:
The following information can be gotten from the question:
Waste = 70kg
Theoretical yield = 280kg
Therefore, the actual yield will be the difference between the theoretical yield and the waste which will be:
= 280kg - 70kg = 210kg
The percent yield will now be:
= Actual yield / Theoretical yield × 100
= 210/280 × 100
= 3/4 × 100
= 75%
Answer:
P= 0.87g/mL or 0.87g/cm^3
Explanation:
P=m/v
P=density
P=17.4g/20mL
P= 0.87g/mL
1mL=1cm^3
LiOH is going to neutralize the acid because it’s a base
Answer:
3.4 M
Explanation:
M = grams/molar mass = ans./volume(L)
M = 919/180 = ans./1.5
