Answer:
c. 20.0332 g to 20,0 g
Explanation:
A significant figure is each of the digits of a number that are used to express it to the required degree of accuracy, starting from the first non-zero digit, with the exception of the trailing zeros.
<em>Which of the following examples illustrates a number that is correctly rounded to three significant figures?
</em>
a. 109 526 g to 109 500 g. NO. The rounded number has 4 significant figures: 109 500.
b. 0.03954 g to 0.040 g. NO. The rounded number has 2 significant figures: 0.040.
c. 20.0332 g to 20.0 g. YES. The rounded number has 3 significant figures: 20.0.
d. 04.05438 g to 4.054 g. NO. The rounded number has 4 significant figures: 4.054.
e. 103.692 g to 103.7g. NO. The rounded number has 4 significant figures: 103.7.
Answer : The reagent present in excess and remains unreacted is, 
Solution : Given,
Moles of 
= 3.00 mole
Moles of = 2.00 mole
Excess reagent : It is defined as the reactants not completely used up in the reaction.
Limiting reagent : It is defined as the reactants completely used up in the reaction.
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
From the balanced reaction we conclude that
As, 2 moles of react with 1 mole of
So, 3.00 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.
Hence, the reagent present in excess and remains unreacted is,
Molar mass of MgCO3 is 84.313 g/mol
You can calculate this from data on the periodic table:
Molar mass Mg = 24.305g/mol
molar mass C = 12.011g/mol
molar mass O = 15.999g/mol mass 3 mol = 47.997g
Total = 84.313g/mol
Mass to be used in 1.2L of 1.5M solution = 84.313g * 1.2L * 1.5mol /L = 151.763g
I have not taken significant figures into account
The balanced equation you provide is not necessary in this calculation
Answer : The pressure of gas will be, 3.918 atm and the combined gas law is used for this problem.
Solution :
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
The combined gas equation is,
where,
= initial pressure of gas = 3 atm
= final pressure of gas = ?
= initial volume of gas = 1.40 L
= final volume of gas = 0.950 L
= initial temperature of gas = 
= final temperature of gas = 
Now put all the given values in the above equation, we get the final pressure of gas.
Therefore, the pressure of gas will be, 3.918 atm and the combined gas law is used for this problem.
Answer:
4.4g
Explanation:
Mass of CaCO3 = 10g
Mass of CaO = 5.6g
Mass of CO2 =?
Mass of CaCO3 = Mass of CaO + Mass of CO2
Mass of CO2 = Mass of CaCO3 — Mass of CaO
Mass of CO2 = 10 — 5.6
Mass of CO2 = 4.4g
