Answer:
Answer choices A, B, and C all contain a certain level <em>(or amount, if you will)</em> of <u>Calcium Carbonate</u>, but <u><em>B. chalk</em></u> contains the most amount out of all three.
Explanation:
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Answer:
= 82%
Explanation:
Percentage purity is calculated by the formula;
% purity = (mass of pure chemical/total mass of sample) × 100
In this case;
1 mole of Ca(NO3)2 = 164 g
but; 164 g of Ca(NO3)2 = 40 g Ca
Therefore; mass of Ca(NO3)2 = 164 /40
= 4.1 g
Thus;
% purity of Ca(NO3)2 = (Mass of Ca(NO3)2/ mass of the sample)× 100
= (4.1 g/ 5 g) × 100
= 82%
Answer:
2NaOH + CO2 -> Na2CO3 + H2O
1) Find the moles of each substance
2) Determine the limitting reagent
∴ Carbon dioxide is limitting as it has a smaller value.
3) multiply the limiting reagent by the mole ratio of unknown over known
n(H2O ) = 0.3976369007 × 1/2
= 0.1988184504 moles
4) Multiply the number of moles by the molar mass of the substance.
m = 0.1988184504 × (1.008 × 2 + 16.00)
= 0.1988184504 × 18.016
= 3.581913202 g
Explanation: