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larisa [96]
3 years ago
15

What is the mass percent of oxygen in sodium bicarbonate (NaHCO3)?

Chemistry
1 answer:
Alenkasestr [34]3 years ago
8 0
The answer is 57.14%.

First we need to calculate molar mass of <span>NaHCO3. Molar mass is mass of 1 mole of a substance. It is the sum of relative atomic masses, which are masses of atoms of the elements.

Relative atomic mass of Na is 22.99 g
</span><span>Relative atomic mass of H is 1 g
</span><span>Relative atomic mass of C is 12.01 g
</span><span>Relative atomic mass of O is 16 g.
</span>
Molar mass of <span>NaHCO3 is:
22.99 g + 1 g + 12.01 g + 3 </span>· <span>16 g = 84 g

Now, mass of oxygen in </span><span>NaHCO3 is:
3 </span>· 16 g = 48 g

mass percent of oxygen in <span>NaHCO3:
48 g </span>÷ 84 g · 100% = 57.14%

Therefore, <span>the mass percent of oxygen in sodium bicarbonate is 57.14%.</span>
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Answer:

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Explanation:

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kramer

Answer/Explanation:

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What volume of 1.00 m hcl in liters is needed to react completely (with nothing left over) with 0.750 l of 0.100 m na2co3?
kotykmax [81]
The balanced equation for the reaction is as follows
Na₂CO₃ + 2HCl --> 2NaCl + CO₂ + H₂O
stoichiometry of Na₂CO₃ to HCl is 1:2
number of Na₂CO₃ moles reacted = molarity x volume
number of Na₂CO₃ moles = 0.100 mol/L x 0.750 L = 0.0750 mol 
according to molar ratio of 1:2
1 mol of Na₂CO₃ reacts with 2 mol of HCl
then 0.0750 mol of Na₂CO₃ mol reacts with - 2 x 0.0750 = 0.150 mol 
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molarity is defined as the number of moles of solute in 1 L of solution 
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therefore there are 0.150 mol in - 0.150 mol / 1.00 mol/L = 0.150 L 
volume of HCl required is 0.150 L 
3 0
3 years ago
How many grams of ca are needed to react completely with 2.20 L of a 4.50 m hcl solution
Dmitry_Shevchenko [17]
Ca + 2HCl = CaCl₂ + H₂

c=4.50 mol/l
v=2.20 l

n(HCl)=cv

m(Ca)/M(Ca)=n(HCl)/2

m(Ca)=M(Ca)cv/2

m(Ca)=40g/mol·4.50mol/l·2.20l/2=198 g

198 grams of Ca are needed

5 0
3 years ago
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In two or more complete sentences explain how to balance the chemical equation and classify its reaction type
Dafna11 [192]

Answer:

Hi! In this case, the reactioncan be correctly balance according to this: 2Al(s) + 3CuSO4(aq) –> Al2(SO4)3(aq) + 3Cu(s).

Explanation:

In this particulary reaction,two semi-reactions happens.

One involving the metallic aluminum that suffers an oxidation reaction:

Al (s) -> Al3 + (aq) + 3e–

and another is a reduction reaction involving copper;

2e– + Cu2 + (aq) -> Cu (s)

4 0
3 years ago
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