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Triss [41]
3 years ago
9

In a 71.4 g sample of nahco 3 3 ​ , how many grams of sodium are present?

Chemistry
1 answer:
sweet [91]3 years ago
4 0

There are 19.5 g Na in 71.4 g NaHCO₃

Calculate the <em>molecular mass of NaHCO₃</em>.

1 Na = 1 × 22.99 u = 22.99   u

1 H   = 1 × 1.008 u =    1.008 u

1 C   = 1 × 12.01  u =  12.01    u

3 O = 3 × 16.00 u =  <u>48.00  u </u>

               TOTAL =  84.008 u

So, there are 22.99 g of Na in 84.008 g NaHCO₃.

∴ Mass of Na = 71.4 g NaHCO₃ × (22.99 g Na/84.008 g NaHCO₃) = 19.5 g Na

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Help pls! thank u (:
alexgriva [62]

Answer:

7.65 moles of silver are produced

Explanation:

Zinc, Zn, reacts with silver nitrate, AgNO3, as follows:

Zn + 2AgNO3 → Zn(NO3)2 + 2Ag

<em>Where 1 mole of Zn reacts with an excess of AgNO3 to produce 2 moles of Ag</em>

To solve this question we must convert the mass of Zn to moles and, using the chemical equation, we can find the moles of Ag as follows:

<em>Moles Zn (Molar mass: 65.38g/mol):</em>

250g Zn * (1mol / 65.38g) = 3.824 moles Zn

<em>Moles Ag:</em>

3.824 moles Zn * (2mol Ag / 1mol Zn) =

<h3>7.65 moles of silver are produced</h3>
7 0
3 years ago
what is the molarity of a RbOH solution if 60.0 mL of the solution is neutralized by 52.8 mL of a 0.5M HCl solution (Hint: Ma x
Dahasolnce [82]
RbOH is a strong base that dissociates completely and HCl is a strong acid that too dissociates completely. the complete reaction between the acid and base is;
RbOH + HCl ---> RbCl + H₂O
stoichiometry of acid to base is 1:1
At neutralisation point
H⁺ mol = OH⁻ mol
mol = molarity x volume 
if Ma - molarity of acid and Va - volume of acid reacted
Mb - molarity of base and Vb - volume of base reacted 
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0.5 M x 52.8 mL = Mb x 60.0 mL 
Mb = 0.44 M 
molarity of base - 0.44 M 

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