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Sonja [21]
3 years ago
7

A 25-mL sample of 0.160M solution of NaOH is titrated with 17 mL of an unknown solution of

Chemistry
1 answer:
leonid [27]3 years ago
8 0
V ( NaOH ) = 25 mL in liters : 25 / 1000 => 0.025 L

M ( NaOH ) = 0.160 M

V ( H2SO4) = 17 mL / 1000 => 0.017 L

M ( H2SO4) = ?

number of moles NaOH:

n = M x V = 0.160 x 0.025 => 0.004 moles NaOH

Mole ratio :

<span>2 NaOH + H2SO4 = Na2SO4 + 2 H2O
</span>
2 moles NaOH ----------- 1 mole H2SO4
0.004 moles NaOH ----- ? moles H2SO4

moles H2SO4 = 0.004 x 1 / 2

= 0.002 moles of H2SO4

M ( H2SO4) = n / V

M = 0.002 / 0.017

 = 0.117 M H2SO4

Answer B

hope this helps!

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Calculate the mass of beryllium (Be) needed to completely react with 18.9 g nitrogen gas (N2) to produce Bez N2, which is the on
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C = 18.29 g  

Explanation:

Given data:

Mass of beryllium needed = ?

Mass of nitrogen = 18.9 g

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Chemical equation:

3Be + N₂    →    Be₃N₂

now we will calculate the number of moles of nitrogen:

Number of moles = mass/molar mass

Number of moles = 18.9 g/ 28 g/mol

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Now we will compare the moles of nitrogen and Be from balance chemical equation.

                     N₂        :       Be        

                       1          :       3

                  0.675       :      3/1×0.675 = 2.03 mol

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