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NeX [460]
3 years ago
10

2KOH + H2SO4 → K2SO4 + 2H20

Chemistry
1 answer:
mars1129 [50]3 years ago
6 0

19.6 × ( 1 mol KOH / 56 grams KOH )

= 0.35 mol KOH

_________________________________

0.35 mol KOH ×( 2 mol H2O / 2 mol KOH )

= 0.35 mol H20

_________________________________

0.35 mol H2O × ( 18 g H2O / 1 mol H2O )

= 6.3 grams H2O

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Answer:

3.4752 moles of water

Explanation:

There are 13.84 mole in one cup of water so,

13.84 divided by 4= 3.4725 :)

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A volume of 90.0 mLmL of aqueous potassium hydroxide (KOHKOH) was titrated against a standard solution of sulfuric acid (H2SO4H2
Alja [10]

Answer:

0.823 M was the molarity of the KOH solution.

Explanation:

H_2SO_4+KOH\rightarrow K_2SO_4+2H_2O (Neutralization reaction)

To calculate the concentration of base , we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is KOH.

We are given:

n_1=2\\M_1=1.50 M\\V_1=24.7 mL\\n_2=1\\M_2=?\\V_2=90.0 mL

Putting values in above equation, we get:

2\times \1.50 M\times 24.7 mL=1\times M_2\times 90.0 mL

M_2=\frac{2\times 1.50M\times 24.7 mL}{1\times 90.0 mL}=0.823 M

0.823 M was the molarity of the KOH solution.

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4 years ago
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Concentration of Ni in 20mL = 5.28ppm x dilution factor = 5.28 x 100/5 = 105.6 ppm = 105.6 mg/L 

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<span>Molarity of Ni = 100.40 x 10^{-3} / 58.6934 = 1.71 x 10^{-3} M = 1.71 mM. </span>
5 0
4 years ago
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