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Ber [7]
3 years ago
10

Question 5 Multiple Choice Worth 4 points)

Chemistry
1 answer:
777dan777 [17]3 years ago
3 0

Answer:

This cannot be determined without knowing the actual mass of the objects.

Explanation:

its like trying to compare the letter A and letter B

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A 50.0-ml sample of 0.200 m sodium hydroxide is titrated with 0.200 m nitric acid. calculate the ph in the titration after you a
attashe74 [19]

Hey there!:

Concentration of NaOH = 0.200 M

Concentration of HNO₃= 0.200 M

Total volume =  50.0 mL + 60.0 mL = 110 mL=> 0.11 L

The neutralization reaction between  NaOH and HNO3 :

OH⁻  + H⁺  ---------->  H₂O

So :

n ( H⁺ ) = 60 mL * 0.200 M / 1000 mL  => 0.012 moles of H⁺

n ( OH⁻ ) = 50 mL 0.200 M / 1000 mL => 0.01 moles of OH⁻

Hence OH⁻ is limiting reagent  .

Remaining moles of  H⁺ = 0.012 - 0.01  =>  0.002 moles

Concentration of H⁺  = 0.002 M / 0.11 L

Concentration of H⁺ = 0.01818 moles/L

Therefore:

pH = - log [ H⁺ ]

pH = - log [ 0.01818 ]

pH = 1.74

Hope that helps!

7 0
3 years ago
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4 years ago
What is the percentage composition for H2O in this compound Al(OH)3.2H2O?
Gelneren [198K]

Answer:

32.73%

Explanation:

To solve this problem, first find the molar mass of Al(OH)₃.2H₂O

 Atomic mass of Al  = 27g/mol

                            O = 16g/mol

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                    = 27 + 51 + 36

                   = 114g/mol

Percentage composition  = \frac{36}{114} x 100 = 32.73%

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I’m pretty sure it’s b
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