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alexgriva [62]
3 years ago
8

Suppose a student titrates a 10.00-ml aliquot of saturated ca(oh)2 solution to the equivalence point with 16.08 ml of 0.0199 m h

cl. what was the initial [oh − ]?
Chemistry
1 answer:
Alborosie3 years ago
6 0
Following chemical reaction is involved upon titration of Ca(OH)2 with HCl,
Ca(OH)2 + 2HCl ↔ CaCL2 + 2H2O

Above is an example of acid-base titration to generate salt and water. Here, H+ ions of acid (HCl) combines with OH- (ions) of base [Ca(OH)2] to generated H2O

Given,
concentration of HCl = 0.0199 M
Total volume of HCl consumed during titration = 16.08 mL = 16.08 X 10^(-3) L

∴, number of moles of H+ consumed = Molarity X Vol. of HCl (in L)
                                                           = 0.0199 X 16.08 X 10^(-3)
                                                           = 3.1999 X 10^-4 mol
Thus, total number of moles of [OH-] ions present initial = 3.1999 X 10-4 mol
So, initial conc. [OH-] ion = 
\frac{number of moles of [OH-]}{volume of solution (L)} = \frac{3.1999 X 10^(^-^4^)}{10 X 10^(^-^3^)} = 0.03199 M
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4 0
3 years ago
Which of the following is the strongest base? ta) CH,ONa (b) NaNH, (c) CH-CH,Li (0) NaOH (6) CHÚCONa
Mashutka [201]

Hey there:

Correct answer is :

(b) NaNH₂

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7 0
3 years ago
What mass of copper is deposited when a current of 10.0a is passed through a solution of copper(ii) nitrate for 30.6 seconds?
asambeis [7]
Data Given:

Time = t = 30.6 s

Current = I = 10 A

Faradays Constant = F = 96500

Chemical equivalent = e = 63.54/2 = 31.77 g

Amount Deposited = W = ?

Solution:
             According to Faraday's Law,

                                          W  =  I t e / F

Putting Values,
 
                            W  =  (10 A × 30.6 s × 31.77 g) ÷ 96500

                            W  =  0.100 g

Result:
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3 0
3 years ago
In the following reaction, how many grams of NaBr will react with 311 grams of Pb(NO3)2?
Mashutka [201]
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</span></span>2 NaBr + 1 Pb(NO3)2 = 2 NaNO3 + 1 PbBr<span>2
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      g NaBr  -------------------> 311 g Pb(NO3)2

331.20  g  =   2*103*311

331.20 g = 64066

mass ( NaBr ) =  64066 / 331.20

mass ( naBr)  = 193,43 g of NaBr

hope this helps!.


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5 0
3 years ago
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vlabodo [156]
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6 0
3 years ago
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