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erik [133]
3 years ago
12

What is the molar mass of lead nitrate (Pb(NO3)2) from the equation below?

Chemistry
1 answer:
AveGali [126]3 years ago
3 0

Answer:

331.2 g/molExplanation:

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Your job is to determine the concentration of ammonia in a commercial window cleaner. In the titration of a 25.0 mL sample of th
ruslelena [56]

Answer:

The initial concentration of ammonia is 0.14 M and the pH of the solution at equivalence point is 5.20

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}      .....(1)

Molarity of HCl solution = 0.164 M

Volume of solution = 23.8 mL = 0.0238 L    (Conversion factor:  1 L = 1000 mL)

Putting values in equation 1, we get:

0.164M=\frac{\text{Moles of HCl}}{0.0238L}\\\\\text{Moles of HCl}=(0.146mol/L\times 0.0238L)=0.0035mol

The chemical equation for the reaction of ammonia and HCl follows:

NH_3+HCl\rightarrow NH_4^++Cl^-

By Stoichiometry of the reaction:

1 mole of HCl reacts with 1 mole of ammonia

So, 0.0035 moles of HCl will react with = \frac{1}{1}\times 0.0035=0.0035mol of ammonia

  • Calculating the initial concentration of ammonia by using equation 1:

Moles of ammonia = 0.0035 moles

Volume of solution = 25 mL = 0.025 L

Putting values in equation 1, we get:

\text{Initial concentration of ammonia}=\frac{0.0035mol}{0.025L}=0.14M

By Stoichiometry of the reaction:

1 mole of ammonia produces 1 mole of ammonium ion

So, 0.0035 moles of ammonia will react with = \frac{1}{1}\times 0.0035=0.0035mol of ammonium ion

  • Calculating the concentration of ammonium ion by using equation 1:

Moles of ammonium ion = 0.0035 moles

Volume of solution = [23.8 + 25] mL = 48.8 mL = 0.0488 L

Putting values in equation 1, we get:

\text{Molarity of ammonium ion}=\frac{0.0035mol}{0.0488L}=0.072M

  • To calculate the acid dissociation constant for the given base dissociation constant, we use the equation:

K_w=K_b\times K_a

where,

K_w = Ionic product of water = 10^{-14}

K_a = Acid dissociation constant

K_b = Base dissociation constant = 1.8\times 10^{-5}

10^{-14}=1.8\times 10^{-5}\times K_a\\\\K_a=\frac{10^{-14}}{1.8\times 10^{-5}}=5.55\times 10^{-10}

The chemical equation for the dissociation of ammonium ion follows:

NH_4^+\rightarrow NH_3+H^+

The expression of K_a for above equation follows:

K_a=\frac{[NH_3][H^+]}{[NH_4^+]}

We know that:

[NH_3]=[H^+]=x

[NH_4^+]=0.072M

Putting values in above expression, we get:

5.55\times 10^{-10}=\frac{x\times x}{0.072}\\\\x=6.32\times 10^{-6}M

To calculate the pH concentration, we use the equation:

pH=-\log[H^+]

We are given:

[H^+]=6.32\times 10^{--6}M

pH=-\log (6.32\times 10^{-6})\\\\pH=5.20

Hence, the initial concentration of ammonia is 0.14 M and the pH of the solution at equivalence point is 5.20

5 0
3 years ago
HELP PLEASE THIS IS MY LAST QUESTION FOR THIS QUIZ
igor_vitrenko [27]

Answer:

Density = 19.3 g/cm³

Explanation:

D = m/v

m = 965 g

v = 50 cm³

so,

965 g / 50 cm³ = 19.3g/cm³

Hope this helps!

6 0
2 years ago
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A buffer solution contains 0.496 M hydrocyanic acid and 0.399 M sodium cyanide . If 0.0461 moles of sodium hydroxide are added t
pochemuha

Answer : The pH of the solution is, 9.63

Explanation : Given,

The dissociation constant for HCN = pK_a=9.31

First we have to calculate the moles of HCN and NaCN.

\text{Moles of HCN}=\text{Concentration of HCN}\times \text{Volume of solution}=0.496M\times 0.225L=0.1116mole

and,

\text{Moles of NaCN}=\text{Concentration of NaCN}\times \text{Volume of solution}=0.399M\times 0.225L=0.08978mole

The balanced chemical reaction is:

                          HCN+NaOH\rightarrow NaCN+H_2O

Initial moles     0.1116       0.0461     0.08978

At eqm.       (0.1116-0.0461)    0       (0.08978+0.0461)

                        0.0655                       0.1359

Now we have to calculate the pH of the solution.

Using Henderson Hesselbach equation :

pH=pK_a+\log \frac{[Salt]}{[Acid]}

Now put all the given values in this expression, we get:

pH=9.31+\log (\frac{0.1359}{0.0655})

pH=9.63

Therefore, the pH of the solution is, 9.63

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