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AnnZ [28]
3 years ago
8

61. Write formulas for the compounds:

Chemistry
1 answer:
OleMash [197]3 years ago
3 0

Answer:

See explanation

Explanation:

(a) aluminum sulfate:

⇒ Al^3+ has oxidation number +3

⇒ SO4^2- has oxidation number -2

→ Al2(SO4)3

(b) ammonium dichromate:

 ⇒ NH4^+ has oxidation number +1

 ⇒ Cr2O7^2- has oxidation number -2

→ (NH4)2Cr2O7

(c) silicon tetrafluoride:

 ⇒ Si^4+ has oxidation number of +4

 ⇒ F- has an oxidation number of -1

→ SiF

 

(d) iron(III) oxide:

 ⇒ Fe^3+ has an oxidation number of +3

 ⇒ O has an oxidation number of -2

→ Fe2O3

(e) tricarbon disulfide:

→ C3S2

(f) cobalt(II) nitrate:

⇒ C has an oxidation number of +2

⇒ NO3- has an oxidation number of -1

→C(NO3)2

(g) strontium nitrite:

 ⇒ NO2- has an oxidation number of -1

 ⇒ Sr has an oxidation number of +2

→Sr(NO2)2   (N2O4Sr)

(h) hydrobromic acid:

 ⇒ H+ has an oxidation number of +1

 ⇒ Br- has an oxidation number of -1

→HBr

(i) iodic acid:

 ⇒ H+ has an oxidation number of +1

 ⇒ IO3- has an oxidation number of -1

→HIO3

(j) phosphorus dichloride trifluoride:

⇒ P has an oxidation number of +5

⇒ F has an oxidation number of -1

⇒ Cl has an oxidation number of -1

→PF3Cl2

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5 0
3 years ago
In a titration of 47.41 mL of 0.3764 M ammonia with 0.3838 M aqueous nitric acid, what is the pH of the solution when 47.41 mL +
Volgvan

<u>Answer:</u> The pH of the solution is 1.136

<u>Explanation:</u>

To calculate the moles from molarity, we use the equation:

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

  • <u>For ammonia:</u>

Molarity of ammonia = 0.3764 M

Volume of ammonia = 47.41 mL = 0.04741 L   (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

0.3764mol/L=\frac{\text{Moles of ammonia}}{0.04741L}\\\\\text{Moles of ammonia}=0.01784mol

  • <u>For nitric acid:</u>

Molarity of nitric acid = 0.3838 M

Volume of ammonia = (47.41 + 10.00) mL = 57.41 mL= 0.05741 L

Putting values in above equation, we get:

0.3838mol/L=\frac{\text{Moles of nitric acid}}{0.05741L}\\\\\text{Moles of nitric acid}=0.02203mol

After the completion of reaction, amount of nitric acid remained = 0.022 - 0.0178 = 0.0042 mol

For the reaction of ammonia with nitric acid, the equation follows:

                       NH_3+HNO_3\rightarrow NH_4NO_3

At t=0             0.0178   0.022

Completion        0     0.0042        0.0178

As, the solution of the reaction is made from strong acid which is nitric acid and the conjugate acid of weak base which is ammonia. So, the pH of the reaction will be based totally on the concentration of nitric acid.

To calculate the pH of the reaction, we use the equation:

pH=-\log[H^+]

where,

[H^+]=\frac{0.0042mol}{0.05741L}=0.0731M

Putting values in above equation, we get:

pH=-\log(0.0731)\\\\pH=1.136

Hence, the pH of the solution is 1.136

8 0
3 years ago
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