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Alchen [17]
3 years ago
9

I really need helpppppppp

Chemistry
1 answer:
stich3 [128]3 years ago
5 0
4, 2, 3,1

brainliest would be fantastic
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Balancing the reaction by oxidation number method k2cr2o7+sncl2+hcl​
Mumz [18]

Answer:

K_2Cr_2O_7 (aq) + 14 HCl (aq) + 3 SnCl_2 (aq)\rightarrow 2 CrCl_3 (aq) + 7 H_2O (l) + 3 SnCl_4 (aq) + 2 KCl (aq)

Explanation:

The products of this reaction are given by:

K_2Cr_2O_7 (aq) + SnCl_2 (aq) + HCl (aq)\rightarrow KCl (aq) + SnCl_4 (aq) + CrCl_3 (aq) + H_2O (l)

Firstly, dichromate anion becomes chromium(III) cation, let's write this change:

Cr_2O_7^{2-} (aq)\rightarrow Cr^{3+} (aq)

The following steps should be taken:

  • balance the main element, chromium: multiply the right side by 2 to get 2 chromium species on both side:

Cr_2O_7^{2-} (aq)\rightarrow 2 Cr^{3+} (aq)

  • balance oxygen atoms by adding 7 water molecules on the right:

Cr_2O_7^{2-} (aq)\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)

  • balance the hydrogen atoms by adding 14 protons on the left:

Cr_2O_7^{2-} (aq) + 14 H^+ (aq)\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)

  • balance the charge (the total net charge on the left is 12+, on the right we have 6+, so 6 electrons are needed on the left):

Cr_2O_7^{2-} (aq) + 14 H^+ (aq) + 6e^-\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)

Similarly, tin(II) cation becomes tin(IV) cation:

Sn^{2+} (aq)\rightarrow Sn^{4+} (aq) + 2e^-

Now that we have the two half-equations, multiply the second one by 3, so that it also has 6 electrons that will be cancelled out upon addition of the two half-equations:

Cr_2O_7^{2-} (aq) + 14 H^+ (aq) + 6e^-\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l)

3 Sn^{2+} (aq)\rightarrow 3 Sn^{4+} (aq) + 6e^-

Add them together:

Cr_2O_7^{2-} (aq) + 14 H^+ (aq) + 3 Sn^{2+} (aq)\rightarrow 2 Cr^{3+} (aq) + 7 H_2O (l) + 3 Sn^{4+} (aq)

Adding the ions spectators:

K_2Cr_2O_7 (aq) + 14 HCl (aq) + 3 SnCl_2 (aq)\rightarrow 2 CrCl_3 (aq) + 7 H_2O (l) + 3 SnCl_4 (aq) + 2 KCl (aq)

7 0
3 years ago
In which pair would both compounds have the same empirical formula? A. H2O and H2O2 B. BaSO4 and BaSO3 C. FeO and Fe2O3 D. C6H12
Kazeer [188]
The pair of both compounds that have the same empirical formula are C6H12O6 and HC2H3O2. The answer is letter D. <span>H2O and H2O2, BaSO4 and BaSO3 and FeO and Fe2O3 do not have the same empirical formula.</span>
4 0
4 years ago
Which of the following are nonelectrolytes? Hc2h302 (acetic acid) CH30H (method alcohol) h2So4 and c12h22011 (sucrose or table s
hoa [83]

Answer:

C₁₂H₂₂O₁₁ and CH₃OH

Explanation:

Sucrose and methyl alcohol are nonelectrolytes. They do not ionize or conduct a current in aqueous solution.

HC₂H₃O₂ is a weak electrolyte. It produces only a few ions and is a poor conductor of electricity in aqueous solution.

HC₂H₃O₂ + H₂O ⇌ H₃O⁺ + C₂H₃O₂⁻

H₂SO₄ is a strong electrolyte. Its first ionization is complete, so it is a good conductor of electricity in aqueous solution.

H₂SO₄ + H₂O ⟶ H₃O⁺ + HSO₄⁻

8 0
3 years ago
Relation between glancing and angle of deviation<br>​
shtirl [24]

Answer:

the Glancing angle is the angle between the incident ray and plane mirror which is 90o in the given case. The angle between the direction of the incident ray and the reflected ray is the angle of deviation. Since the angle of deviation for a plane mirror is twice the glancing angle, the angle of deviation is 1800.

5 0
3 years ago
Give the number of significant figures indicated 78
Mama L [17]

2 significant figures

6 0
3 years ago
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