1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Irina18 [472]
3 years ago
14

Complete and balance the following redox reaction in basic solution

Chemistry
1 answer:
inn [45]3 years ago
8 0

Answer:

balanced in ACID not BASE

Cr2O7^2-(aq) +3Hg(l) +14 H^1+ ----> 3Hg^2+ + 2Cr^3+(aq) + 7H2O

Answer

Cr2O7^2-(aq) +3Hg(l) +14 H^1+ ----> 3Hg^2+ + 2Cr^3+(aq) + 7H2O

Explanation:

Cr2O7^2-(aq) + Hg(l) ----> Hg^2+(aqH) + Cr^3+(aq)

add H^1+ (acid) to capture the O and make 7 water molecules

Cr2O7^2-(aq) + Hg(l) + H^1+ ----> Hg^2+(aqH) + Cr^3+(aq) + 7H2O

Cr goes from +6 to +3 by gaining 3 e

Hg goes from 0 to +2 by losing 2 e

we need 3 Hg for every 2 Cr

so

Cr2O7^2-(aq) +3Hg(l) +14 H^1+ ----> 3Hg^2+ + 2Cr^3+(aq) + 7H2O

2 Cr on the right and left

Net 12 positive charges on the right and the left

3 Hg on the right and left

14 H on the right and left

the equation is balanced

we cannot balance the equation in a basic solution with OH^1-

we have plenty of O in the dichromate ion. we need to convert it to water which take free H^1+ from the acid

You might be interested in
In order to complete his research project, Roger needs to make a mixture of 86 mL of a 36% acid solution from a 39% acid
Zina [86]

The volume of the 39% acid solution that Roger needs to use to make a mixture of 86 mL of a 36% acid solution is 74.27 mL.

The mixture of the acids solutions is given by:

CV = C_{1}V_{1} + C_{2}V_{2}    (1)      

Where:

C: is the concentration if the mixture = 36%  

V: is the total volume of the mixture = 86 mL

C₁: is the concentration of acid 1 = 39%

V₁: is the volume if acid 1 =?

C₂: is the concentration of acid 2 = 17%

V₂: is the volume of acid 2

The sum of V₁ and V₂ must be equal to V, so:

V = V_{1} + V_{2}

V_{2} = V - V_{1}  (2)

By entering equation (2) into (1), we have:

CV = C_{1}V_{1} + C_{2}(V - V_{1})

36\%*86 mL = 39\%*V_{1} + 17\%(86 mL - V_{1})

Changing the percent values to decimal ones:

0.36*86 mL = 0.39*V_{1} + 0.17(86 mL - V_{1})  

Now, by solving the above equation for V₁:

V_{1} = 74.27 mL  

Therefore, the volume of the 39% acid solution is 74.27 mL.

       

To learn more about mixture and solutions, go here: brainly.com/question/6358654?referrer=searchResults

I hope it helps you!  

         

7 0
3 years ago
Read 2 more answers
Asteroids are mostly made up of which of the following? Select the two correct
Marina86 [1]

Answer:

The answer is below!!

Explanation:

The C-type (chondrite) asteroids are most common. They probably consist of clay and silicate rocks, and are dark in appearance. They are among the most ancient objects in the solar system. The S-types ("stony") are made up of silicate materials and nickel-iron.

Hope I Helped!!

:)

7 0
2 years ago
Gravity on Earth is 9.8 m/s2, and gravity on the Moon is 1.6 m/s2.
Ivanshal [37]

Answer:

Mass is the same but it weights 64 Newtons

Explanation:

First of Mass is the same in any sort of gravity. Now let's calculate weight

W = MG

where W = Weight

M = Mass

G = Gravity

W = (40kg)(1.6)

W = 64

Sorry for the spelling mistakes, hope this helps

7 0
3 years ago
Read 2 more answers
There are the same number of atoms in a mole of calcium (Ca) than in a
zmey [24]

Answer:

True

Explanation:

A mole is defined as a certain amount of a unit. Such as atoms or molecules.

5 0
3 years ago
For the reaction Na2CO3+Ca(NO3)2⟶CaCO3+2NaNO3 how many grams of calcium carbonate, CaCO3, are produced from 79.3 g of sodium car
Alexus [3.1K]

Answer:

74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.

Explanation:

The balanced reaction is:

Na₂CO₃ + Ca(NO₃)₂ ⟶ CaCO₃ + 2 NaNO₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

  • Na₂CO₃: 1 mole
  • Ca(NO₃)₂: 1 mole
  • CaCO₃: 1 mole
  • NaNO₃: 2 mole

Being the molar mass of the compounds:

  • Na₂CO₃: 106 g/mole
  • Ca(NO₃)₂: 164 g/mole  
  • CaCO₃: 100 g/mole
  • NaNO₃: 85 g/mole

then by stoichiometry the following quantities of mass participate in the reaction:

  • Na₂CO₃: 1 mole* 106 g/mole= 106 g
  • Ca(NO₃)₂: 1 mole* 164 g/mole= 164 g
  • CaCO₃: 1 mole* 100 g/mole= 100 g
  • NaNO₃: 2 mole* 85 g/mole= 170 g

You can apply the following rule of three: if by stoichiometry 106 grams of Na₂CO₃ produce 100 grams of  CaCO₃, 79.3 grams of Na₂CO₃ produce how much mass of  CaCO₃?

mass of CaCO_{3} =\frac{79.3 grams of Na_{2} CO_{3} *100 grams of of CaCO_{3}}{106 grams of Na_{2} CO_{3}}

mass of CaCO₃= 74.81 grams

<u><em>74.81 grams of calcium carbonate are produced from 79.3 g of sodium carbonate.</em></u>

6 0
3 years ago
Other questions:
  • NO3 + 12 + 103 + NO2<br> What is being oxidized?
    10·2 answers
  • Determine the reagents needed and the synthetic intermediate that is formed in the conversion of the secondary alcohol into the
    9·1 answer
  • What is one benefit of using a chemical formula?
    12·1 answer
  • Why 5.67 + 0.00032 = 5.67 (and not 5.7 or 5.67032 or any other number).
    6·1 answer
  • The chemical formula for glucose is C6H1206. Therefore, four molecules of glucose will have() carbon atoms()
    7·1 answer
  • A tank containing both HF and HBr gases developed a leak. The ratio of the rate of effusion of HF to the rate of effusion of HBr
    10·2 answers
  • Can someone please answer my questions it’s urgent!
    14·2 answers
  • How do you find the ideal and actual MAs of the pulley systems?
    13·1 answer
  • What does the electron configuration for the noble gases tell us about the elements in the<br> group
    5·1 answer
  • How many grams of water will be produced when 14.9 L of H2 reacts according to the following equation?
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!