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Zielflug [23.3K]
3 years ago
10

Balance the Equation with the steps please:)​

Chemistry
2 answers:
vivado [14]3 years ago
8 0

Explanation:

Fe2o3+ 3co ---------. 2fe + 3co2

Dmitrij [34]3 years ago
5 0

Answer:  3 CO2

Explanation:  1Fe2O3 +3CO = 2Fe +<u> 3</u>CO2

    Reactants   Products

O         6                 6

Fe       2                  2

C          3                 3

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I need it done asap as fast as possible
Lubov Fominskaja [6]

Answer:

1. B

2. D. the form of a substance changes but not its identity

3. C

4. D

6 0
3 years ago
How many grams of cupric sulfate pentahydrate are needed to prepare 50.00 mL of 0.0800M CuSO4× 5H2O?
shepuryov [24]

Explanation:

Molarity is defined as number of moles per liter of solution.

Mathematically,         molarity = \frac{no. of moles}{Volume (in L) of solution}

It is given that molarity is 0.0800 M and volume is 50.00 mL or 0.05 L.

           molarity = \frac{no. of moles}{Volume of solution in liter}

            0.0800 M = \frac{no. of moles}{0.05 L}

            no. of moles = 1.6 mol

Therefore, molar mass of cupric sulfate pentahydrate is 249.68 g/mol. So, calculate the mass as follows.

                No. of moles = \frac{mass in grams}{molar mass}

             mass in grams = no. of moles \times molar mass of CuSO_{4}.5H_{2}O

                                       = 1.6 mol \times 249.68 g/mol

                                       = 399.488 g

Thus, we can conclude that 399.488 g of cupric sulfate pentahydrate are needed to prepare 50.00 mL of 0.0800M CuSO4× 5H2O.

4 0
3 years ago
Dilute 0.407M &amp; 2.56L solution to 7.005L. CF?
Alenkinab [10]

Answer:

C₂ = 0.149 M

Explanation:

Given data:

Initial concentration = 0.407 M

Initial volume = 2.56 L

Final volume = 7.005 L

Final concentration = ?

Solution:

Formula:

C₁V₁ = C₂V₂

C₁ = Initial concentration

V₁ = Initial volume

C₂ = Final concentration

V₂ =Final volume

Now we will put the values.

0.407 M × 2.56 L =  C₂ × 7.005 L

1.042 =  C₂ × 7.005 L

C₂ = 1.042 M.L / 7.005 L

C₂ = 0.149 M

5 0
3 years ago
Use calc to determine whether it is possible to remove 99.99% Cu2 by converting it to Cu(s) in a solution mixture containing 0.1
inessss [21]

Answer:

it is possible to remove 99.99% Cu2 by converting it to Cu(s)

Explanation:

So, from the question/problem above we are given the following ionic or REDOX equations of reactions;

Cu2+ + 2e- <--------------------------------------------------------------> Cu (s) Eo= 0.339 V

Sn2+ + 2e- <---------------------------------------------------------------> Sn (s) Eo= -0.141 V

In order to convert 99.99% Cu2 into Cu(s), the equation of reaction given below is needed:

Cu²⁺ + Sn ----------------------------------------------------------------------------> Cu + Sn²⁺.

Therefore, E°[overall] = 0.339 - [-0.141] = 0.48 V.

Therefore, the change in Gibbs' free energy, ΔG° = - nFE°. Where E° = O.48V, n= 2 and F = 96500 C.

Thus, ΔG° = - 92640.

This is less than zero[0]. Therefore,  it is possible to remove 99.99% Cu2 by converting it to Cu(s) because the reaction is a spontaneous reaction.  

7 0
3 years ago
What is an example of conduction?
pickupchik [31]

Answer:

The correct answer is A.

5 0
3 years ago
Read 2 more answers
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