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nataly862011 [7]
3 years ago
6

PLEASE HELP!!!

Chemistry
1 answer:
kaheart [24]3 years ago
3 0

2MnO4-(aq) + 16H+ + 10Cl-(aq) ⇔ 2Mn2+(aq) + 8H2O + 5Cl2 (g)

<h3>Further explanation</h3>

Given

Reaction(unbalanced)

Mno4- (aq) + Cl- (aq) → Mn2+ + Cl2 (g)

Required

Half reaction

Solution

1. Add coefficient(equalizing atoms in reaction)

2. Adding H₂O on the O-deficient side.  

3. Adding H⁺ on the H-deficient side.  

4. Adding e⁻

5. Equalizing the number of electrons and sum the all the half-reaction

1.MnO₄⁻(aq) = Mn²⁺(aq) reduction

2.MnO4(aq) = Mn2+(aq) + 4H2O

3. MnO4-(aq) + 8H+ = Mn2+(aq) + 4H2O

4. MnO4-(aq) + 8H+ + 5e- = Mn2+(aq) + 4H2O

Cl⁻(aq) = Cl₂(g) oxidation

1. 2Cl-(aq) = Cl2 (g)

2-3 none

4. 2Cl-(aq) = Cl2 (g) + 2e-

5.

MnO4-(aq) + 8H+ + 5e- = Mn2+(aq) + 4H2O x2

2Cl-(aq) = Cl2 (g) + 2e-                                    x5

2MnO4-(aq) + 16H+ + 10e- = 2Mn2+(aq) + 8H2O

10Cl-(aq) = 5Cl2 (g) + 10e-    

<em>2MnO4-(aq) + 16H+ + 10Cl-(aq) ⇔ 2Mn2+(aq) + 8H2O + 5Cl2 (g)</em>

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Write a balanced equation for rusting. (Assume rust becomes Fe+2)
Stells [14]

Answer:

4Fe + 3O2 + 6H2O → 4Fe(OH)3

Explanation:

The chemical formula for rust is Fe2O3 and is commonly known as ferric oxide or iron oxide. The final product is a series of chemical reactions simplified below as- The rusting of the iron formula is simply 4Fe + 3O2 + 6H2O → 4Fe(OH)3. The rusting process requires both the elements of oxygen and water.

8 0
3 years ago
If a system performs 147 kJ of work while receiving 47 kJ of heat, what is change in its internal energy?
Verdich [7]

Answer:

-100 kJ

Explanation:

We can solve this problem by applying the first law of thermodynamics, which states that:

\Delta U = Q-W

where:

\Delta U is the change in internal energy of a system

Q is the heat absorbed/released by the system (it is positive if absorbed by the system, negative if released by the system)

W is the work done by the system (it is positive if done by the system, negative if done on the system)

For the system in this problem we have:

W = +147 kJ is the work done by the system

Q = +47 kJ is the heat absorbed by the system

So , its change in internal energy is:

\Delta U = +47 - (+147) =-100 kJ

6 0
3 years ago
1-Bromopropane is treated with each of the following reagents. Draw the major substitution product if the reaction proceeds in g
sammy [17]

Answer:

Explanation:

If we look at the structure of 1-Bromopropane; we will see that it is a derivative of alkane family by the the substitution of an alkyl group. The position of the Bromine in the propane is 1, making 1-Bromopropane  a primary alkyl-halide.

Primary alkyl - halide undergo SN2 mechanism. This nucleophilic reaction needs to be a strong alkyl halide , such as 1-Bromopropane used  otherwise it will result to a reactive mechanism if a weak electrophile is used.

However, the critical and the main objective here is to Draw the major substitution product if the reaction proceeds in good yield. If no reaction is expected or yields will be poor, draw the starting material in the box. If a charged product is formed, be sure to draw the counterion.

The attached diagrams portraying this notions is shown in the attached file below.

5 0
3 years ago
How much positive charge is in 0.7 kg of lithium? with each atom having 3 protons and 3 electrons. The elemental charge is 1.602
hichkok12 [17]

Explanation:

As a neutral lithium atom contains 3 protons and its elemental charge is given as 1.602 \times 10^{-19} C. Hence, we will calculate its number of moles as follows.

          Moles = \frac{mass}{\text{molar mass}}

                     = \frac{0.7 \times 1000 g}{7 g/mol}

                     = 100 mol

According to mole concept, there are 6.023 \times 10^{23} atoms present in 1 mole. So, in 100 mol we will calculate the number of atoms as follows.

        No. of atoms = 100 \times 6.023 \times 10^{23}

                               = 6.023 \times 10^{25} atoms

Since, it is given that charge on 1 atom is as follows.

                     3 \times 1.602 \times 10^{-19}C

                    = 4.806 \times 10^{-19}C

Therefore, charge present on 6.023 \times 10^{25} atoms will be calculated as follows.

    6.023 \times 10^{25} atoms \times 4.806 \times 10^{-19} C

            28.95 \times 10^{6}C

Thus, we can conclude that a positive charge of 28.95 \times 10^{6}C is in 0.7 kg of lithium.

3 0
3 years ago
True or false. (Pls help)
grigory [225]
I believe they’re both true.
5 0
3 years ago
Read 2 more answers
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