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OLga [1]
2 years ago
8

Can someone help me on this one?

Chemistry
1 answer:
ryzh [129]2 years ago
3 0

Answer:

gaand mai danda lauda sanda maadarchod gu ka kida

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What is a potential benefit of fuel cell cars?
77julia77 [94]

Answer:

no recharging necessary

3 0
3 years ago
Read 2 more answers
Convert Au(clo3)2 to s name
Sholpan [36]
Since Au is a symbol for Gold, and once you split the name into to giving each ion its charge... you'll see that this compound has Au+2 and Cl03- .... so  the name would be 

Gold(II) Chlorate

Hope this helps!

3 0
3 years ago
g 2BrO3- + 5SnO22-+ H2O5SnO32- + Br2+ 2OH- In the above reaction, the oxidation state of tin changes from to . How many electron
Archy [21]

Answer:

In the above reaction, the oxidation state of tin changes from 2+ to 4+.

10 moles of electrons are transferred in the reaction

Explanation:

Redox reaction is:

2BrO₃⁻ + 5SnO₂²⁻ + H2O ⇄ 5SnO₃²⁻ + Br₂ + 2OH⁻

SnO₂²⁻ → SnO₃²⁻

Tin changes the oxidation state from +2 to +4. It has increased it so this is the oxidation from the redox (it released 2 e⁻). We are in basic medium, so we add water in the side of the reaction where we have the highest amount of oxygen. We have 2 O on left side and 3 O on right side so we add 1 water on the right and we complete with OH⁻ in the opposite side to balance the H.  

SnO₂²⁻ + 2OH⁻ → SnO₃²⁻ + 2e⁻ + H₂O <u>Oxidation</u>

BrO₃⁻ →  Br₂

First of all, we have unbalance the bromine, so we add 2 on the BrO₃⁻. We have 6 O in left side and there are no O on the right, so we add 6 H₂O on the left. To balance the H, we must complete with 12OH⁻. Bromate reduces to bromine at ground state, so it gained 5e⁻. We have 2 atoms of Br, so finally it gaines 10 e⁻.

6H₂O + 10 e⁻ + 2BrO₃⁻ →  Br₂ + 12OH⁻ <u>Reduction</u>

In order to balance the main reaction and balance the electrons we multiply  (x5) the oxidation and (x1) the reduciton

(SnO₂²⁻ + 2OH⁻ → SnO₃²⁻ + 2e⁻ + H₂O) . 5

(6H₂O + 10 e⁻ + 2BrO₃⁻ →  Br₂ + 12OH⁻) . 1

5SnO₂²⁻ + 10OH⁻ + 6H₂O + 10 e⁻ + 2BrO₃⁻ → Br₂ + 12OH⁻ + 5SnO₃²⁻ + 10e⁻ + 5H₂O

We can cancel the e⁻ and we substract:

12OH⁻ - 10OH⁻ = 2OH⁻ (on the right side)

6H₂O - 5H₂O = H₂O (on the left side)

2BrO₃⁻ + 5SnO₂²⁻ + H2O ⇄ 5SnO₃²⁻ + Br₂ + 2OH⁻

6 0
3 years ago
The temperature of a city during a week was 35° C, 36°C, 34°C, 38°C, 40°C, 39°C and 44°C. What was the average daily temperature
Zigmanuir [339]

Answer:38°

Explanation:

8 0
2 years ago
What happens when The vapour obtained by dropping conc. H2SO4 in a mixture of KI and MnO2 is treated with hypo solution​
Shalnov [3]

Iodine is decolorized.

The first reaction stated in the question occurs as follows;

2 KI (aq) + 2 H2SO4 (aq) + MnO2 (s) → MnSO4 (aq) + K2SO4 (aq) + I2 (s) + 2 H2O (l)

The reaction here is the formation of iodine from MnO2 and KI in the presence of dropwise H2SO4.

Hypo is the common name of sodium thio-sulphate or sodium hypo-sulfite.

The equation of the titration reaction is;

2Na2S2O3 + I2→ Na2S4O6 + 2NaI

When this reaction takes place, iodine is decolorized due to its reduction to I^-.

6 0
2 years ago
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