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Taya2010 [7]
3 years ago
8

What are the products obtained in the electrolysis of molten nai?

Chemistry
1 answer:
yanalaym [24]3 years ago
4 0
Answer is: sodium (Na) and iodine (I₂).

<span> First ionic bonds in this salt are separeted because of heat: 
</span>NaI(l) → Na⁺(l) + I⁻(l).

Reaction of reduction at cathode(-): Na⁺(l) + e⁻ → Na(l) /×2.

2Na⁺(l) + 2e⁻ → 2Na(l).

Reaction of oxidation at anode(+): 2I⁻(l) → I₂(l) + 2e⁻.

The anode is positive and the cathode is negative.


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Why do atoms form blonds​
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An atom that contains 22 protons,
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Recall that mass number = number of protons + number of neutrons. Hence, the mass number of the atom

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6 0
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Read 2 more answers
How many liters of water vapor are in 36.21 g?
Elden [556K]

Answer:

45.02 L.

Explanation:  

  • Firstly, we need to calculate the no. of moles of water vapor.
  • n = mass / molar mass = (36.21 g) / (18.0 g/mol) = 2.01 mol.
  • We can calculate the volume of knowing that 1.0 mole of a gas at STP occupies 22.4 L.

<em><u>Using cross multiplication:</u></em>

1.0 mole of CO occupies → 22.4 L.

2.01 mole of CO occupies → ??? L.

∴ The volume of water vapor in 36.21 g = (22.4 L)(2.01 mole) / (1.0 mole) = 45.02 L.

4 0
2 years ago
How many half-lives will pass by the time 1.56% of I-131 is present? B. Approximately how many days does that equal? *
serg [7]

Answer: Hmmmmm that's crazy....

There are a couple of equations one could use for this type of problem, but I find the following to be the easiest to use and to understand.

Fraction remaining (FR) = 0.5n

n = number of half lives that have elapsed

In this problem, we need to find n and are given the FR, which is 1.56% or 0.0156 (as a fraction).

0.0156 = 0.5n

log 0.0156 = n log 0.5

-1.81 = -0.301 n

n = 6.0 half lives have elapsed

Explanation:

Just wanted to help. Hopefully it's correct wouldn't want to waster your time ;)

6 0
2 years ago
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