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Katyanochek1 [597]

3 years ago

13

Balance the following chemical equation using the smallest set of integers. I₂ + Cl₂ + H₂O ---> HIO₂ + HCl The coefficient on

H₂O is:
A. 1

B. 2

C. 5

D. 6

1 answer:

Arisa [49]3 years ago

4 0

Answer:

d. 6

Explanation:

Chemical equation:

I₂ + Cl₂ + H₂O → HIO₃ + HCl

Balanced chemical equation:

I₂ + 5Cl₂ + 6H₂O → 2HIO₃ + 10HCl

Step 1:

I₂ + Cl₂ + H₂O → HIO₃ + HCl

Reactant Product

I = 2 I = 1

Cl = 2 Cl = 1

H = 2 H = 2

O = 1 O = 2

Step 2:

I₂ + Cl₂ + H₂O → 2 HIO₃ + HCl

Reactant Product

I = 2 I = 2

Cl = 2 Cl = 1

H = 2 H = 3

O = 1 O = 6

Step 3:

I₂ + Cl₂ + 6H₂O → 2 HIO₃ + HCl

Reactant Product

I = 2 I = 2

Cl = 2 Cl = 1

H = 12 H = 3

O = 6 O = 6

Step 4:

I₂ + Cl₂ + 6H₂O → 2HIO₃ + 10HCl

Reactant Product

I = 2 I = 2

Cl = 2 Cl = 10

H = 12 H = 12

O = 6 O = 6

Step 5:

I₂ + 5Cl₂ + 6H₂O → 2HIO₃ + 10HCl

Reactant Product

I = 2 I = 2

Cl = 10 Cl = 10

H = 12 H = 12

O = 6 O = 6

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Explanation:

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$V_s$ = volume of solution in L

Given : 59.4 g of $H_2SO_4$ in 100 g of solution

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Volume of solution = $\frac{\text {mass of solution}}{\text {density of solution}}=\frac{100g}{1.83g/ml}=54.6ml$

Now put all the given values in the formula of molality, we get

$Molality=\frac{0.61\times 1000}{54.6ml}=11.2M$

To calculate the volume of acid, we use the equation given by neutralisation reaction:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of stock acid which is $H_2SO_4$

$M_2\text{ and }V_2$ are the molarity and volume of dilute acid which is $H_2SO_4$

We are given:

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Putting values in above equation, we get:

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2 years ago

The metallic radius of a potassium atom is 231 pm. What is the volume of a potassium atom in cubic meters?

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3 years ago

What is the composition, in atom percent, of an alloy that consists of a) 5.5 wt% Pb and b) 94.5 wt% of Sn? Assume that the atom

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Answer : The percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

Explanation :

First we have to calculate the number of atoms in 5.5 wt% Pb and 94.5 wt% of Sn.

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So, 5.5 g of lead contains $\frac{5.5}{207.2}\times 6.022\times 10^{23}=1.59\times 10^{22}$ atoms

and,

As, 118.71 g of lead contains $6.022\times 10^{23}$ atoms

So, 94.5 g of lead contains $\frac{94.5}{118.71}\times 6.022\times 10^{23}=4.79\times 10^{23}$ atoms

Now we have to calculate the percent composition of Pb and Sn in atom.

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and,

$\% \text{Composition of Sn}=\frac{\text{Atoms of Sn}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100$

$\% \text{Composition of Sn}=\frac{4.79\times 10^{23}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=96.8\%$

Thus, the percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.

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Answer:

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Explanation:

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$n=\frac{86g/mol}{43g/mol}=2$

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