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jekas [21]
3 years ago
11

Balance the following equation:

Chemistry
1 answer:
Firdavs [7]3 years ago
8 0

Answer: a) 2K_2CrO_4+3Na_2SO_3+10HCl\rightarrow 4KCl+3Na_2SO_4+2CrCl_3+5H_2O

b) 1 mole of SO_2 is produced.

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

The skeletal equation is:

K_2CrO_4+Na_2SO_3+HCl\rightarrow KCl+Na_2SO_4+CrCl_3 +H_2O

The balanced equation will be:

2K_2CrO_4+3Na_2SO_3+10HCl\rightarrow 4KCl+3Na_2SO_4+2CrCl_3+5H_2O

Thus the coefficients are 2, 3 , 10 , 4 , 3 , 2 and 5.

b) Oxidation: 2I-^-\rightarrow I_2+2e-^-

Reduction: SO_4^{2-}+2e^-+4H^+\rightarrow SO_2+2H_2O

Net reaction:  2I-^-+SO_4^{2-}+4H^+\rightarrow I_2+SO_2+2H_2O

When 1 mole of I_2 is produced, 1 mole of SO_2 is produced.

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Answer:

86.3 g  of N₂ are in the room

Explanation:

First of all we need the pressure from the N₂ in order to apply the Ideal Gases Law and determine, the moles of gas that are contained in the room.

We apply the mole fraction:

Mole fraction N₂ = N₂ pressure / Total pressure

0.78 . 1 atm = 0.78 atm → N₂ pressure

Room temperature → 20°C → 20°C + 273 = 293K

Let's replace data: 0.78 atm . 95L = n . 0.082 . 293K

(0.78 atm . 95L) /0.082 . 293K = n

3.08 moles = n

Let's convert the moles to mass → 3.08 mol . 28g /1mol = 86.3 g  

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3 years ago
Boyle's law states that the pressure of a gas is always:
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Answer:

d. inversely proportional to the volume of its container.

Explanation:

Boyle's law  states that at constant temperature and number of moles, the pressure of the gas is inversely proportional to the volume of the gas.

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P is the pressure

T is the temperature

For two gases at same temperature, the law can be written as:-

{P_1}\times {V_1}={P_2}\times {V_2}

<u>Thus, according to the question, the answer is:- d. inversely proportional to the volume of its container.</u>

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B. Add the second reactant slower.

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Answer:

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