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nikklg [1K]
2 years ago
8

The number of Al atoms in 0.25 mol of Al2(SO4)3​

Chemistry
1 answer:
Harrizon [31]2 years ago
4 0

Answer:

3.011 × 10²³ atoms of Al

Explanation:

Given data:

Number of moles of Al₂(SO₄)₃ = 0.25 mol

Number of atoms of Al = ?

Solution:

In one mole of Al₂(SO₄)₃  there are 2 moles of Al.

In 0.25 mol × 2 = 0.5 mol

1 mole of any substance contain 6.022 × 10²³ atoms.

In 0.5 moles of Al:

1 mole contain 6.022 × 10²³ atoms.

0.5 mol × 6.022 × 10²³ atoms/ 1 mol

3.011 × 10²³ atoms

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Natalka [10]

Answer:

Option A

Explanation:

Leguminous plants like pulses etc. have root nodules comprising of rhizobacterium which live in a symbiotic relationship with the roots of the plant and in turn fix the nitrogen in the soil in the roots of the leguminous plants.

Hence, option A is correct

8 0
2 years ago
Consider the reaction between iodine gas and chlorine gas to form iodine monochloride. A reaction mixture at 298.15k initially c
uranmaximum [27]

Step 1 : Write balanced chemical equation

The balanced chemical equation for the reaction between iodine gas and chlorine gas is given below.

I_{2}  (g) + Cl_{2}  (g) ----->  2 ICl (g)

Step 2 : Set up ICE table

We will set up an ICE table for the above reaction

Following points are considered while drawing ICE table

- The initial concentration of product is assumed as 0

- The change in concentration (C) is assumed as x. Change (x) is negative for reactants and positive for products

- The coefficients in balanced equation are considered while writing C values

Check attached file for ICE table

Step 3 : Set up equilibrium constant equation

The equation for equilibrium constant can be written as

K_{eq} = \frac{[ICl]^{2}}{[I_{2}][Cl_{2}]}

Step 4 : Solving for x

Keq at 298.15 K is given as 81.9

Let us plug in the equilibrium values (E) for I₂, Cl₂ and ICl from ICE table

81.9 = \frac{(2x)^{2}}{(0.437-x) (0.269-x)}

81.9 = \frac{(2x)^{2}}{x^{2} -0.706x + 0.118}

(2x)^{2} = 81.9 [ x^{2} -0.706x + 0.118]

4x^{2} = 81.9x^{2} -57.8x +9.66

77.9x^{2} -57.8x+9.66 = 0

Solving the above equation using quadratic formula we get

x = 0.488 or x = 0.254

The value 0.488 cannot be used because the change (C) cannot be greater that initial concentration of the reactants.

Therefore the change in concentration of the gases during the reaction is 0.254 M

Hence, x = 0.254 M

From the ICE table, we know that the equilibrium concentration of ICl is 2x

[ICl]_{eq} = 2 ( 0.254) = 0.508 M

The concentration of ICl when the reaction reaches equilibrium is 0.508 M

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