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2 years ago

Calculate the no. of each atoms presents in 10 grams of calcium carbonate.

Chemistry

1 answer:

raketka [301]2 years ago

8 0

Molar Mass of Calcium carbonate:-

$\\ \sf\longmapsto CaCO_3$

$\\ \sf\longmapsto 40u+12u+3(16u)$

$\\ \sf\longmapsto 52u+48u$

$\\ \sf\longmapsto 100u$

$\\ \sf\longmapsto 100g/mol$

Given Mass=10g

$\boxed{\sf No\:of\;moles=\dfrac{Given\:Mass}{Molar\:Mass}}$

$\\ \sf\longmapsto No\:of\;moles=\dfrac{10}{100}$

$\\ \sf\longmapsto No\:of\:moles=0.1mol$

Now

$\boxed{\sf No\:of\:molecules=No\;of\:moles\times Avagadro\: No}$

$\\ \sf\longmapsto 0.1\times 6.023\times 10^{23}$

$\\ \sf\longmapsto 6.023\times 10^{22}molecules$

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Answer: Maximum work that can be obtained by given amount of methanol is -343kJ.

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By Stoichiometry of the reaction:

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2 years ago

Read 2 more answers

BH+ClO4- is a salt formed from the base B (Kb = 1.00e-4) and perchloric acid. It dissociates into BH+, a weak acid, and ClO4-, w

Len [333]

Answer:

The pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44

Explanation:

Given: The base dissociation constant: $K_{b}$ = 1 × 10⁻⁴, Concentration of salt: BH⁺ClO₄⁻ = 0.1 M

Also, water dissociation constant: $K_{w}$ = 1 × 10⁻¹⁴

The acid dissociation constant ( $K_{a}$ ) for the weak acid (BH⁺) can be calculated by the equation:

$K_{a}. K_{b} = K_{w}$

$\Rightarrow K_{a} = \frac{K_{w}}{K_{b}}$

$\Rightarrow K_{a} = \frac{1\times 10^{-14}}{1\times 10^{-4}} = 1\times 10^{-10}$

Now, the acid dissociation reaction for the weak acid (BH⁺) and the initial concentration and concentration at equilibrium is given as:

Reaction involved: BH⁺ + H₂O ⇌ B + H₃O+

Initial: 0.1 M x x

Change: -x +x +x

Equilibrium: 0.1 - x x x

The acid dissociation constant: $K_{a} = \frac{\left [B \right ] \left [H_{3}O^{+}\right ]}{\left [BH^{+} \right ]} = \frac{(x)(x)}{(0.1 - x)} = \frac{x^{2}}{0.1 - x}$

$\Rightarrow K_{a} = \frac{x^{2}}{0.1 - x}$

$\Rightarrow 1\times 10^{-10} = \frac{x^{2}}{0.1 - x}$

$As, x$

$\Rightarrow 0.1 - x = 0.1$

$\therefore 1\times 10^{-10} = \frac{x^{2}}{0.1 }$

$\Rightarrow x^{2} = (1\times 10^{-10})\times 0.1 = 1\times 10^{-11}$

$\Rightarrow x = \sqrt{1\times 10^{-11}} = 3.16 \times 10^{-6}$

Therefore, the concentration of hydrogen ion: x = 3.6 × 10⁻⁶ M

Now, pH = - ㏒ [H⁺] = - ㏒ (3.6 × 10⁻⁶ M) = 5.44

Therefore, the pH of 0.1 M BH⁺ClO₄⁻ solution is 5.44

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