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3 years ago

10

How many grams are there in 2.00 moles of copper (ii) oxide

1 answer:

tatuchka [14]3 years ago

7 0

Answer: 159 grams

Explanation:

Copper (ii) oxide has the chemical formula CuO.

Now given that:

Mass of CuO in grams = ? (let unknown value be Z)

Number of moles = 2.00 moles

Molar mass of CuO = ?

For the molar mass of CuO: Atomic mass of Copper = 63.5g ; Oxygen = 16g

= 63.5g + 16g

= 79.5 g/mol

Apply the formula:

Number of molecules = (mass in grams/molar mass)

2.00 moles = (Z / 79.5 g/mol)

Z = 79.5 g/mol x 2.00 moles

Z = 159g

Thus, there are 159 grams in 2.00 moles of copper (ii) oxide

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3 years ago

We want to calculate the concentrations of all species in a 0.58 M Na 2 SO 3 (sodium sulfite) solution. The ionization constants

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Explanation:

Reaction equation is as follows.

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As, $K_{b} = \frac{K_{w}}{K_{a_{2}}}$

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According to the ICE table for the given reaction,

$SO^{2-}_{3} + H_{2}O \rightleftharpoons HSO^{-}_{3} + OH^{-}$

Initial: 0.58 0 0

Change: -x +x +x

Equilibrium: 0.58 - x x x

So,

$K_{b} = \frac{[HSO^{-}_{3}][OH^{-}]}{[SO^{2-}_{3}]}$

$1.59 \times 10^{-7} = \frac{x^{2}}{0.58 - x}$

$x^{2} = 1.59 \times 10^{-7} \times (0.58 - x)$

x = 0.0003 M

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= 0.579 M

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The reaction will be as follows.

$HSO^{2-}_{3} + H_{2}O \rightleftharpoons H_{2}SO_{3} + OH^{-}$

Initial: 0.0003

Equilibrium: 0.0003 - x x x

$K_{b} = \frac{x^{2}}{0.0003 - x}$

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= $\frac{10^{-14}}{1.4 \times 10^{-2}}$

= $7.14 \times 10^{-13}$

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As, x <<<< 0.0003. So, we can neglect x.

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= $\frac{10^{-14}}{0.0003}$

= $3.33 \times 10^{-11}$ M

Thus, we can conclude that the concentration of spectator ion is $3.33 \times 10^{-11}$ M.

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3 years ago