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2 years ago

How do you predict a reaction outcome for Ni(s) + H2O(L)?

Chemistry

1 answer:

ElenaW [278]2 years ago

7 0

Answer:

Ni(s) + H₂O(l) -------------> [Ni(H₂O)₆]²⁺

Explanation:

Data Given:

Reactants:

Ni(s) + H₂O(l)

Product = ?

Solution:

Names of the Reactants

Ni = Nickel

H₂O = water

Reaction:

Normally nickel directly does not react with water under normal condition. But indirectly in acidic or neutral condition it form complex ion.

For this it first dissolve slowly in dilute acid, in this reaction it liberate Ni²⁺ ions these nickel ions form light green complex ions in aqueous solution.

Ni(s) + H₂SO₄(aq) -------> Ni²⁺(aq) + SO₄²⁻(aq) + H₂

This is a type of complex formation in which Nickel react with water and produced a light green color new complex ion or product.

Complete reaction is as under

Ni(s) + H₂O(l) -------------> [Ni(H₂O)₆]²⁺

Balance Reaction:

Ni(s) + 6 H₂O(l) -------------> [Ni(H₂O)₆]²⁺

So.

by this reaction one product is formed that is [Ni(H₂O)₆]²⁺ named as hexaqua nickel ion

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A monoprotic weak acid when dissolved in water is 0.66% dissociated and produces a solution with a pH of 3.04. Calculate the Ka

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Answer:

Ka = 6.02x10⁻⁶

Explanation:

The equilibrium that takes place is:

HA ⇄ H⁺ + A⁻

Ka = [H⁺][A⁻]/[HA]

We calculate [H⁺] from the pH:

pH = -log[H⁺]

[H⁺] = $10^{-pH}$

[H⁺] = 9.12x10⁻⁴ M

Keep in mind that [H⁺]=[A⁻].

As for [HA], we know the acid is 0.66% dissociated, in other words:

[HA] * 0.66/100 = [H⁺]

We calculate [HA]:

[HA] = 0.138 M

Finally we calculate the Ka:

Ka = $\frac{[9.12x10^{-4}]*[9.12x10^{-4}]}{[0.138]}$ = 6.02x10⁻⁶

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Vaselesa [24]

Answer:

(See explanation for further details)

Explanation:

1) The quantity of moles of sulfur is:

$n = \frac{1.20\times 10^{24}\,atoms}{6.022\times 10^{23}\,\frac{atoms}{mol} }$

$n = 1.993\,moles$

2) The number of atoms of helium is:

$x = (1.5\,moles)\cdot \left(6.022\times 10^{23}\,\frac{atoms}{mole} \right)$

$x = 9.033\times 10^{23}\,atoms$

3) The quantity of moles of carbon monoxide is:

$n = \frac{4.15\times 10^{23}\,molecules}{6.022\times 10^{23}\,\frac{molecules}{mol} }$

$n = 0.689\,moles$

4) The number of molecules of sulfur dioxide is:

$x = (2.25\,moles)\cdot \left(6.022\times 10^{23}\,\frac{molecules}{mole} \right)$

$x = 1.355\times 10^{24}\,molecules$

5) The quantity of moles of sodium chloride is:

$n = \frac{2.4\times 10^{23}\,molecules}{6.022\times 10^{23}\,\frac{molecules}{mol} }$

$n = 0.399\,moles$

6) The number of formula units of magnesium iodide is:

$x = (1.8\,moles)\cdot \left(6.022\times 10^{23}\,\frac{f.u.}{mole} \right)$

$x = 1.084\times 10^{24}\,f.u.$

7) The quantity of moles of potassium permanganate is:

$n = \frac{3.67\times 10^{23}\,f.u.}{6.022\times 10^{23}\,\frac{f.u.}{mol} }$

$n = 1.214\,moles$

8) The number of molecules of carbon tetrachloride is:

$x = (0.25\,moles)\cdot \left(6.022\times 10^{23}\,\frac{molecules}{mole} \right)$

$x = 1.506\times 10^{23}\,molecules$

9) The quantity of moles of aluminium is:

$n = \frac{3.67\times 10^{23}\,atoms}{6.022\times 10^{23}\,\frac{atoms}{mol} }$

$n = 0.609\,moles$

10) The number of molecules of oxygen difluoride is:

$x = (3.52\,moles)\cdot \left(6.022\times 10^{23}\,\frac{molecules}{mole} \right)$

$x = 2.120\times 10^{24}\,molecules$

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