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4 years ago

What is the net ionic equation for the reaction that occurs when aqueous solutions of koh and zncl2 are mixed?

Chemistry

1 answer:

Mademuasel [1]4 years ago

5 0

Answer:

Net ionic equation:

Zn²⁺(aq) + 2OH⁻(aq) → Zn(OH)₂(s)

Explanation:

Chemical equation:

ZnCl₂ + KOH → KCl + Zn(OH)₂

Balanced chemical equation:

ZnCl₂ + 2KOH → 2KCl +Zn(OH)₂

Ionic equation;

Zn²⁺(aq) + 2Cl⁻(aq) + 2K⁺(aq) + 2OH⁻(aq) → 2K⁺(aq) + 2Cl⁻(aq) +Zn(OH)₂(s)

Net ionic equation:

Zn²⁺(aq) + 2OH⁻(aq) → Zn(OH)₂(s)

The K⁺ and Cl⁻ are spectator ions that's why these are not written in net ionic equation. The Zn(OH)₂ can not be splitted into ions because it is present in solid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.

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The value of the equilibrium constant for a reaction is 2.65 x 10-6 at 35° Calculate the value of LaTeX: \DeltaΔG°rxn.

stealth61 [152]

Answer:

The answer is "33.95 $\bold{ \frac{kj}{mol}}\\$ ".

Explanation:

Formula:

$\bigtriangleup G_0= -R \times T l_n \times K_{eq}\\$

where

$K_{eq} = \text{equilibrium constant}\\$

Given value:

$T =35^{\circ} C\\$

convert temperature celsius (°C) to Kelvin (K):

$= (273+45) \ kelvin \\\\= 318 \ Kelvin \\\\= 318 \ K$

$R = 8.314 \ \ \frac{J}{ mol \cdot K}$

$\bigtriangleup G_0= -(8.314 ) \times 31.8 \times l_n (2.65\times 10^{-6})\\$

$=-(8.314 ) \times 31.8T l_n \times (2.65\times 10^{-6})\\\\$

After solving the value it will give:

= 33.95 $\bold{ \frac{kj}{mol}}\\$

3 0

3 years ago

12 = 3X X = <br> 7X = 57 X = <br> 11X = 27.7 X =

dybincka [34]

Answer:
1. $x=4$
2. $x=\frac{57}{7}$
3. $x=\frac{27.7}{11}$

Explanation:

Given the following questions:

$12=3x$
$7x=57$
$11x=27.7$

In order to find the answers to these questions, we have to recognize that in order to isolate the variable we need to divide on both sides.

Question one:
$12=3x$
$\frac{3x}{3} =\div3$
$12\div3=4$
$x=4$

Question two:
$7x=57$
$\frac{7x}{7} =\div7$
$57\div7=8.14285714=\frac{57}{7}$
$x=\frac{57}{7}$

Question three:
$11x=27.7$
$\frac{11x}{11} \div11$
$27.7\div11=2.51818182=\frac{27.7}{11}$
$x=\frac{27.7}{11}$

Hope this helps.