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2 years ago

_Al(NO3)3 +_(NH4)3 P04_AlPO4+_ NH, NO3balancing equations

Chemistry

1 answer:

astra-53 [7]2 years ago

4 0

Answer:

Al(NO₃)₃ + (NH₄)₃PO₄ —> AlPO₄ + 3NH₄NO₃

The coefficients are: 1, 1, 1, 3

Explanation:

__Al(NO₃)₃ + __(NH₄)₃PO₄ —> __AlPO₄ + __NH₄NO₃

The above equation can be balance as illustrated below:

Al(NO₃)₃ + (NH₄)₃PO₄ —> AlPO₄ + NH₄NO₃

There are 12 atoms of H on the left side and 4 atoms on the right side. It can be balance by writing 3 before NH₄NO₃ as shown below:

Al(NO₃)₃ + (NH₄)₃PO₄ —> AlPO₄ + 3NH₄NO₃

Now the equation is balanced.

The coefficients are: 1, 1, 1, 3

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Trava [24]

Answer: 2.52 M

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The product of molarity (moles/litre) and volume in litres yields moles, and the numbers of moles in two solutions means dilute and concentrated are equal, which is expressed by the following equation:

$M_1V_1=M_2V_2$

$$Given that\\M $1=12.0 \mathrm{M}$ or mole $/ \mathrm{L}$\\$\mathrm{V} 1=420 \mathrm{ml}$\\$\mathrm{M} 2=$ ?\\$\mathrm{V} 2=2.0 \mathrm{~L}$ or $2000 \mathrm{ml}$\\\\$\mathrm{M} 1 \mathrm{~V} 1=\mathrm{M} 2 \mathrm{~V} 2$\\$\mathrm{M} 2=\mathrm{M} 1 \mathrm{~V} 1 / \mathrm{V} 2$\\$=12.0 * 420 / 2000$\\$=2.52 \ \mathrm{M}$

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1 year ago

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2 years ago

9. Given the following reaction:CO (g) + 2 H2(g) CH3OH (g)In an experiment, 0.45 mol of CO and 0.57 mol of H2 were placed in a 1

WITCHER [35]

Answer:

$Keq=11.5$

Explanation:

Hello,

In this case, for the given reaction at equilibrium:

$CO (g) + 2 H_2(g) \rightleftharpoons CH_3OH (g)$

We can write the law of mass action as:

$Keq=\frac{[CH_3OH]}{[CO][H_2]^2}$

That in terms of the change $x$ due to the reaction extent we can write:

$Keq=\frac{x}{([CO]_0-x)([H_2]_0-2x)^2}$

Nevertheless, for the carbon monoxide, we can directly compute $x$ as shown below:

$[CO]_0=\frac{0.45mol}{1.00L}=0.45M\\$

$[H_2]_0=\frac{0.57mol}{1.00L}=0.57M\\$

$[CO]_{eq}=\frac{0.28mol}{1.00L}=0.28M\\$

$x=[CO]_0-[CO]_{eq}=0.45M-0.28M=0.17M$

Finally, we can compute the equilibrium constant:

$Keq=\frac{0.17M}{(0.45M-0.17M)(0.57M-2*0.17M)^2}\\\\Keq=11.5$

Best regards.

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3 years ago

How many valence electrons should you use to draw a Lewis structure of ammonia (NH3)?

Mekhanik [1.2K]

Hello there,

Your correct answer would be 8. There are 8 valence electrons available for the Lewis structure for NH3.

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kotegsom [21]

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_Al(NO3)3 +_(NH4)3 P04_AlPO4+_ NH, NO3balancing equations​

_Al(NO3)3 +_(NH4)3 P04_AlPO4+_ NH, NO3balancing equations