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Alja [10]
3 years ago
15

Write the balanced NET IONIC equation for the reaction that occurs when ammonium nitrate and potassium hydroxide are combined. N

H4 OH- This reaction is classified as . A. Strong Acid Strong Base B. Weak Acid Strong Base C. Strong Acid Weak Base D. Weak Acid Weak Base The extent of this reaction is: . A. ... Below 50% B. ... 50% C. ... Above 50% D. ... 100%
Chemistry
1 answer:
vova2212 [387]3 years ago
8 0

Answer:

Net Ionic equation

NH₄⁺ + OH⁻ → NH₃ + H₂O

Option B is correct.

Weak Acid Strong Base

Check Explanation for the extent of the reaction.

Explanation:

Ammonium nitrate = NH₄NO₃

Potassium Hydroxide = KOH

Ammonium salts combine with alkalis to liberate NH₃ and form water.

The two reactants combine to give

NH₄NO₃ + KOH → KNO₃ + NH₃ + H₂O

In ionic form,

- NH₄NO₃ exists as NH₄⁺ and NO₃⁻

- KOH exists as K⁺ and OH⁻

- KNO₃ as K⁺ and NO₃⁻

And NH₃ and H₂O stay as they are, as per covalent compounds.

So, we have

NH₄⁺ + NO₃⁻ + K⁺ + OH⁻ → K⁺ + NO₃⁻ + NH₃ + H₂O

Eliminating the ions that exist on both sides, we have the net ionic equation to be

NH₄⁺ + OH⁻ → NH₃ + H₂O

which shows that this reaction is essentially a neutralization reaction in which the Bronsted Lowry acid, NH₄⁺, loses its proton to the base, OH⁻ and gives conjugate base, NH₃ and conjugate acid, H₂O.

This reaction is classified as a Weak acid versus Strong Base reaction as NH₄⁺ is from a Weak acid and OH⁻ is from a strong base.

Since this reaction is between a Weak base and a strong acid, the ionization isn't expected to be 100%, Hence, the extent of this reaction will be any option that is not 100%, a couple pieces of information might be required for the correct estimate, but above 50% seems correct.

Hope this Helps!!!

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Calculate the unit cell edge length for an 85 wt% fe-15 wt% v alloy. All of the vanadium is in solid solution, and, at room temp
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Answer is 0.289nm.

Explanation: The wt % of Fe and wt % of V is given for a Fe-V alloy.

wt % of Fe in Fe-V alloy = 85%

wt % of V in Fe-V alloy = 15%

We need to calculate edge length of the unit cell having bcc structure.

Using density formula,

\rho_{ave}=\frac{Z\times M_{ave}}{a^3\times N_A}

For calculating edge length,

a=(\frac{Z\times M_{ave}}{\rho_{ave}\times N_A})^{1/3}

For calculating M_{ave}, we use the formula

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Similarly for calculating (\rho)_{ave}, we use the formula

\rho_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{\rho_{Fe}}+\frac{(wt\%)_{V}}{\rho_V}}

From the periodic table, masses of the two elements can be written

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a=\left [\frac{Z\left (\frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{Fe}}{M_{Fe}}}  \right )}{N_A\left (\frac{100}{\frac{(wt\%)_V}{\rho_V}+\frac{(wt\%)_V}{\rho_V}}  \right )}  \right ]^{1/3}

a=\left [\frac{2atoms/\text{unit cell}\left (\frac{100}{\frac{85\%}{55.85g/mol}+\frac{15\%}{50.941g/mol}}  \right )}{(6.023\times10^{23}atoms/mol)\left (\frac{100}{\frac{85\%}{7.874g/cm^3}+\frac{15\%}{6.10g/cm^3}}  \right )}  \right ]^{1/3}

By calculating, we get

a=2.89\times10^{-8}cm=0.289nm

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