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4 years ago

Are these correct ?! Please help !!!!!

Chemistry

1 answer:

julsineya [31]4 years ago

6 0

All part B is right.

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Which of the following ions has the largest radius? Br- S2- B3+ Mg2+ Li+

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Answer:

Br - has the largest radius

Explanation:

These ions each have achieved the nearest noble gas configuration. Br - has a like electron configuration to that of Krypton, S2- has a like electron configuration to that of Argon, B3+ to Helium, Mg2+ to Neon, and Li+ to Helium.

_______________________________________________________

Traveling down a group, the radius tends to increase as atom gains another shell. Here we are comparing elements of like electron configuration to the noble gases Helium, Neon, Argon, Krypton. Krypton, being the last element present in the group out of the 4, has the greatest radius as it has the most shells. Thus, Br - has the largest radius.

<u><em>Hope that helps!</em></u>

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Read 2 more answers

What is the numerical value of the equilibrium constant Kc?

Nezavi [6.7K]

ANSWER

EXPLANATION

Given that;

The number of moles of NH3 is 3 moles

The number of moles of H2 is 1 mole

The number of moles of N2 is 2 moles

At equilibrium, the concentration of ammonia is 1.4 moles/L

To find the value of Kc, follow the steps below

Step 1: Write the balanced equation of the reaction

$\text{ N}_{2(g)}\text{ + 3H}_{2(g)}\text{ }\rightleftarrows\text{ 2NH}_{3(g)}$

Step 2: Write the equation of the reaction in terms of Kc

$\text{ K}_C\text{ = }\frac{[\text{ NH}_3]^2}{[N_2]\text{ \lbrack H}_2]^3}$

Step 3: Find the concentration of each reactant at equilibrium using a stoichiometry ratio

From the reaction above, you will see that 1 mole of Nitrogen reacts with 3 moles of hydrogen to give 2 moles of ammonia.

Let x represents the concentration of nitrogen at equilibrium

Recall, that the concentration of ammonia at equilibrium is 1.4 moles/L

$\begin{gathered} \text{ 2 mole NH}_3\text{ }\rightarrow\text{ 1 mole N}_2 \\ \text{ 1.04 mole/L NH}_3\text{ }\rightarrow\text{ x moles/L N}_2 \\ \text{ cross multiply} \\ \text{ 2 moles NH}_3\times\text{ x moles/L N}_2\text{ = 1 mole N}_2\times1.4\text{ mol/L} \\ \text{ Isolate x} \\ \text{ x mol/L N}_2\text{ = }\frac{1\text{ moles N}_2\times1.41\cancel{\frac{mol}{L}}}{2\cancel{moles}} \\ \text{ x mol/L = }\frac{1.04}{2} \\ \text{ x= 0.52 mole/L} \end{gathered}$

Since the initial number of moles of nitrogen is 1 mole, hence, the concentration of nitrogen at equilibrium is calculated below as

Concentration at equilibrium = 1 -0.52

Concentration of nitrogen at equilibrium = 0.48 mole/L

The next step is to find the concentration of hydrogen at equilibrium

Let y represent the mole of hydrogen at equilibrium

$\begin{gathered} \text{ 2 moles NH}_3\rightarrow\text{ 3 moles H}_2 \\ \text{ 1.04 moles/L NH}_3\text{ }\rightarrow\text{ y moles/L H}_2 \\ \text{ cross multiply} \\ \text{ 2 moles NH}_3\times\text{ y moles/L H}_2\text{ = 1.04 moles/L NH}_3\times3\text{ moles H}_2 \\ \text{ Isolate y} \\ \text{ y moles/L H}_2\text{ = }\frac{1.04\cancel{\frac{moles}{L}}NH_3\times3mole\text{ H}_2}{2\cancel{molsNH_3}} \\ y\text{ = }\frac{1.40\times3}{2} \\ \text{ y = }\frac{3.12}{2} \\ \text{ y = 1.56 moles} \end{gathered}$

Since the initial concentration of hydrogen is 2 moles, hence the concentration of hydrogen at equilibrium can be calculated below as

Concentration at equilibrium = 2 - 1.56

Concentration at equilibrium = 0.44 mole/L

Step 4: Find the value of Kc using the equation in step 2

$\text{ Kc = }\frac{[NH_3]^2}{[N_2]\text{ \lbrack H}_2]^3}$ $\begin{gathered} \text{ Kc = }\frac{1.04^2}{0.48\times0.44^3} \\ \\ \text{ K}_c=\text{ }\frac{1.0816}{0.48\times\text{ 0.085184}} \\ \\ \text{ kc = }\frac{1.0816}{0.0408832} \\ \text{ kc = 26.4558547276} \end{gathered}$

Hence, the value of Kc is 26

6 0

2 years ago