2. Match each diagram with its correct description. Diagrams will be used once.

The highest energy occupied molecular orbital in the f-f bond of the f2 molecule is _____

Morgarella [4.7K]

The basis of finding the answer to this problem is to know the electronic configuration of Fluorine. That would be: [He] 2s² 2p⁵. The valence electrons, which are the outermost electrons of the atom, are the ones that participate in bonding. Since the highest orbital for F is 2p, that means the highest energy occupied would be 2.

3 0

2 years ago

Jamal wants to make a model of a hill near his house to test the way the slope affects how rain runs down the hill. Which type o

Nikitich [7]

a scale-model mound made of the same materials that make the real hill

4 0

2 years ago

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2. If 2.50g of sodium hydroxide is being reacted with 4.30g of magnesium chloride, how many grams of magnesium hydroxide would b

Virty [35]

Answer:

1.822 g of magnesium hydroxide would be produced.

Explanation:

Balanced reaction: $2NaOH+MgCl_{2}\rightarrow Mg(OH)_{2}+2NaCl$

Compound Molar mass (g/mol)

NaOH 39.997

$MgCl_{2}$ 95.211

$Mg(OH)_{2}$ 58.3197

So, 2.50 g of NaOH = $\frac{2.50}{39.997}$ mol of NaOH = 0.0625 mol of NaOH

4.30 g of $MgCl_{2}$ = $\frac{4.30}{95.211}$ mol of $MgCl_{2}$ = 0.0452 mol of $MgCl_{2}$

According to balanced equation-

2 mol of NaOH produce 1 mol of $Mg(OH)_{2}$

So, 0.0625 mol of NaOH produce $(\frac{0.0625}{2})$ mol of NaOH or 0.03125 mol of NaOH

1 mol of $MgCl_{2}$ produces 1 mol of $Mg(OH)_{2}$

So, 0.0452 mol of $MgCl_{2}$ produce 0.0452 mol of $Mg(OH)_{2}$

As least number of moles of $Mg(OH)_{2}$ are produced from NaOH therefore NaOH is the limiting reagent.

So, amount of $Mg(OH)_{2}$ would be produced = 0.03125 mol

= $(0.03125\times 58.3197)$ g

= 1.822 g

6 0

2 years ago

A 0.72-mol sample of PCl5 is put into a 1.00-L vessel and heated. At equilibrium, the vessel contains 0.40 mol of PCl3(g) and 0.

Sonja [21]

Answer:

Equilibrium constant for $PCl_5$ is 0.5

Equilibrium constant for decomposition of $NO_2$ is $1.79 \times 10^{-14}$

Explanation:

$PCl_5$ dissociates as follows:

$PCl_5 \rightleftharpoons PCl_3+Cl_2$

initial 0.72 mol 0 0

at eq. 0.72 - 0.40 0.40 0.40

Expression for the equilibrium constant is as follows:

$k=\frac{[PCl_3][Cl_2]}{[PCl_5]}$

Substitute the values in the above formula to calculate equilibrium constant as follows:

$k=\frac{[0.40/1][0.40/1]}{0.32/1} \\=\frac{0.40 \times 0.40}{0.32} \\=0.5$

Therefore, equilibrium constant for $PCl_5$ is 0.5

Now calculate the equilibrium constant for decomposition of $NO_2$

It is given that $3.3 \times 10^{-3} \%$ is decomposed.

$NO_2$ decomposes as follows:

$2NO_2 \rightleftharpoons 2NO + O_2$

initial 1.0 M 0 0

at eq. concentration of $NO_2$ is:

$[NO_2]_{eq}=1-(0.000066) = 0.999934\ M$

$[NO]_{eq}=6.6 \times 10^{-5}\ M$

$[O_2]_{eq}=3.3\times 10^{-5} = 3.3\times 10^{-5}\ M$

Expression for equilibrium constant is as follows:

$K=\frac{[NO]^2[O_2]}{[NO_2]^2}$

Substitute the values in the above expression

$K=\frac{[6.6\times 10^{-5}]^2[3.3 \times 10^{-5}]}{[0.999934]^2} \\=1.79\times 10^{-14}$

Equilibrium constant for decomposition of $NO_2$ is $1.79 \times 10^{-14}$

8 0

2 years ago

How many hydrogen atoms are in 11C2H6

tresset_1 [31]

Answer:

6

Explanation:

You will see H6 and the H stands for helium and the 6 is how many of that atom is there

5 0

2 years ago

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