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3 years ago

Use the Lewis structure to predict the molecular geometry of each molecule.

Chemistry

2 answers:

slava [35]3 years ago

4 0

Answer:

Trigonal planar

Explanation:

The geometry will be trigonal planar because B has 3 valence electron and all three each of it will form single bond with with electron of chlorine

solmaris [256]3 years ago

3 0

Answer:

trigonal planar ; bent

Explanation:

got it right on edge

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Hydrogen gas (a potential future fuel) can be formed by the reaction of methane with water according to the following equation:

rusak2 [61]

Answer:

60.42% is the percent yield of the reaction.

Explanation:

Moles of methane gas at 734 Torr and a temperature of 25 °C.

Volume of methane gas = V = 26.0 L

Pressure of the methane gas = P = 734 Torr = 0.9542 atm

Temperature of the methane gas = T = 25 °C = 298.15 K

Moles of methane gas = n

$PV=nRT$

$n=\frac{PV}{RT}=\frac{0.9542 atm\times 26.0L}{0.0821 atm L/mol K\times 298.15 K}=1.0135 mol$

Moles of water vapors at 700 Torr and a temperature of 125 °C.

Volume of water vapor = V' = 23.0 L

Pressure of water vapor = P' = 700 Torr = 0.9100 atm

Temperature of water vapor = T' = 125 °C = 398.15 K

Moles of water vapor gas = n'

$P'V'=n'RT'$

$n'=\frac{PV}{RT}=\frac{0.9100 atm\times 23.0L}{0.0821 atm L/mol K\times 398.15 K}=0.6402 mol$

$CH_4(g)+H_2O(g)\rightarrow CO(g)+3H_2(g)$

According to reaction , 1 mol of methane reacts with 1 mol of water vapor. As we can see that moles of water vapors are in lessor amount which means it is a limiting reagent and formation of hydrogen gas will depend upon moles of water vapors.

According to reaction 1 mol of water vapor gives 3 moles of hydrogen gas.

Then 0.6402 moles of water vapor will give:

$\frac{3}{1}\times 0.6402 mol=1.9208 mol$ of hydrogen gas

Moles of hydrogen gas obtained theoretically = 1.9208 mol

The reaction produces 26.0 L of hydrogen gas measured at STP.

At STP, 1 mole of gas occupies 22.4 L of volume.

Then 26 L of volume of gas will be occupied by:

$\frac{1}{22.4 L}\times 26 L= 1.1607 mol$

Moles of hydrogen gas obtained experimentally = 1.1607 mol

Percentage yield of hydrogen gas of the reaction:

$\frac{Experimental}{Theoretical}\times 100$

$\%=\frac{ 1.1607 mol}{1.9208 mol}\times 100=60.42\%$

60.42% is the percent yield of the reaction.

8 0

4 years ago

A 100.0 mL sample of a 0.200 M aqueous solution of K2CrO4 was added to 100.0 mL of a 0.100 M aqueous solution of BaCl2. The mixt

stepan [7]

The correct answer is 2.53 g of precipitate, BaCrO4.

5 0

3 years ago

Write both answers to at least two decimal places. Calculate the pH of a 0.160 M solution of KOH.Part 2 (1 point) Calculate the

Evgen [1.6K]

To calculate the pH of a solution, we first need to find the concentration of hydronium ions in the solution. Since KOH is a strong base, it dissociates completely in water to produce hydroxide ions (OH-) and potassium ions (K+).

The concentration of hydronium ions in a solution of KOH can be calculated using the concentration of hydroxide ions and the equilibrium constant for water, which is equal to 1.00 x 10^-14 at 25 degrees Celsius.

The concentration of hydroxide ions in a 0.160 M solution of KOH is equal to the concentration of KOH, which is 0.160 M. The concentration of hydronium ions in the solution can be calculated using the equation below:

[H3O+] = (1.00 x 10^-14) / [OH-]

Substituting the concentration of hydroxide ions into the equation above, we get:

[H3O+] = (1.00 x 10^-14) / (0.160 M) = 6.25 x 10^-13 M

To calculate the pH of the solution, we need to take the negative logarithm of the concentration of hydronium ions. This can be done using the equation below:

pH = -log([H3O+])

Substituting the concentration of hydronium ions into the equation above, we get:

pH = -log(6.25 x 10^-13) = 12.20

The pH of a 0.160 M solution of KOH is 12.20.

To calculate the pOH of a solution, we first need to find the concentration of hydroxide ions in the solution. Since we already calculated this value above, we can simply use the concentration of hydroxide ions we found earlier: 0.160 M.

To calculate the pOH of the solution, we need to take the negative logarithm of the concentration of hydroxide ions. This can be done using the equation below:

pOH = -log([OH-])

Substituting the concentration of hydroxide ions into the equation above, we get:

pOH = -log(0.160 M) = 1.80

The pOH of a 0.160 M solution of KOH is 1.80.

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