How much does one mole of NH3 weigh?

Calculate the percent dissociation of benzoic acid C6H5CO2H in a 2.4mM aqueous solution of the stuff. You may find some useful d

Musya8 [376]

Answer:

34%

Explanation:

5 0

2 years ago

using the equation you wrote determine how many moles of butane c4h10 are needed to react with 5.5 moles of oxygen

GaryK [48]

Answer:

0.846 moles.

Explanation:

This is a stichiometric problem.

The balanced equation of complete combustion of butane is:

C₄H₁₀ + 6.5 O₂ → 4 CO₂ + 5 H₂O

It is clear from the stichiometry of the balanced equation that complete combustion of 1.0 mole of butane needs 6.5 moles of O₂ to produce 4 moles of CO₂ and 5 moles of H₂O.

Using cross multiplication:

1.0 mole of C₄H₁₀ reacts with → 6.5 moles of O₂

??? moles of C₄H₁₀ are needed to react with → 5.5 moles of O₂

The number of moles of C₄H₁₀ that are needed to react with 5.5 moles of O₂ = (1.0 x 5.5 moles of O₂) / (6.5 moles of O₂) = 0.846 moles.

3 0

2 years ago

The public is not yet able to purchase cars powered by hydrogen fuel cells because engineers have to determine how the cars perf

Leya [2.2K]

Answer:

It’s B just did it

Explanation:

6 0

3 years ago

A sample of gallium Bromide GaBr2,weighing 0.165 g was dissolved in water and treated with silver nitrate AgNO3, and resulting t

tresset_1 [31]

Answer: The percent gallium in gallium bromide is 30.30 %.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of gallium bromide = 0.165 g

Molar mass of titanium gallium bromide = 229.53 g/mol

Putting values in equation 1, we get:

$\text{Moles of gallium bromide}=\frac{0.165g}{229.53g/mol}=0.00072mol$

The chemical equation for the reaction of gallium bromide and silver nitrate follows:

$GaBr_2+2AgNO_3\rightarrow 2AgBr(s)+Ga(NO_3)_2$

By Stoichiometry of the reaction:

1 moles of gallium bromide produces 1 mole of gallium nitrate

So, 0.00072 moles of gallium bromide will produce = $\frac{1}{1}\times 0.00072=0.00072moles$ of gallium nitrate

Now, calculating the mass of gallium nitrate from equation 1, we get:

Molar mass of gallium nitrate = 193.73 g/mol

Moles of gallium nitrate = 0.00072 moles

Putting values in equation 1, we get:

$0.00072mol=\frac{\text{Mass of gallium nitrate}}{193.73g/mol}\\\\\text{Mass of gallium nitrate}=0.139g$

Calculating the mass of gallium in the reaction, we use unitary method:

In 1 mole of gallium nitrate, 1 mole of gallium atom is present.

In 193.73 grams of gallium nitrate, 69.72 g of gallium atom is present.

So, in 0.139 grams of gallium nitrate, the mass of gallium present will be = $\frac{69.72}{193.73}\times 0.139=g$

To calculate the percentage composition of gallium in gallium bromide, we use the equation:

$\%\text{ composition of gallium}=\frac{\text{Mass of gallium}}{\text{Mass of gallium bromide}}\times 100$

Mass of gallium bromide = 0.165 g

Mass of gallium = 0.050 g

Putting values in above equation, we get:

$\%\text{ composition of gallium}=\frac{0.050g}{0.165g}\times 100=30.30\%$

Hence, the percent gallium in gallium bromide is 30.30 %.

3 0

2 years ago

A solution that contains a large amount of salt and a small amount of water is said to be a _______ solution. A. diffused B. uns

Pachacha [2.7K]

Answer:

D. Concentrated

Explanation:

5 0

2 years ago

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