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Alexus [3.1K]
3 years ago
8

What is the of 0.068 M of OH^ - ion?

Chemistry
1 answer:
Marina CMI [18]3 years ago
4 0

Answer:

reeeeeeeeeeeeee

Explanation:

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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]

<u>Answer:</u> The percent gallium in gallium bromide is 30.30 %.

<u>Explanation:</u>

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
3 years ago
Hydrogen cyanide, HCN, is prepared from ammonia, air, and natural gas (CH₄), by the following process:
kykrilka [37]

Answer:

6.75 g of HCN can be produced by the reaction

Explanation:

Complete reaction is:

2NH₃ (g) + 3O₂ (g) + 2CH₄ (g) → 2HCN (g) + 6H₂O (g)

Let's determine the moles of each reactant:

11.5 g . 1mol / 17g = 0.676 moles of ammonia

12 g . 1 mol / 32g = 0.375 moles of oxygen

10.5 g . 1mol/ 16 g =  0.656 moles of methane

Now is all about rules of three:

2 moles of ammonia reacts with 3 moles of O₂ and 2 moles of methane

0.676 moles of NH₃ may react with:

(0.676 . 3) /2 = 1.014 moles of O₂

(0.676 . 2) / 2 = 0.676 moles of methane

Both can be the limiting reactant.

3 moles of O₂ react with 2 moles of NH₃ and 2 moles of methane

0.375 moles of O₂ will react with:

(0.375 .2) / 3  = 0.375 moles

The same amount for methane, 0.375 moles

2 moles of CH₄ reacts with 3 moles of O₂ and 2 moles of NH₃

0.656 moles of methane would react with 0.656 moles of NH₃

(0.656 . 3 ) /2 = 0.437 moles of O₂   I do not have enough O₂

Oxygen is the limiting reactant → We can work with the reaction now.

Ratio is 3:2. 3 moles of oxygen produce 2 moles of cyanide

0.375 moles of O₂ may produce (0.375 .2 ) / 3 = 0.250 moles

If we convert the moles to mass → 0.250 mol . 27 g / 1mol = 6.75 g

4 0
3 years ago
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