Producers are generally found at the beginning of a food chain. Which

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

3 years ago

What occurs at one of the electrodes in both an electrolytic cell and a voltaic cell?

rjkz [21]

Anode- oxidization

Cathode-reduction

3 0

2 years ago

A graduated cylinder had 20mL of water. A sample of pyrite was placed in the cylinder and the volume moved to 35 mL. What is the

Ksju [112]

15mL because if it started as 20 mL and went to 35mL when u put in pyrite, the pyrite takes up 15mL

4 0

2 years ago

A solution is created by dissolving 13.0 grams of ammonium chloride in enough water to make 295 mL of solution. How many moles o

Masja [62]

Answer:

(a) Moles of ammonium chloride = 0.243 moles

(b) $Molarity_{ammonium\ chloride}=0.824\ M$

(c) 60.68 mL

Explanation:

(a) Mass of ammonium chloride = 13.0 g

Molar mass of ammonium chloride = 53.491 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{13.0\ g}{53.491\ g/mol}$

Moles of ammonium chloride = 0.243 moles

(b) Moles of ammonium chloride = 0.243 moles

Volume = 295 mL = 0.295 L ( 1 mL = 0.001 L)

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$Molarity_{ammonium\ chloride}=\frac{0.243}{0.295}$

$Molarity_{ammonium\ chloride}=0.824\ M$

(c) Moles of ammonium chloride = 0.0500 moles

Volume = ?

Molarity = 0.824 M

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

$0.824\ M=\frac{0.0500}{Volume}$

Volume = 0.05 / 0.824 L = 0.06068 L = 60.68 mL

6 0

3 years ago

7) Which element would have the lowest first ionization energy? * Li Be Na Mg

olga_2 [115]

Answer:

Dos quintos al cuadrado dividido cinco medios a la menos 1.

Explanation:

7 0

2 years ago

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