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Tems11 [23]
3 years ago
13

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

he precipitation of 0.299 g AgBr. Use this data to compute the %Ga (by mass) GaBr2
Chemistry
1 answer:
tresset_1 [31]3 years ago
3 0

<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 %.

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A container is filled with Neon gas. It has a volume of 1.5L and a pressure of 101.3 kPa. If the volume of the container is incr
Marta_Voda [28]

Answer:

37.98 kPa.

Explanation:

  • We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R  is the general gas constant,

T is the temperature of the gas in K.

  • If n and T are constant, and have different values of P and V:

<em>(P₁V₁) = (P₂V₂)</em>

<em></em>

  • Knowing that:

P₁ = 101.3 kPa, V₁ = 1.5 L,

P₂ = ??? kPa, V₂ = 4.0 L.

  • Applying in the above equation

<em>(P₁V₁) = (P₂V₂)</em>

<em></em>

<em>∴ P₂ = (P₁V₁)/V₂</em> = (101.3 kPa)(1.5 L)/(4.0 L) = <em>37.98 kPa.</em>

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A farmer has a 10 acre plot to feed his family. What is the most efficient way to use the plot?
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3 years ago
(40 Points) Complete, balance, compute the amounts of the products assuming 100% yield. 12g Zinc Solid + 24g Silver Nitrate
VARVARA [1.3K]

Answer:

Zn + 2AgNO₃ → Zn(NO₃)₂ +2Ag

13.34g Zn(NO₃)₃ and 15.24g Ag are formed if reaction is 100% complete.

Explanation:

molar mass of Zn=65 g

molar mass of AgNO₃=170 g

molar mass of Zn(NO₃)₂ =189  g

molar mass of Silver = 108 g

Zn + 2  AgNO₃ → Zn(NO₃)₂ +2 Ag            eq(1)

1 mole Zn  reacts with 2 moles Silver nitrate to give 1 mole Zinc nitrate and 2 moles Silver

or

65 g Zn reacts with 2×170 g of AgNO₃  → 189 g Zn(NO₃)₂ and 2×108 g Ag    - eq(2)

First find limiting reagent of the reaction

65g Zn reacts with 2×170g of AgNO₃

12 g Zn reacts with (12÷65)×2×170 g AgNO₃

=62.7 g AgNO₃

For the reaction to go to 100% yield 12 g Zn will need 62.7 g AgNO₃

but amount of AgNO₃ is 24 g

So the reaction yields is limited by amount of AgNO₃.

AgNO₃ is the limiting reagent.

So calculate the yield of products with the amount of AgNO₃

by eq(2)

2× 170 gAgNO₃ gives 189 g Zn(NO₃)₃

=340 g AgNO₃ gives 189 g Zn(NO₃)₃

24 g AgNO₃ gives (24÷340) ×189 g Zn(NO₃)₃

=13.34g Zn(NO₃)₃

again by eq2

2×170 g AgNO₃ gives 2×108 g Ag

= 340 g AgNO₃ gives 216 g Ag

24 g AgNO₃ gives (24÷340)×216 g Ag

= 15.24g Ag

6 0
3 years ago
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