How do you do Number 4

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

What nuclear particle determines the isotope of an atom?

chubhunter [2.5K]

I believe the answer would be Neutrons

4 0

3 years ago

Read 2 more answers

Isotopes that decay with a nuclear break-up and emit a significant amount of energy are said to be ________.

podryga [215]

Answer to this is Radioactive isotopes.

Isotopes are the species of the same element having different atomic masses that means the number of protons remains the same but number of neutrons do differ. For example $_{1}^{2}\textrm{H}$ and $_{1}^{3}\textrm{H}$ are the two isotopes of Hydrogen ( $_{1}^{1}\textrm{H}$ ).

Radioactive isotopes are the isotopes which release some kind of energy in the form of alpha particles, beta particles or gamma radiation. Examples of each of the decay processes are :

Alpha Decay: In this decay one alpha particle having atomic mass 4 and atomic number 2 or we can say a He molecule will come out. $_{Z}^{A}\textrm{X}\rightarrow _{Z-2}^{A-4}\textrm{Y}+_{2}^{4}\alpha$

Beta Decay: In this decay a $\beta$ particle is emitted increasing the atomic number of the reactant by 1 unit.

$_{Z}^{A}\textrm{X}\rightarrow _{Z+1}^{A}\textrm{Y}+_{-1}^{0}\beta$

Gamma Radiation: In this type of reaction only radiation is emitted out which does not change the original molecule.

$_{Z}^{A}\textrm{X}\rightarrow _{Z}^{A}\textrm{X}+\gamma\text{ radiation}$

3 0

2 years ago

Why sodium chloride is a poor conductor of electricity in solid state?

aev [14]

Answer:

Because the electrons in this ionic compound arent free to move and so cannot carry charge. For an iconic compound to conduct electricity it must be a liquid, either in a molten form or dissolved in water.

Explanation:

Is this clear?

5 0

2 years ago

The hydroboration of an alkene occurs in ___________ which places the boron of the borane on the ___________ carbon of the doubl

Nana76 [90]

Answer: The hydroboration of an alkene occurs in TWO CONCERTED STEP which places the boron of the borane on the LESS SUBSTITUTED carbon of the double bond. The oxidizing agent then acts as a nucleophile, attacking the electrophilic BORON and resulting in the placement of a hydroxyl group on the attached carbon. Thus, the major product of the hydroboration oxidation reaction DOES NOT follow Markovnikov's rule.

Explanation:

Hydroboration is defined as the process which allows boron to attain the octet structure. This involves a two steps pathway which leads to the production of alcohol.

--> The first step: this involves the initiation of the addittion of borane to the alkene and this proceeds as a concerted reaction because bond breaking and bond formation occurs at the same time.

--> The second step: this involves the addition of boron which DOES NOT follow Markovnikov's rule( that is, Anti Markovnikov addition of Boron). This is so because the boron adds to the less substituted carbon of the alkene, which then places the hydrogen on the more substituted carbon.

Note: The Markovnikov rule in organic chemistry states that in alkene addition reactions, the electron-rich component of the reagent adds to the carbon atom with fewer hydrogen atoms bonded to it, while the electron-deficient component adds to the carbon atom with more hydrogen atoms bonded to it.

8 0

2 years ago

How do you do Number 4​

How do you do Number 4