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3 years ago

Most common energy source in cellular respiration

Chemistry

2 answers:

erma4kov [3.2K]3 years ago

8 0

Answer:

Light

Explanation:

I know because we are going over the same thing

topjm [15]3 years ago

7 0

Answer: cytoplasm

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]

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

Which numerical setup can be used to calculate the atomic mass of the element bromine?

Gekata [30.6K]

Answer:

From the numerical steps highlighted under explanation, the average atomic mass of bromine is 79.91 u

Explanation:

The steps to be taken will involve;

1) Find the number of isotopes of bromine.

2) Identify the atomic mass and relative abundance of each of the isotopes.

3) Multiply the atomic mass of each of the isotopes by their corresponding values relative abundance value.

4) Add the value in step 3 above to get the average atomic mass of bromine.

Now;

Bromine has 2 isotopes namely;

Isotope 1: Atomic mass = 78.92amu and a relative abundance of 50.69%.

Isotope 2: Atomic mass = 80.92amu and a relative abundance of 49.31%.

Using step 3 above, we have;

(78.92 × 50.69%)

And (80.92 × 49.31%)

Using step 4 above, we have;

(78.92 × 50.69%) + (80.92 × 49.31%) ≈ 79.91 u

6 0

3 years ago

If I log out of Brainly and delete the Brainly app from my phone, will my account still be up and active?

Darya [45]

Yes and you can log in to it later as well

4 0

3 years ago

A large atom decays and emits a particle. After the reaction is complete, the atom’s mass has changed substantially. What kind o

hoa [83]

The answer is alpha particles

5 0

3 years ago

Read 2 more answers

For the decomposition of ammonia on a platinum surface at 856 °C

Crazy boy [7]

$\\ \tt\leadsto \dfrac{d[NH_3]}{dt}=1.50\times 10^{-6}$

dt remains same for reaction

$\\ \tt\leadsto \dfrac{d[H_2]}{dt}=\dfrac{3}{2}\dfrac{d[NH_3]}{dt}$

$\\ \tt\leadsto \dfrac{d[H_2]}{dt}=\dfrac{3}{2}(1.5\times 10^{-6})$

$\\ \tt\leadsto \dfrac{d[H_2]}{dt}=2.25\times 10^{-6}Ms^{-1}$

M is molarity here not metre

5 0

2 years ago