Please help me I will give you extra points and the brain thing please actually do it.

Calculate the percent of each component in the mixture. Show your calculations. Circle final answers.

Colt1911 [192]

Answer:

See Explanation

Explanation:

The question is incomplete; as the mixtures are not given.

However, I'll give a general explanation on how to go about it and I'll also give an example.

The percentage of a component in a mixture is calculated as:

$\%C_E = \frac{E}{T} * 100\%$

Where

E = Amount of element/component

T = Amount of all elements/components

Take for instance:

In $(Ca(OH)_2)$

The amount of all elements is: (i.e formula mass of $(Ca(OH)_2)$ )

$T = 1 * Ca + 2 * H + 2 * O$

$T = 1 * 40 + 2 * 1 + 2 * 16$

$T = 74$

The amount of calcium is: (i.e formula mass of calcium)

$E = 1 * Ca$

$E = 1 * 40$

$E = 40$

So, the percentage component of calcium is:

$\%C_E = \frac{E}{T} * 100\%$

$\%C_E = \frac{40}{74} * 100\%$

$\%C_E = \frac{4000}{74}\%$

$\%C_E = 54.05\%$

The amount of hydrogen is:

$E = 2 * H$

$E = 2 * 1$

$E = 2$

So, the percentage component of hydrogen is:

$\%C_E = \frac{E}{T} * 100\%$

$\%C_E = \frac{2}{74} * 100\%$

$\%C_E = \frac{200}{74}\%$

$\%C_E = 2.70\%$

Similarly, for oxygen:

The amount of oxygen is:

$E = 2 * O$

$E = 2 * 16$

$E = 32$

So, the percentage component of oxygen is:

$\%C_E = \frac{E}{T} * 100\%$

$\%C_E = \frac{32}{74} * 100\%$

$\%C_E = \frac{3200}{74}\%$

$\%C_E = 43.24\%$

5 0

2 years ago

From what materials are fossil fuels made?

Temka [501]

Answer:

It is buried combustible geologic deposits of organic materials, formed from decayed plants and animals that have been converted to crude oil, coal, natural gas, or heavy oils by exposure to heat and pressure in the earth's crust over hundreds of millions of years.

Explanation:

5 0

3 years ago

a certain country, pivate enterprises provide various goods and services But the govemment provides tree public transportation a

puteri [66]

Mixed economy. Wherever the private and public sector have an active role, they are determined to be a mixed economy.

8 0

3 years ago

4. Ammonium nitrate (NH4NO3) fertilizer decomposes explosively into gaseous nitrogen, oxygen, and water, resulting in huge therm

AURORKA [14]

Answer:

280 g

Explanation:

Let's consider the decomposition of ammonium nitrate.

NH₄NO₃(s) ⇒ N₂(g) + 0.5 O₂(g) + 2 H₂O(g)

We can establish the following relations:

The molar mass of NH₄NO₃ is 80.04 g/mol.
The molar ratio of NH₄NO₃ to N₂ is 1:1.
The molar mass of N₂ is 28.01 g/mol.

The mass of N₂ produced when 800 g of NH₄NO₃ react is:

$800gNH_{4}NO_{3}.\frac{1molNH_{4}NO_{3}}{80.04gNH_{4}NO_{3}} .\frac{1molN_{2}}{1molNH_{4}NO_{3}} .\frac{28.01gN_{2}}{1molN_{2}} =280gN_{2}$

6 0

3 years ago

1. If a solution containing 48.99 g of mercury(II) perchlorate is allowed to react completely with a solution containing 8.564 g

Vitek1552 [10]

1. Chemical eqn:

Hg(ClO4)2 + Na2S -> HgS + 2NaClO4

mercury perchlorate has a molar mass of 399.5g/mol (1d.p) and for Na2S it is 78.0g/mol (1d.p.)

no. of moles of mercury perchlorate= 48.99÷399.5= 0.12263mol(5s.f.)

no. of moles of Na2S= 8.564÷78.0= 0.10979mol ( 5s.f.)

so no. of moles of Hg(ClO4)2/ no. of moles of Na2S= 1/1 according to the eqn

so no. of moles of Hg(ClO4)2 needed if all Na2S is used up is 1/1×0.10979= 0.10979 mols

since no. of moles of mercury perchlorate needed < no. of moles of it provided, it is in excess and Na2S is the limiting factor.

since HgS is a solid and NaClO4 is aqueous, solid ppt formed will only be HgS.

no. of moles of HgS/ no. of moles of NaClO4= 1/1

no. of moles of HgS produced= 0.10979mols

molar mass of HgS= 232.7g/mol 1d.p.

grams of solid produced= 232.7×0.10979= 25.5g (3s.f.)

2. reactant in excess is Hg(ClO4)2,

no. of excess moles= 0.12263-0.10979= 0.01284mols

grams of excess reactant= 0.01284×399.5= 5.13g (3s.f.)

3 0

3 years ago