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OleMash [197]
3 years ago
14

When trying to clean muddy, dirty river water, which method would work

Chemistry
1 answer:
Llana [10]3 years ago
6 0
When trying to clean muddy, dirty river water, FILTRATION would work best
You might be interested in
How many moles of carbon are there in a 0.70 g (3.5 carat) diamond?
maksim [4K]

Answer:

3.5 x 10^22 moles of carbon are there in a 0.70 g (3.5 carat) diamond.

Explanation:

In case of diamond, the number of atoms is the same as the number of molecules.

7 0
3 years ago
What is the mass of 4 moles of helium, He?
aniked [119]
I am pretty sure it is 4.002602
4 0
3 years ago
A compound is composed of C, H and O. A 1.621 g sample of this compound was combusted, producing 1.902 g of water and 3.095 g of
vlada-n [284]

Answer: The molecular of the compound is, C_2H_3O

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=3.095g

Mass of H_2O=1.902g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 3.095g of carbon dioxide, \frac{12}{44}\times 3.095=0.844g of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 1.902g of water, \frac{2}{18}\times 1.092=0.121g of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = (1.621)-[(0.844)+(0.121)]=0.656g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.844g}{12g/mole}=0.0703moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.121g}{1g/mole}=0.121moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.656g}{16g/mole}=0.041moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.041 moles.

For Carbon = \frac{0.0703}{0.041}=1.71\approx 2

For Hydrogen  = \frac{0.121}{0.041}=2.95\approx 3

For Oxygen  = \frac{0.041}{0.041}=1

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 3 : 1

Hence, the empirical formula for the given compound is C_2H_3O_1=C_2H_3O

The empirical formula weight = 2(12) + 3(1) + 1(16) = 43 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}

n=\frac{46.06}{43}=1

Molecular formula = (C_2H_3O_1)_n=(C_2H_3O_1)_1=C_2H_3O

Therefore, the molecular of the compound is, C_2H_3O

6 0
3 years ago
Write the names of stages in the life cycle of butterfly
stepladder [879]

Answer:

There are four stages in the metamorphosis of butterflies and moths: egg, larva, pupa, and adult.

Hope this helps :)

3 0
3 years ago
According to the law of conservation of mass, the total mass of the reactants is
yarga [219]

Answer:

is equal to the total number of products

Explanation:

3 0
3 years ago
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