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3 years ago

Explain how you would find the mass percent of each element in water.

Chemistry

1 answer:

nydimaria [60]3 years ago

3 0

Answer:

finding the mass percentage oven element in a compound might sound complicated, but the calculation is simple. For example, to determine the mass percentage of hydrogen in water H2O, divide the major mass of hydrogen by the total molar mass of water and then multiply the result by 100

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Covalent bonds are formed when electrons are

lubasha [3.4K]

Covalent bonds formed when two atoms share an electron, as opposed to an Ionic bond where one atom takes an electron(or two) from another atom.

5 0

4 years ago

3.5 grams is equivalent to how many kilograms? 3500 kg 35,000 kg 0.35 kg 0.0035 kg

bagirrra123 [75]

There are 1000 grams in a kg.
To convert g to kg, dovide by 1000.

3.5/1000= 0.0035 kg

Final answer: D

4 0

3 years ago

Two descriptions for a sub-atomic particle are listed below:

Dmitrij [34]

Can not be D or B and A is not enough so C is right.

4 0

3 years ago

Read 2 more answers

Could some some one help with 2 4 5 7 8

sesenic [268]

Answer:

For number 2, your answer is C.

Explanation:

The nucleus is positively charged because the proton is positive and a neutron is neutral. An electron has a negative charge.

6 0

3 years ago

Calculate the density of O2(g) at 415 K and 310 bar using the ideal gas and the van der Waals equations of state. Use a numerica

Lera25 [3.4K]

Answer:

Explanation:

From the given information:

The density of O₂ gas = $d_{ideal} = \dfrac{P\times M}{RT}$

here:

P = pressure of the O₂ gas = 310 bar

= $310 \ bar \times \dfrac{0.987 \ atm}{1 \ bar}$

= 305.97 atm

The temperature T = 415 K

The rate R = 0.0821 L.atm/mol.K

molar mass of O₂ gas = 32 g/mol

∴

$d_{ideal} = \dfrac{305.97 \ \times 32}{0.0821 \times 415}$

$d_{ideal}$ = 287.37 g/L

To find the density using the Van der Waal equation

Recall that:

the Van der Waal constant for O₂ is:

a = 1.382 bar. L²/mol² &

b = 0.0319 L/mol

The initial step is to determine the volume = Vm

The Van der Waal equation can be represented as:

$P =\dfrac{RT}{V-b}-\dfrac{a}{V^2}$

where;

R = gas constant (in bar) = 8.314 × 10⁻² L.bar/ K.mol

Replacing our values into the above equation, we have:

$310 =\dfrac{0.08314\times 415}{V-0.0319}-\dfrac{1.382}{V^2}$

$310 =\dfrac{34.5031}{V-0.0319}-\dfrac{1.382}{V^2}$

$310V^3 -44.389V^2+1.382V-0.044=0$

After solving;

V = 0.1152 L

∴

$d_{Van \ der \ Waal} = \dfrac{32}{0.1152}$

$d_{Van \ der \ Waal}$ = 277.77 g/L

We say that the repulsive part of the interaction potential dominates because the results showcase that the density of the Van der Waals is lesser than the density of ideal gas.

5 0

3 years ago