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2 years ago

Which of these types of changes is a physical change?

Chemistry

2 answers:

Degger [83]2 years ago

7 0

The answer is a because the other ones are chemical

Leto [7]2 years ago

6 0

The correct answer is A. Can I get the Brainliest?

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Which law relates to the ideal gas law?

Kamila [148]

The equation of state for a hypothetical ideal gas is known as the ideal gas law, sometimes known as the general gas equation. i.e. PV = nRT or P1V1 = P2V2.

According to the ideal gas law, the sum of the absolute temperature of the gas and the universal gas constant is equal to the product of the pressure and volume of one gram of an ideal gas.
Robert Boyle, Gay-Lussac, and Amedeo Avogadro's observational work served as the basis for the ideal gas law. The Ideal gas equation, which simultaneously describes every relationship, is obtained by combining all of their observations into a single statement.
When applying the gas constant R = 0.082 L.atm/K.mol, pressure, volume, and temperature should all be expressed in units of atmospheres (atm), litres (L), and kelvin (K).
At high pressure and low temperature, the ideal gas law basically fails because molecule size and intermolecular forces are no longer negligible but rather become significant considerations.

Learn more about ideal gas law here:

brainly.com/question/26040104

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3 0

1 year ago

What is the mass of 100 mL of corn oil?

prisoha [69]

Answer:

m=(0.92g/ml)(100ml)

=92g

hope this helps

5 0

2 years ago

Severus Snape explains that if you have the molar mass of a compound and the empirical formula of a compound you can determine t

KengaRu [80]

Answer:

The molecular formula of the compound : $C_4H_6O_2$

Explanation:

The empirical formula of the compound = $C_2H_3O$

The molecular formula of the compound = $C_{2n}H_{3n}O_n$

The equation used to calculate the valency is :

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

We are given:

Mass of molecular formula = 86 g/mol

Mass of empirical formula = 43 g/mol

Putting values in above equation, we get:

$n=\frac{86g/mol}{43g/mol}=2$

The molecular formula of the compound :

$C_{2\times 2}H_{3\times 2}O_2=C_4H_6O_2$

7 0

3 years ago

Ammonia is produced by the following reaction. 3H2(g) N2(g) Right arrow. 2NH3(g) When 7. 00 g of hydrogen react with 70. 0 g of

harkovskaia [24]

In the ammonia production process given by the reaction 3H₂(g) + N₂(g) → 2NH₃(g), when 7.00 g of hydrogen react with 70.0 g of nitrogen, hydrogen is considered the limiting reactant because 7.5 moles of hydrogen would be needed to consume the available nitrogen (option 1).

The reaction is the following:

3H₂(g) + N₂(g) → 2NH₃(g) (1)

To know why hydrogen is considered the limiting reactant, we need to calculate the number of moles of nitrogen and hydrogen with the following equation:

$n = \frac{m}{M}$

Where:

m: is the mass

M: is the molar mass

For hydrogen we have:

$n_{H_{2}} = \frac{m}{M} = \frac{7.00 g}{2.016 g/mol} = 3.47 \:moles$

And for nitrogen:

$n_{N_{2}} = \frac{m}{M} = \frac{70.0 g}{28.013 g/mol} = 2.50 \:moles$

We can see in reaction (1) that 3 moles of hydrogen react with 1 mol of nitrogen, so the number of hydrogen moles needed to react nitrogen is:

$n_{H_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*n_{N_{2}} = \frac{3\:moles\:H_{2}}{1\:moles\:N_{2}}*2.50 \:moles = 7.50 \:moles$

Since we have 3.47 moles of hydrogen and we need 7.50 moles to react with all the mass of nitrogen, the limiting reactant is hydrogen.

We can find the number of ammonia moles produced with the limiting reactant (hydrogen) konwing that 3 moles of hydrogen produces 2 moles of ammonia, so:

$n_{NH_{3}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*n_{H_{2}} = \frac{2\:moles\:NH_{3}}{3\:moles\:H_{2}}*3.47 \:moles = 2.31 \:moles$