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2 years ago

Solid calcium carbonate reacts with hydrochloric acid, forming aqueous calcium chloride, carbon dioxide

Chemistry

1 answer:

12345 [234]2 years ago

6 0

Answer:

When hydrochloric acid comes into contact with calcium carbonate, the following chemical reaction ensues: CaCO3 + 2HCl → CaCl + CO2 + H2O, which provides acid neutralization alongside the formation of byproducts

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When of a certain molecular compound X are dissolved in of benzene , the freezing point of the solution is measured to be . Calc

puteri [66]

The question is incomplete. Here is the complete question.

When 2.10 g of a certain molecular compound X are dissolved in 65.0 g of benzene (C₆H₆), the freezing point of the solution is measured to be 3.5°C. Calculate the molar mass of X. If you need any additional information on benzene, use only what you find in the ALEKS Data resource. Also, be sure your answer has a unit symbol, and is rounded to 2 significant digits.

Answer: MM = 47.30 g/mol.

Explanation: There is a relationship between freezing point depression and molality. With this last one, is possible to calculate molar mass or molar weight of a compound.

Freezing Point Depression occurs when a solute is added to a solvent: the freezing point of the solvent decreases when a non-volatile solute is incremented.

Molality or molal concentration is a quantity of solute dissolved in a certain mass, in kg, of solvent. Its symbol is m and it's defined as

$m=\frac{moles(solute)}{kg(solvent)}$

Freezing point depression and molal are related as the following:

$\Delta T_{f}=K_{f}.m$

where

$\Delta T_{f}$ is freezing point depression of solution

$K_{f}$ is molal freezing point depression constant

m is molality

Now, to determine molar mass, first, find molality of the mixture:

$\Delta T_{f}=K_{f}.m$

$m=\frac{\Delta T_{f}}{K_{f}}$

For benzene, constant is 5.12°C/molal. Then

$m=\frac{3.5}{5.12}$

m = 0.683 molal

Second, knowing the relationship between molal and moles of solute, determine the last one:

$m=\frac{moles(solute)}{kg(solvent)}$

$mol(solute)=m.kg(solvent)$

mol(solute) = 0.683(0.065)

mol(solute) = 0.044 mol

The definition for Molar mass is the mass in grams of 1 mol of substance:

$n(moles)=\frac{m(g)}{MM(g/mol)}$

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

In the mixture, there are 0.044 moles of X, so its molecular mass is

$MM=\frac{2.1}{0.044}$

MM = 47.30 g/mol

The molecular compound X has molecular mass of 47.30 g/mol.

7 0

3 years ago

Name the subatomic particles contained in the nucleus of the atom

lawyer [7]

Neutron, proton
form those are both quarks

7 0

3 years ago

The density of a metal is 11.4 g/cm3. How much volume (in cm3) would a sample of 30.5 g have?

julia-pushkina [17]

The Concept:

We are given the density of a sample of the metal = 11.4 grams / cm³

and we need to find the volume occupied by a sample of 30.5 grams

For this solution, we will use dimensional analysis

from the given information, we can also say that the density of the metal is:

1 cm³ / 11.4 grams

If we multiply this value by 30.5 grams, the 'grams' in the numerator and the denominator will cancel out and we will be left with the volume occupied by 30.5 grams of the metal

Solving for the volume:

$\frac{ 1 cm^{3} }{11.4 grams}$ X 30.5 grams = (30.5 / 11.4) cm³

Volume of 30.5 grams of the sample = 2.68 cm³

6 0

3 years ago

In plants, this is period in which a seed or plant does not grow, waiting for the necessary environmental conditions like the ri

Lunna [17]

Answer:

The answer is Germination.

Explanation:

During the stage of germination, the plant needs the right temperature, day length, amount of water, and nutrients.

8 0

4 years ago

Calculate the number of moles of Cu in 3.8×10213.8×1021 atoms of Cu.

rosijanka [135]

Answer:

6.3×10⁻³ moles of Cu

Explanation:

The relation is, that 1 mol of particles are contained by 6.02×10²³ particles.

This is Avogadro's number that explains that: one mole determines the number of fundamental units that are contained in a constant number but they do not depend on the type of material or the type of particle, and this quantity is 6.02 × 10²³

We can make a rule of three:

6.02×10²³ atoms are contained 1 in mol of Cu

Therefore, 3.8×10²¹ atoms will be contained in (3.8×10²¹ . 1) / NA = 6.3×10⁻³ moles

7 0

4 years ago