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3 years ago

How many molecules are present in a drop of ethanol, c2h5oh, of mass 2.310^-2.3g? (NA=6.010^23mol^-1)

Chemistry

1 answer:

sdas [7]3 years ago

5 0

Answer:

3.0 × 10²⁰ molecules

Explanation:

Given data:

Mass of ethanol = 2.3 × 10⁻²°³ g

Number of molecules = ?

Solution:

Number of moles of ethanol:

Number of moles = mass/ molar mass

Number of moles = 2.3 × 10⁻²°³ g / 46.07 g/mol

Number of moles = 0.05 × 10⁻²°³ mol

Number of molecules:

One mole = 6.022 × 10²³ molecules

0.05 × 10⁻²°³ mol × 6.022 × 10²³ molecules / 1 mol

0.30 × 10²⁰°⁷ molecules

3.0 × 10¹⁹°⁷ molecules which is almost equal to 3.0 × 10²⁰ molecules.

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HELP! PLEASE! ASAP! Manganese(III) fluoride, MnF3, can be prepared by the following reaction:

Crank

Answer:

14.336 g MnF₂

Explanation:

number of moles = mass / molecular weight

number of moles of MnI₂ = 55 / 309 = 0.178 moles

number of moles of F₂ = 55 / 38 = 1.447 moles

From the reaction and the number of moles calculated we deduce that the fluorine F₂ is a limiting reactant.

So:

if 13 moles of F₂ reacts to produce 2 moles of MnF₃

then 1.447 moles of F₂ reacts to produce X moles of MnF₃

X = (1.447 × 2) / 13 = 0.223 moles of MnF₃ (100% yield)

For 57.2% yield we have:

number of moles of MnF₃ = (57.2 / 100) × 0.223 = 0.128 moles

mass = number of moles × molecular weight

mass of MnF₃ = 0.128 × 112 = 14.336 g

3 0

3 years ago

Compounds have two or more

frozen [14]

Answer: B

Explanation:

6 0

3 years ago

Explain, in terms of intermolecular forces, why ammonia has a higher boiling point than the other compounds in the ta

Kobotan [32]

Answer is: ammonia has a higher boiling point because it has stronger intermolecular forces.

Intermolecular forces are the forces between molecules. The stronger are intermolecular forces, the higher is boiling point of compound, because more energy is needed to break interaction between molecules.
There are several types of intermolecular forces: hydrogen bonding, ion-induced dipole forces, ion-dipole forces andvan der Waals forces.

7 0

3 years ago

How many electrons are there in 2 g of H2

KiRa [710]

Answer:

6.02 x 10²³ electrons

Explanation:

Given parameters:

Mass of H₂ = 2g

Unknown:

Number of electrons = ?

Solution:

To find the number of electrons, we must determine the number of moles of H₂ first.

Number of moles = $\frac{mass}{molar mass}$

Molar mass of H₂ = 2(1) = 2g/mol

Number of moles = $\frac{2}{2}$ = 1mol

1 mole of a substance contains 6.02 x 10²³ particles

the particles can be protons, neutrons, electrons

So,

2g of H₂ will contain 6.02 x 10²³ electrons.

4 0

3 years ago

A water solution contains 1.704 [kg] of HNO3 per [kg] of water, and has a specific gravity of 1.382 at 20 [°C]. Please, express

Rudik [331]

Answer:

(a) The weight percent HNO3 is 63%.

(b) Density of HNO3 = 111.2 lb/ft3

Explanation:

(a) Weight percent HNO3

To calculate a weight percent of a component of a solution we can express:

$wt = \frac{mass \, of \, solute}{mass \, of \, solution}=\frac{mass\,HNO_3}{mass \, HNO_3+mass \, H_2O}\\ \\wt=\frac{1.704}{1.704+1} =\frac{1.704}{2.704}= 0.63$

(b) Density of HNO3, in lb/ft3

In this calculation, we use the specific gravity of the solution (1.382). We can start with the volume balance:

$V_s=V_{HNO3}+V_w\\\\\frac{M_s}{\rho_s}=\frac{M_{HNO3}}{\rho_{HNO3}}+\frac{M_w}{\rho_w}\\\\\frac{M_{HNO3}}{\rho_{HNO3}} = \frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}\\\\\rho_{HNO3}=\frac{M_{HNO3}}{\frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{\frac{2.704}{1.382*\rho_w}+\frac{1}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{2.956/ \rho_w}= 0.576*\rho_w[tex]V_s=V_{HNO3}+V_w\\\\\frac{M_s}{\rho_s}=\frac{M_{HNO3}}{\rho_{HNO3}}+\frac{M_w}{\rho_w}\\\\\frac{M_{HNO3}}{\rho_{HNO3}} = \frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}\\\\\rho_{HNO3}=\frac{M_{HNO3}}{\frac{M_s}{\rho_s}-\frac{M_w}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{\frac{2.704}{1.382*\rho_w}-\frac{1}{\rho_w}} \\\\ \rho_{HNO3}=\frac{1.704}{0.956/\rho_w}= 1.782*\rho_w$

The density of HNO3 is 1.782 times the density of water (Sp Gr of 1.782). If the density of water is 62.4 lbs/ft3,

$\rho_{HNO3}= 1.782*\rho_w=1.782*62.4 \, lbs/ft3=111.2\, lbs/ft3$

If we use the molar mass of HNO3: 63.012 g/mol, we can say that in 1,704 kg (or 1704 g) of HNO3 there are 1704/63.012=27.04 mol HNO3.

When there are 1.704 kg of NHO3 in solution, the total mass of the solution is (1.704+1)=2.704 kg.

If the specific gravity of the solution is 1.382 and the density of water at 20 degC is 998 kg/m3, the volume of the solution is

$Vol=\frac{M_{sol}}{\rho_{sol}}=\frac{2.704\, kg}{1.382*998 \, kg/m3} = 0.00196m3$

We can now calculate the molarity as

$Molarity HNO_3=\frac{MolHNO3}{Vol}=\frac{27.04mol}{0.00196m3} =13792 \frac{molHNO3}{m3}$

8 0

3 years ago

How many molecules are present in a drop of ethanol, c2h5oh, of mass 2.3*10^-2.3g? (NA=6.0*10^23mol^-1)

How many molecules are present in a drop of ethanol, c2h5oh, of mass 2.310^-2.3g? (NA=6.010^23mol^-1)