Does anybody know the answer to these two questions ?

When 551. mg of a certain molecular compound X are dissolved in 100 g of benzonitrile (CH,CN), the freezing point of the solutio

Ludmilka [50]

Answer:

M1 = 49.04 g/mol

Explanation:

The pure benzonitrile has freezing point -12.8°C. By adding a nonvolatile compound, the freezing point will be changed, a process called cryoscopy. The freezing point will be reduced. In this case, the new freezing point is -13.4°C. The variation at the temperature can be calculated by the equation:

ΔT = Kc*W*i

Where ΔT is the variation at the freezing temperature (without the solute less with the solute), Kc is the cryoscopy constant (5.34 for benzonitrile), W is the molality, and i the Van't Hoff correction factor, which is 1 for benzonitrile.

((-12.8-(-13.4)) = 5.34*W

5.34W = 0.6

W = 0.1124 mol/kg

W = m1/M1*m2

Where m1 is the mass of the solute (in g), M1 is the molar mass of the solute (in g/mol), and m2 is the mass of the solvent (in kg).

m1 = 0.551 g, m2 = 0.1 kg

0.1124 = 0.551/M1*0.1

0.01124M1 = 0.551

M1 = 49.04 g/mol

7 0

3 years ago

Combustion analysis of 0.600 g of an unknown compound containing carbon, hydrogen, and oxygen produced 1.043 g of CO2 and 0.5670

Aliun [14]

Answer : The empirical formula of the compound is $C_3H_8O_2$

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 $CO_2=1.043g$

Mass of $H_2O=0.5670g$

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 1.043 g of carbon dioxide, $\frac{12}{44}\times 1.043=0.284g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 0.5670 g of water, $\frac{2}{18}\times 0.5670=0.063g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.600) - (0.284 + 0.063) = 0.253 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.284g}{12g/mole}=0.0237moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.063g}{1g/mole}=0.063moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.253g}{16g/mole}=0.0158moles$

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 0.0158 moles.

For Carbon = $\frac{0.0237}{0.0158}=1.5$

For Hydrogen = $\frac{0.063}{0.0158}=3.98\approx 4$

For Oxygen = $\frac{0.0158}{0.0158}=1$

The ratio of C : H : O = 1.5 : 4 : 1

To make in a whole number we are multiplying the ratio by 2, we get:

The ratio of C : H : O = 3 : 8 : 2

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 8 : 2

Hence, the empirical formula for the given compound is $C_3H_8O_2$

6 0

3 years ago

Uniformitarianism is the idea that

Hunter-Best [27]

Answer:

Uniformitarianism is the idea that the natural processes operating today are the same as the natural processes that operated in the past

3 0

3 years ago

Who are the scientists that contributed in arranging of the periodic table?

Irina-Kira [14]

In 1869 Russian chemist DIMITRI MENDELEEV started the development of the periodic table,arranging chemical elements by atomic mass. He predicted the discovery of other elements and left spaces open in his periodic table for them. HOPE THIS HELPSS HAVE A GREAT DAY <333

7 0

3 years ago

How did the Bohr model of the atom affect scientific thought?

Vadim26 [7]

Answer:

C. Scientists accepted the model at first but later rejected it.

Explanation:

Scientists accepted the model at first because it explained the hydrogen emission spectrum.

However, with the development of quantum mechanics, scientists had to modify the model (not reject it).

Electrons still had specific energies, but they no longer travelled in fixed orbits.

Instead, electrons had a probability of being found in a given region of space.

5 0

3 years ago

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