The strongest bases have pH values close to

The freezing point of ethanol, CH3CH2OH, is -117.300 °C at 1 atmosphere. Kf(ethanol) = 1.99 °C/m

MAXImum [283]

Answer : The molecular weight of this compound is 891.10 g/mol

Explanation : Given,

Mass of compound = 12.70 g

Mass of ethanol = 216.5 g

Formula used :

$\Delta T_f=i\times K_f\times m\\\\T_f^o-T_f=i\times T_f\times\frac{\text{Mass of compound}\times 1000}{\text{Molar mass of compound}\times \text{Mass of ethanol}}$

where,

$\Delta T_f$ = change in freezing point

$T_f^o$ = temperature of pure ethanol = $-117.300^oC$

$T_f$ = temperature of solution = $-117.431^oC$

$K_f$ = freezing point constant of ethanol = $1.99^oC/m$

i = van't hoff factor = 1 (for non-electrolyte)

m = molality

Now put all the given values in this formula, we get

$(-117.300)-(-117.431)=1\times 1.99^oC/m\times \frac{12.70g\times 1000}{\text{Molar mass of compound}\times 216.5g}$

$\text{Molar mass of compound}=891.10g/mol$

Therefore, the molecular weight of this compound is 891.10 g/mol

7 0

2 years ago

Draw the product formed when the compound shown below undergoes a reaction with hcl in ch2cl2.

Taya2010 [7]

The product formed when HCl and CH2Cl2 react is CH4
H
H-C- H methane structure
H

HCl react with CH2Cl2 to form methane (CH4) and chlorine gas(Cl2)
that is,

2HCl(g) + CH2Cl2(l) = CH4 (g) +2Cl2 (g)

6 0

3 years ago

Which of the following quantities is conserved in a chemical reaction

Nezavi [6.7K]

Answer:

In both nuclear and chemical reactions, two physical quantities are seen to be conserved and unchanging: the number of particles and the total charge. A constant number of particles in nuclear reactions does not imply that mass is conserved.

Explanation:

7 0

3 years ago

Antoinc Laurent Lavoisicr's contribution to chemistry was. a.the law of conservation of mass.

Ulleksa [173]

The answer for the following question is explained below.

The option for the following answer is "d".

Explanation:

The phlogiston theory of combustion:(Greek word phlogiston means <u><em>"BURN"</em></u>)

Phlogiston theory states that phlogisticated substances are the substances that contain phlogiston and dephlogisticate when burned.

Dephlogisticating is the process of releasing stored phlogiston,which is absorbed by the air.

Growing plants then absorb this phlogiston,which is why air does not spontaneously combust and also plant matte burns as well as it does.

The prevailing theory was that flammable materials contained a substance called phlogiston that was released during the combustion.

For example:

phlogiston was transferring into the surrounding air.

8 0

3 years ago

A 14.60g sample of an unknown compound, composed only of carbon, hydrogen, and oxygen, produced 28.6g of CO2 and 14.6g of H2O in

o-na [289]

Answer: The empirical formula for the given compound is $C_{2}H_{5}O$

Explanation:

The chemical equation for the combustion of compound 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 $CO_2=28.6g$

Mass of $H_2O=14.6g$

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 28.6 g of carbon dioxide, $\frac{12}{44}\times 28.6=7.8g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 14.6 g of water, $\frac{2}{18}\times 14.6=1.6$ of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = (14.60) - (7.8 + 1.6) = 5.2 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{7.8g}{12g/mole}=0.65moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.6g}{1g/mole}=1.6moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{5.2g}{16g/mole}=0.32moles$

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 0.32 moles.

For Carbon = $\frac{0.65}{0.32}=2.03\approx 2$

For Hydrogen = $\frac{1.6}{0.32}=5$

For Oxygen = $\frac{0.32}{0.32}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 2 : 5 : 1

Hence, the empirical formula for the given compound is $C_{2}H_{5}O_{1}=C_{2}H_{5}O$

8 0

3 years ago

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