Write the difference between isomerism and allotropy with one example each.

Chemistry

1 answer:

fredd [130]3 years ago

7 0

Answer:

hope it helps ..

Explanation:

Allotropes can be defined as different types of compounds made out of the same single element but in different chemical formulas and different arrangements. Isomers can be defined as the chemical compounds that have a similar molecular formula but with different structural formulae. Graphite and Diamond.

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Calculate the molarity of the sodium acetate solution as described below.

Airida [17]

Answer:

This question is incomplete.

Explanation:

This question is incomplete because of the absence of given mass and volume, however, the steps below will help solve the completed question. The molarity (M) of a solution is the number of moles of solute per liter of solvent. The formula is illustrated below;

Molarity = number of moles (n) / volume (in liter or dm³)

To calculate the number of moles of NaC₂H₃O₂, we say

number of moles (n) =

given or measured mass of NaC₂H₃O₂ ÷ molar mass of NaC₂H₃O₂

The volume of the solvent must be in liter (same as dm³). Thus, to convert mL to liter, we divide by 1000

The unit for Molarity is M (Molar concentration), mol/L or mol/dm³

4 0

2 years ago

of an unknown protein are dissolved in enough solvent to make 5.00mL of solution. The osmotic pressure of this solution is measu

krok68 [10]

The question is incomplete . The complete question is :

100 mg of an unknown protein are dissolved in enough solvent to make 5.00mL of solution. The osmotic pressure of this solution is measured to be 0.107atm at 25.0°C. Calculate the molar mass of the protein. Round your answer to 3 significant digits.

Answer: The molar mass of the protein is $4.57\times 10^3g/mol$

Explanation:

$\pi =CRT$

$\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT$

where,

$\pi$ = osmotic pressure of the solution = 0.107 atm

i = Van't hoff factor = 1 (for non-electrolytes)

Mass of solute (protein) = 100 mg = 0.1 g (Conversion factor: 1 g = 1000 mg)

Volume of solution = 5.00 mL

R = Gas constant = $0.0821\text{ L.atm }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[273+25]=298K$

Putting values in above equation, we get:

$0.107=1\times \frac{0.1\times 1000}{\text{Molar mass of insulin}\times 5.00}\times 0.0821\text{ Latm }mol^{-1}K^{-1}\times 298K\\\\\text{molar mass of protein}$