Which of the following molecules is non polar? A.HF. B.HCN. C.NH3. D.CF4

What is the molecular formula of a compound with the empirical formula

ipn [44]

Answer: $C_2H_2O_2$

Explanation:

Molecular formula is the chemical formula which depicts the actual number of atoms of each element present in the compound.

Empirical formula is the simplest chemical formula which depicts the whole number of atoms of each element present in the compound.

The empirical formula is $CHO$

The empirical weight of $CHO$ = 1(12)+1(1)+1(16)= 29 g.

The molecular weight = 60 g/mole

Now we have to calculate the molecular formula:

$n=\frac{\text{Molecular weight}}{\text{Equivalent weight}}=\frac{60}{29}=2$

The molecular formula will be= $2\times CHO=C_2H_2O_2$

Thus molecular formula will be $C_2H_2O_2$

5 0

3 years ago

Determine the molality of a solution of methanol dissolved in ethanol for which the mole fraction of methanol is 0.135. Give you

Alja [10]

Answer: The molality of the solution is 0.11 m

Explanation:

We are given:

Mole fraction of methanol = 0.135

This means that 0.135 moles of methanol is present in 1 mole of a solution

Moles of ethanol = 1 - 0.135 = 0.865 moles

To calculate the mass for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of ethanol = 0.865 moles

Molar mass of ethanol = 46 g/mol

$0.865mol=\frac{\text{Mass of ethanol}}{46g/mol}\\\\\text{Mass of ethanol}=(0.865mol\times 46g/mol}=39.79g$

To calculate the molality of solution, we use the equation:

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

Where,

$m_{solute}$ = Given mass of solute (methanol) = 0.135 g

$M_{solute}$ = Molar mass of solute (methanol) = 32 g/mol

$W_{solvent}$ = Mass of solvent (ethanol) = 39.79 g

Putting values in above equation, we get:

$\text{Molality of methanol}=\frac{0.135\times 1000}{32\times 39.79}\\\\\text{Molality of methanol}=0.106m\approx 0.11m$

Hence, the molality of the solution is 0.11 m

6 0

3 years ago

Which skeleton equation represents a neutralization reaction? Mg + O2 MgO AlCl3 Al + Cl2 Li + HNO3 H2 + LiNO3 HBr + LiOH LiBr +

creativ13 [48]

Neutralization reaction is the reaction between acid and base.
There the neutralization reaction is:

HBr + LiOH ----------> LiBr + H2O.

Hope this helps!

5 0

2 years ago

Read 2 more answers

Organic chemistry involves which type of study?

Mariulka [41]

Answer: option A

Explanation: Organic chemistry deals with the study of carbon and hydrogen containing compounds, drugs and pharmaceuticals etc.

7 0

2 years ago

A 150.0 mL sample of an aqueous solution at 25°C contains 15.2 mg of an unknown nonelectrolyte compound. If the solution has an

inysia [295]

Answer: The molar mass of the unknown compound is 223.2 g/mol

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

where,

$\pi$ = osmotic pressure of the solution = 8.44 torr

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

M = molarity of solute = ?

R = Gas constant = $62.3637\text{ L torr }mol^{-1}K^{-1}$

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

Putting values in above equation, we get:

$8.44torr=1\times M\times 62.3637\text{ L. torr }mol^{-1}K^{-1}\times 298K\\\\M=\frac{8.44}{1\times 62.3637\times 298}=4.54\times 10^{-4}M$

To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:

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

We are given:

Molarity of solution = $4.54\times 10^{-4}M$

Given mass of unknown compound = 15.2 mg = 0.0152 g (Conversion factor: 1 g = 1000 mg)

Volume of solution = 150.0 mL

Putting values in above equation, we get:

$4.54\times 10^{-4}M=\frac{0.0152\times 1000}{\text{Molar mass of unknown compound}\times 150.0}\\\\\text{Molar mass of unknown compound}=\frac{0.0152\times 1000}{150.0\times 4.54\times 10^{-4}}=223.2g/mol$

Hence, the molar mass of the unknown compound is 223.2 g/mol

5 0

3 years ago