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2 years ago

Based on your knowledge of intermolecular forces and trends, select the statement that is most accurately compares HF and HCl.

Chemistry

1 answer:

tangare [24]2 years ago

8 0

Answer:

(A) The intermolecular attraction between HF molecules are stronger than between HCl molecules mainly to due hydrogen bonding.

Explanation:

Since Flourine is highly electronegative and as such, when it bonds with Hydrogen it forms a hydrogen bond. Whereas the HCL molecule is a polar molecule whose inter-molecular forces are dipole dipole interactions. Although a Hydrogen bond is a type of dipole dipole interaction it is stronger than the traditional dipole dipole forces and London dispersion forces. HF also has a shorter bond length which makes the bond and inter-molecular forces stronger as compared to HCL.

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The attractions that allow molecules of krypton to exist in the solid phase are due to:

OLga [1]

Answer:

4) Van der waals forces

Explanation:

Krypton (Kr) belongs to the noble gas group and has fully filled valence orbitals. In the solid phase, Kr exists as a white solid with a face centered cubic structure.

Intermolecular forces of attraction from the strongest to the weakest include:

Ionic > hydrogen bonding > dipole-dipole > london dispersion

Kr is monoatomic and non-polar. When fully filled (stable) valence orbitals of 2 Kr atoms approach each other in close proximity they experience a repulsive force which prevents the formation of strong bonds. Thus, the only force of attraction in Kr is the long range weak Van Der Waals force also known as the london dispersion force.

7 0

2 years ago

mixture of N 2 And H2 Gases weighs 13.22 g and occupies a volume of 24.62 L at 300 K and 1.00 atm.Calculate the mass percent of

anygoal [31]

Answer: The mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

Explanation:

To calculate the number of moles, we use the equation given by ideal gas equation:

PV = nRT

where,

P = Pressure of the gaseous mixture = 1.00 atm

V = Volume of the gaseous mixture = 24.62 L

n = number of moles of the gaseous mixture = ?

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

T = Temperature of the gaseous mixture = 300 K

Putting values in above equation, we get:

$1.00atm\times 24.62L=n_{mix}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 300K\\\\n_{mix}=\frac{1.00\times 24.62}{0.0821\times 300}=0.9996mol$

We are given:

Total mass of the mixture = 13.22 grams

Let the mass of nitrogen gas be 'x' grams and that of hydrogen gas be '(13.22 - x)' grams

To calculate the number of moles, we use the equation:

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

For nitrogen gas:

Molar mass of nitrogen gas = 28 g/mol

$\text{Moles of nitrogen gas}=\frac{x}{28}mol$

For hydrogen gas:

Molar mass of hydrogen gas = 2 g/mol

$\text{Moles of hydrogen gas}=\frac{(13.22-x)}{2}mol$

Equating the moles of the individual gases to the moles of mixture:

$0.9996=\frac{x}{28}+\frac{(13.22-x)}{2}\\\\x=12.084g$

To calculate the mass percentage of substance in mixture we use the equation:

$\text{Mass percent of substance}=\frac{\text{Mass of substance}}{\text{Mass of mixture}}\times 100$

Mass of the mixture = 13.22 g

For nitrogen gas:

Mass of nitrogen gas = x = 12.084 g

Putting values in above equation, we get:

$\text{Mass percent of nitrogen gas}=\frac{12.084g}{13.22g}\times 100=91.41\%$

For hydrogen gas:

Mass of hydrogen gas = (13.22 - x) = (13.22 - 12.084) g = 1.136 g

Putting values in above equation, we get:

$\text{Mass percent of hydrogen gas}=\frac{1.136g}{13.22g}\times 100=8.59\%$

Hence, the mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

5 0

2 years ago

A student pours Liquid A through filter paper. the liquid passes through the filter paper, but a residue remains on the filter.

AVprozaik [17]

Answer:

I think it is A.Element maybe

4 0

2 years ago

Read 2 more answers

The chemical formula for magnesium oxide is MgO. A chemist determined by measurements that 0.030 moles of magnesium oxide partic

tamaranim1 [39]

Answer:

1.209g of MgO participates

Explanation:

In this problem, we have 0.030 moles of MgO that participates in a particular reaction.

And we are asked to solve for the mass of MgO that participates, that means, we need to convert moles to grams.

To convert moles to grams we need to use molar mass of the compound:

1 atom of Mg has a molar mass of 24.3g/mol

1 atom of O has a molar mass of 16g/mol

That means molar mass of MgO is 24.3g/mol + 16g/mol = 40.3g/mol

And mass of 0.030 moles of MgO is:

0.030 moles MgO * (40.3g/mol) =

<h3>1.209g of MgO participates</h3>

3 0

2 years ago

The rate constant k for a certain reaction is measured at two different temperatures temperature 376.0 °c 4.8 x 108 280.0 °C 2.3

9966 [12]

Answer:

The activation energy for this reaction = 23 kJ/mol.

Explanation:

Using the expression,

$\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )$

Where,

$k_1\ is\ the\ rate\ constant\ at\ T_1$

$k_2\ is\ the\ rate\ constant\ at\ T_2$

$E_a$ is the activation energy

R is Gas constant having value = 8.314×10⁻³ kJ / K mol

$k_2=2.3\times 10^8$

$T_2=280\ ^0C$

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (280 + 273.15) K = 553.15 K

$T_2=553.15\ K$

$k_1=4.8\times 10^8$

$T_1=376\ ^0C$

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (376 + 273.15) K = 649.15 K

$T_1=649.15\ K$

So,

$\left(\ln \left(\:\frac{4.8\times \:\:\:10^8}{2.3\times \:\:\:10^8}\right)\right)\:=-\frac{E_a}{8.314\times \:10^{-3}\ kJ/mol.K}\times \:\left(\frac{1}{649.15\ K}-\frac{1}{553.15\ K}\right)$

$E_a=-\frac{10^{-3}\times \:8.314\ln \left(\frac{10^8\times \:4.8}{10^8\times \:2.3}\right)}{-\frac{96}{359077.3225}}\ kJ/mol$

$E_a=-\frac{\frac{8.314\ln \left(\frac{4.8}{2.3}\right)}{1000}}{-\frac{96}{359077.3225}}\ kJ/mol$

$E_a=22.87\ kJ/mol$

The activation energy for this reaction = 23 kJ/mol.

6 0

2 years ago