All categories

3 years ago

The average strength of a hydrogen bond is approximately what percent of the average strength of a covalent bond?

1 answer:

8 0

<span>Hydrogen bonds are approximately 5% of the bond strength of covalent bonds, for example (C-C or C-H bonds).
Hydrogen bonds strength in water is approximately 20 kJ/mol, strenght of carbon-carbon bond is approximately 350 kJ/mol and strengh of carbon-hydrogen bond is approximately 340 kJ/mol.
20 kJ/350 kJ = 0,057 = 5,7 %.</span>

You might be interested in

A solution is made by mixing 55.g of thiophene C4H4S and 65.g of acetyl bromide CH3COBr.

EleoNora [17]

Answer:

Mole fraction of C₄H₄S = 0.55

Explanation:

Mole fraction is moles of solute / Total moles

Total moles are the sum of moles of solute + moles of solvent.

Let's find out the moles of our solute and our solvent.

Mass of solute: 55g

Mass of solvent: 65g

Mol = Mass / molar mass

55 g / 84.06 g/mol = 0.654 moles of C₄H₄S

65 g /123 g/mol = 0.529 moles of C₂H₃BrO

Total moles = 0.654 + 0.529 = 1.183 moles

Mole fraction of thiophene = Moles of tiophene / Total moles

0.654 / 1.183 = 0.55

4 0

3 years ago

A sample of nitrogen occupies 10.0 liters at 25°C what would be the new volume at 20°C? 7.9 L Ob 9.8 L 10.2 L 10.6 L

PtichkaEL [24]

Answer:

9.4 liter

Explanation:

1) Data:

V₁ = 10.0 L

T₁ = 25°C = 25 + 273.15 K = 298.15 K

P₁ = 98.7 Kpa

T₂ = 20°C = 20 + 273.15 K = 293.15 K

P₂ = 102.7 KPa

V₂ = ?

2) Formula:

Used combined law of gases:

PV / T = constant

P₁V₁ / T₁ = P₂V₂ / T₂

3) Solution:

Solve the equation for V₂:

V₂ = P₁V₁ T₂ / (P₂ T₁)

Substitute and compuite:

V₂ = P₁V₁ T₂ / (P₂ T₁)

V₂ = 98.7 KPa × 10.0 L × 293.15 K / (102.7 KPa × 298.15 K)

V₂ = 9.4 liter ← answer

You can learn more about gas law problems reading this other answer on

Explanation:

7 0

3 years ago

Convert 9.90km to mm

sukhopar [10]

Answer:

9900000

add 20 LeTtErS or mOrE

3 0

3 years ago

Consider the second-order reaction:

kirza4 [7]

Answer:

Initial concentration of HI is 5 mol/L.

The concentration of HI after $4.53\times 10^{10} s$ is 0.00345 mol/L.

Explanation:

$2HI(g)\rightarrow H_2(g)+I_2(g)
$

Rate Law: $k[HI]^2
$

Rate constant of the reaction = k = $6.4\times 10^{-9} L/mol s$

Order of the reaction = 2

Initial rate of reaction = $R=1.6\times 10^{-7} Mol/L s$

Initial concentration of HI = $[A_o]$

$1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2$

$[A_o]=5 mol/L$

Final concentration of HI after t = [A]

t = $4.53\times 10^{10} s$

Integrated rate law for second order kinetics is given by:

$\frac{1}{[A]}=kt+\frac{1}{[A_o]}$

$\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}$

$[A]=0.00345 mol/L$

The concentration of HI after $4.53\times 10^{10} s$ is 0.00345 mol/L.

5 0

4 years ago

Find the normality of 0.321 g sodium carbonate in a 250 mL solution.

Radda [10]

Answer:

the normality of the given solution is 0.0755 N

Explanation:

The computation of the normality of the given solution is shown below:

Here we have to realize the two sodiums ions per carbonate ion i.e.

N = 0.321g Na_2CO_3 × (1mol ÷ 105.99g)×(2eq ÷ 1mol)

= 0.1886eq ÷ 0.2500L

= 0.0755 N

Hence, the normality of the given solution is 0.0755 N

5 0

3 years ago