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valina [46]
3 years ago
14

What is bond dissociation?

Chemistry
1 answer:
ElenaW [278]3 years ago
5 0
Bond dissociation is the measure/amount of strength in a chemical bond. The reaction and the products and what the strength is between them. 
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I think it’s c I could be wrong
3 0
3 years ago
What type of bond occurs when two atoms share one pair of valence electrons?
mestny [16]

A covalent bond occurs when two atoms share one pair of valence electrons

Brainliest answer please? :)

5 0
3 years ago
Read 2 more answers
Please help: A 0.200 M NaOH solution was used to titrate a 18.25 mL HF
algol [13]

The molar concentration of the original HF  solution : 0.342 M

Further explanation

Given

31.2 ml of 0.200 M NaOH

18.2 ml of HF

Required

The molar concentration of HF

Solution

Titration formula

M₁V₁n₁=M₂V₂n₂

n=acid/base valence (amount of H⁺/OH⁻, for NaOH and HF n =1)

Titrant = NaOH(1)

Titrate = HF(2)

Input the value :

\tt 0.2\times 31.2\times 1=M_2\times 18.25\times 1\\\\M_2=0.342

7 0
3 years ago
Acetylene, C2H2, can be converted to ethane, C2H6, by a process known as hydrogenation. The reaction is C2H2(g) + 2H2(g) ⇌ C2H6(
sukhopar [10]

Answer:

-255.4 kJ

Explanation:

The free energy of a reversible reaction can be calculated by:

ΔG = (ΔG° + RTlnQ)*n

Where R is the gas constant (8.314x10⁻³ kJ/mol.K), T is the temperature in K, n is the number of moles of the products (n =1), and Q is the reaction quotient, which is calculated based on the multiplication of partial pressures by the partial pressure of the products elevated by their coefficient divide by the multiplication of the partial pressure of the reactants elevated by their coefficients.

C₂H₂(g) + 2H₂(g) ⇄ C₂H₆(g)

Q = pC₂H₆/[pC₂H₂ * (pH₂)²]

Q = 0.261/[8.58*(3.06)²]

Q = 3.2487x10⁻³

ΔG = -241.2 + 8.314x10⁻³x298*ln(3.2487x10⁻³)

ΔG = -255.4 kJ

4 0
4 years ago
A 0.020 M solution of niacin has a pH of 3.26. (a) What percentage of the acid is ionized in this solution
mariarad [96]

Answer: 2.75%

Explanation:

pH=-log [H+]

3.26 = -log [H+]

[H+] = 5.495\times 10^{-4} M

HA\rightleftharpoons H^++A^-

initial      0.020     0           0

eqm        0.020 -x    x      x

K_a=\frac{[H+][A-]}{[HA]}

K_a=\frac{[x][x]}{[0.020-x]}

x=5.495\times 10^{-4}

K_a=\frac{[5.495\times 10^{-4}]^2}{[0.020-5.495\times 10^{-4}]}

K_a =1.553\times 10^{-5}

percent dissociation = \frac{[H^+_eqm]}{[Acid_{initial}]}\times 100

percent dissociation=\frac{5.495\times 10^{-4}}{0.020}\times 100

Thus percent dissociation= 2.75 %

5 0
3 years ago
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