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ozzi
3 years ago
11

Formic acid, from the Latin formica, is the acid present in ants sting. What is the

Chemistry
1 answer:
leonid [27]3 years ago
3 0

For the reaction of deprotonation of formic acid, the concentration of HCOO⁻ at equilibrium is 0.0151 M if the initial concentration of formic acid is 1.35 M.  

The reaction of deprotonation of formic acid is the following:

CHOOH + H₂O ⇄ HCOO⁻ + H₃O⁺

At the equilibrium we have:

CHOOH + H₂O ⇄ HCOO⁻ + H₃O⁺    (1)

1.35 - x                         x           x

The acid <em>equilibrium </em>constant for this reaction is:

K_{a} = \frac{[HCOO^{-}][H_{3}O^{+}]}{[CHOOH]} = 1.7 \cdot 10^{-4}  (2)

Entering the values of [CHOOH] = 1.35-x, [HCOO⁻] = [H₃O⁺] = x, into equation (2) we have:

1.7 \cdot 10^{-4} = \frac{[HCOO^{-}][H_{3}O^{+}]}{[CHOOH]} = \frac{x^{2}}{(1.35 - x)}  

1.7 \cdot 10^{-4}(1.35 - x) - x^{2} = 0

After solving the above <em>quadratic equation</em> and taking the positive value for x (<u>concentrations cannot be negative</u>), we have:

x = [HCOO^{-}] = [H_{3}O^{+}] = 0.0151 M

Therefore, the concentration of HCOO⁻ at equilibrium is 0.0151 M.

Learn more about the equilibrium constant here:

  • brainly.com/question/7145687?referrer=searchResults
  • brainly.com/question/9173805?referrer=searchResults

I hope it helps you!

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3 0
2 years ago
The thallium Subscript 81 Superscript 208 Baseline Tl nucleus is radioactive, with a half-life of 3.053 min. At a given instant,
Genrish500 [490]

Answer: (E) 300 bq

Explanation:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

Radioactive decay process is a type of process in which a less stable nuclei decomposes to a stable nuclei by releasing some radiations or particles like alpha, beta particles or gamma-radiations. The radioactive decay follows first order kinetics.

Half life is represented by t_{\frac{1}{2}

Half life of Thallium-208 = 3.053 min

Thus after 9 minutes , three half lives will be passed, after ist half life, the activity would be reduced to half of original i.e. \frac{2400}{2}=1200, after second  half life, the activity would be reduced to half of 1200 i.e. \frac{1200}{2}=600,  and after third half life, the activity would be reduced to half of 600 i.e. \frac{600}{2}=300,

Thus the activity 9 minutes later is 300 bq.

7 0
3 years ago
Why cant magnesium ions be detected in a flam test?
Dmitry [639]
When magnesium ion doesn't give any characteristics colour with the flame test as electronic transisitons do not give out visible light.
6 0
3 years ago
Which of the followings is true about G0'? A. G0' can be determined using Keq' B. G0' indicates if a reaction can occur under no
nekit [7.7K]

Answer:

A and D are true , while B and F statements are false.

Explanation:

A) True.  Since the standard gibbs free energy is

ΔG = ΔG⁰ + RT*ln Q

where Q= [P1]ᵃ.../([R1]ᵇ...) , representing the ratio of the product of concentration of chemical reaction products P and the product of concentration of chemical reaction reactants R

when the system reaches equilibrium ΔG=0 and Q=Keq

0 = ΔG⁰ + RT*ln Q → ΔG⁰ = (-RT*ln Keq)

therefore the first equation also can be expressed as

ΔG = RT*ln (Q/Keq)

thus the standard gibbs free energy can be determined using Keq

B) False. ΔG⁰ represents the change of free energy under standard conditions . Nevertheless , it will give us a clue about the ΔG around the standard conditions .For example if ΔG⁰>>0 then is likely that ΔG>0 ( from the first equation) if the temperature or concentration changes are not very distant from the standard conditions

C) False. From the equation presented

ΔG⁰ = (-RT*ln Keq)

ΔG⁰>0 if Keq<1 and ΔG⁰<0 if Keq>1

for example, for a reversible reaction  ΔG⁰ will be <0 for forward or reverse reaction and the ΔG⁰ will be >0 for the other one ( reverse or forward reaction)

D) True. Standard conditions refer to

T= 298 K

pH= 7

P= 1 atm

C= 1 M for all reactants

Water = 55.6 M

5 0
3 years ago
The sediment deposited by debris-laden melt water is called _______.
AlekseyPX
Answer: ( Outwash )
hope this helps!
8 0
3 years ago
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