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olya-2409 [2.1K]
3 years ago
7

Would you predict that the C−O bond lengths in the formate ion would be longer or shorter relative to those in CO2?

Chemistry
1 answer:
ivolga24 [154]3 years ago
3 0
The Lewis structures for formate ion and carbon dioxide are shown in the attached pictures. As you can see, the formate ion consists of one C=O bond and one C-O bond, while carbon dioxide has two C=O bonds. Double bonds are weaker, hence, longer compared to single bonds. Therefore, the C-O bonds in the formate ion is shorter than those in carbon dioxide.

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Which is the correct molar mass for the compound h2SO4
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Explanation: just add the atomic masses on the periodic table together for each atom.

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A scientist prepared an aqueous solution of a 0.45 M weak acid. The pH of the solution was 2.72. What is the percentage ionizati
aleksandr82 [10.1K]

Answer:

0.42%

Explanation:

<em>∵ pH = - log[H⁺].</em>

2.72 = - log[H⁺]

∴ [H⁺] = 1.905 x 10⁻³.

<em>∵ [H⁺] = √Ka.C</em>

∴ [H⁺]² = Ka.C

∴ ka = [H⁺]²/C = (1.905 x 10⁻³)²/(0.45) = 8.068 x 10⁻⁶.

<em>∵ Ka = α²C.</em>

Where, α is the degree of dissociation.

<em>∴ α = √(Ka/C) </em>= √(8.065 x 10⁻⁶/0.45) = <em>4.234 x 10⁻³.</em>

<em>∴ percentage ionization of the acid = α x 100</em> = (4.233 x 10⁻³)(100) = <em>0.4233% ≅ 0.42%.</em>

4 0
3 years ago
Read 2 more answers
Non metal are usually poor conductor of hart and electricity . The are non -lustourous,non-sonours,non-malleable and are coloure
vekshin1
Check the solution in the attachment.

7 0
3 years ago
Carbon-14 is a radioactive isotope that decays according to first-order kinetics in a process that has a half-life of 5730 years
Sliva [168]

Answer : The time passed in years is 2.83\times 10^3\text{ years}

Explanation :

Half-life = 5730 years

First we have to calculate the rate constant, we use the formula :

k=\frac{0.693}{t_{1/2}}

k=\frac{0.693}{5730\text{ years}}

k=1.21\times 10^{-4}\text{ years}^{-1}

Now we have to calculate the time passed.

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 1.21\times 10^{-4}\text{ years}^{-1}

t = time passed by the sample  = ?

a = let initial amount of the reactant  = X g

a - x = amount left after decay process = 71\% \times (x)=\frac{71}{100}\times (X)=0.71Xg

Now put all the given values in above equation, we get

t=\frac{2.303}{1.21\times 10^{-4}}\log\frac{X}{0.71X}

t=2831.00\text{ years}=2.83\times 10^3\text{ years}

Therefore, the time passed in years is 2.83\times 10^3\text{ years}

8 0
3 years ago
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