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olga55 [171]
3 years ago
5

If 15.00 mL of 0.0100 M Ca(IO3)2 solution are mixed with 0.500 g KI, what is the theoretical yield (in grams) of I2?

Chemistry
1 answer:
n200080 [17]3 years ago
8 0

The theoretical yield of I2 in the reaction would be 0.23 g

<h3>Theoretical yield</h3>

This refers to the stoichiometric yield of a reaction.

From the equation of the reaction:

Ca(IO3)2 + 10 KI + 12 HCl → 6 I2 + CaCl2 + 10 KCl + 6 H2O

The mole ratio of Ca(IO3)2 and I2 is 1: 6

Mole of 15.00 mL, 0.0100 M Ca(IO3)2 = 15/1000 x 0.0100

                                                              = 0.00015 mole

Equivalent mole of I2 = 0.00015 x 6

                                      = 0.009 mole

mass of 0.0009 I2 = 0.0009 x 253.809

                                = 0.23 g

More on stoichiometric calculations can be found here: brainly.com/question/6907332

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2) what is the mass of 6.02 x 1023 atoms of arsenic?
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1 mole of arsenic → 75g ------------ 6,02*10²³ molecules
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Rank these acids according to their expected pKa values.
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Answer:

According to their expected pKa values, the order of those acids should be:

1- Cl2CHCOOH is the strongest acid and the lowest pKa.

2- ClCH2COOH is a strong acid, but no more than the first. Medium pKa value.

3- ClCH2CH2COOH is a strong acid, but no more than the two previous acids. High pKa value.

4- CH3CH2COOH  is the weakest acid, so the highest pKa value.

Explanation:

The pKa values are the negative logarithm of dissociation constant. It represents the relative strengths of the acids. Stronger acids show smaller pKa values and weak acids present larger pKa value. The stronger the acid, the weaker it's the conjugate base. The larger the pKa of the conjugate base, the stronger the acid. The strength of an acid is inversely related to the strength of its conjugate.

Conjugate bases are the substance that has one less proton than the parent acid. The conjugate base of the acid presented in the problem are:

ClCH2COOH -> ClCH2COO-  + H+

ClCH2CH2COOH -> ClCH2CH2COO- + H+

CH3CH2COOH -> CH3CH2COO- + H+

Cl2CHCOOH -> Cl2CHCOO - + H+

Cl2CHCOOH. The negative charge presented on its conjugate base is by resonance and inductive effect. This is the strongest acid.

ClCH2COOH. A negative charge is stabilized by resonance and electron-withdrawing but only one atom is present. So this acid is less strong than the first one.

ClCH2CH2COOH. The negative charge is stabilized by resonance and electron-withdrawing atom but the effect is less compared to the two acids showed previously.

CH3CH2COOH. The negative charge is stabilized by resonance and destabilized due to CH3 group. This is the weakest acid among the problem.

Stronger acids have smaller pKa values and weak acids have larger pKa values. Due to the information present in this problem, Cl2CHCOOH is the strongest acid and the lowest pKa. CH3CH2COOH is the weakest acid, so the highest pKa value.

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3- ClCH2CH2COOH is a strong acid, but no more than the two previous acids. High pKa value.

4- CH3CH2COOH  is the weakest acid, so the highest pKa value.

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