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Karo-lina-s [1.5K]
3 years ago
11

What is the formula for a hydrate which consists of 90.7 % SrC2O4 and 9.30 % H20

Chemistry
2 answers:
jekas [21]3 years ago
5 0
Carbon dioxide could possibly be the answer unfortunately idk wat im talking about
padilas [110]3 years ago
5 0
I’m assuming that this is based on percent mass.

Strontium Oxalate has a formula weight of 175.64 g/mol, and water has a f.w. of 18.02 g/mol. Because SrC2O4 makes up nearly 91% of the weight by mass, we can assume that the number of H2O molecules in this formula are very low.

You can test SrC2O4•H2O (f.w. 193.66) and the role that the strontium oxalate plays by mass by dividing the weight of the strontium oxalate by the weight of the strontium oxalate monohydrate. In this case:

175.64/193.66 = 0.9069, or 90.69% by weight.

Because the weight calculated here matches the one requested in the problem, SrC2O4•H2O, or strontium oxalate monohydrate, is the formula for the hydrate.
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Answer:

[C₆H₅COO⁻][H₃O⁺]/[C₆H₅COOH] = Ka

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The reaction of dissociation of the benzoic acid in water is given by the following equation:

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<em>Where the dissociation constant of the acid (Ka) is equal to the ratio of the concentration of the dissociated forms of the acid, [A⁻][H⁺], and the concentration of the acid, [HA].     </em>

So, starting from the equations (2) and (3), the constant equation for the dissociation reaction of benzoic acid in water, of the equation (1), is:

K_{a} = \frac{[C_{6}H_{5}COO^{-}][H_{3}O^{+}]}{[C_{6}H_{5}COOH]}

I hope it helps you!          

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