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kobusy [5.1K]
3 years ago
6

At 25.0∘c, the molar solubility of barium chromate in water is 1.10×10−5 m . calculate the solubility in grams per liter.

Chemistry
1 answer:
Masja [62]3 years ago
6 0
When in water, barium chromate dissociates into ions. The dissociation equation would be as follows:

<span>BaCrO4 = Ba^2+ + CrO4^2+
</span>
For this case, we are to convert only from moles per kg to grams per liter. We do as follows:

<span>1.10×10^−5 mol / kg ( 253.37 g /mol) ( 1 kg/L ) = 2.79x10^-3 kg / L = 2.79 g/L</span>
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B. Better ways to recycle plastics

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How many grams of hydrogen chloride can be produced from 1g of hydrogen and 55g of chlorine? What is the limiting reactant?
vova2212 [387]

Answer:

The limiting reactant is hydrogen, and the grams HCl produced is 36.175 g.

Explanation:

Balanced equation is 2 H + Cl2 = 2 HCl.

First thing, convert grams to moles via using molar mass.

Molar mass for hydrogen is 1.0079 g/mol. 1g x 1 mol / 1.0079 g = 0.99216 mol.

Molar mass for chlorine is 70.906 g/mol. 55g x 1 mol / 70.906 g = 0.7756748 mol.

Next, determine which is the limiting reactant - probably the fastest way to do it is just to take one of the reactants, say it's the limiting one, and calculate how much of the other reactant would be needed if that really was the limiting reactant, and then compare it to the actual moles of reactant available.

If hydrogen was the limiting reactant at 0.992 mol, you'd need .496 mol of Cl2 to complete the reaction.

If chloride was the limiting reactant at 0.776 mol, you'd need 1.55 mol of H to complete the reaction.

Comparing these numbers to the amounts we actually have available, the limiting reactant is hydrogen.

Once you've determined that, just plug in the amounts to the balanced equation to get the number of moles of HCL produced, which in this case, is just 0.992 mol.

Now, reverse the process that you took to get the moles of reactant, and you have the grams of product produced.

0.992 mol x 36.4609 g / 1 mol = 36.175 g.

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3 years ago
Propanoic acid, ch3ch2cooh, has a pka =4.9. draw the structure of the conjugate base of propanoic acid and give the ph above whi
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Conjugate base of Propanoic acid (CH_{3}CH_{2}COOH is propanoate where -COOH group gets converted to -COO^{1-}. The structure of conjugate base of Propanoic acid is shown in the diagram.

The p^{H} above which 90% of the compound will be in this conjugate base form can be determined using Henderson's equation as propanoic acid is weak acid and it can form buffer solution on reaction with strong base.

p^{H}= p^{k_{a}+ log\frac{[Conjugate base]}{[acid]}=4.9+log\frac{90}{10}=5.85

As 90% conjugate base is present, so propanoic acid present 10%.

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<h2><em>can't understand your question..... can u explain me May </em><em>b</em><em>e</em><em> I can help u!</em></h2><h2 /><h2 />
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