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vovikov84 [41]
2 years ago
11

Calculate the number of grams in 0.278 moles of H3PO4

Chemistry
1 answer:
stellarik [79]2 years ago
7 0
27.2 grams = 0.278 (mols) x 97.994 (g/mol)
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zhuklara [117]

Answer:

K_{sp} of Zn(CH_{3}COO)_{2} is 0.0513

Explanation:

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Zn(CH_{3}COO)_{2}\rightleftharpoons Zn^{2+}+2CH_{3}COO^{-}

Solubility product of Zn(CH_{3}COO)_{2} (K_{sp}) is written as-            K_{sp}=[Zn^{2+}][CH_{3}COO^{-}]^{2}

Where [Zn^{2+}] and [CH_{3}COO^{-}] represents equilibrium concentration (in molarity) of Zn^{2+} and CH_{3}COO^{-} respectively.

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So, solubility of Zn(CH_{3}COO)_{2} = \frac{43.0}{183.48}M = 0.234M

1 mol of Zn(CH_{3}COO)_{2} gives 1 mol of Zn^{2+} and 2 moles of CH_{3}COO^{-} upon dissociation.

so,   [Zn^{2+}] = 0.234 M and [CH_{3}COO^{-}] = (2\times 0.234)M=0.468M

so, K_{sp}=(0.234)\times (0.468)^{2}=0.0513          

8 0
3 years ago
What Group is Zn (zinc) in?<br> O A. 4<br> O B. 2A<br> O C. 12<br> O D. 30
Oxana [17]
I believe it’s (C.12)
3 0
3 years ago
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