describe how 250 cm³ of 0.2 mol/dm³ H2SO4 could be prepared from 150 cm³ of 1.0mol/dm³ stock solution of the acid
1 answer:
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Answer: The correct answer is: " 0.54 g " .
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Explanation:
Note that "hydrogen gas" is:
H₂ (g) ; that is: a "diatomic element" (diatomic gas) ;
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The molecular weight of "H" is: 1.00794 g ;
(From the Periodic Table of Elements).
So, the molecular weight of: H₂ (g) is:
" 1.00794 g * 2 = 2.01588 g ; {use calculator) ;
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Note the conversion for a gas at STP:
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1 mol of a gas = 22.4 L gas;
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i.e. " 1 mol / 22.4 L " ;
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So: " 5.7 L H₂ (g)
The "L" ("literes" cancel out to "1" ; since "L/L = 1 ;
The "mol" (moles) cancel out to "1" ; since "mol/mol = 1 ;
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and we are left with:
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[5.7 * 2.104588 g ] / 22.4 = ? g ;
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→ [ 11.9961516 g ] / 22.4 =
0.53554248214 g ;l
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We round this value to: " 0.54 g " ;
→ since "5.7 L " has 2 (two) significant figures;
22.4 is an exact number conversion;
and "5.7 L" has fewer significant figures than:
" 2.104588 " ; or: " 1.00794 " .
→ as such: We round to "2 (two) significant figures."
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Hope this is helpful. Wishing you the best in your academic endeavors!
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Answer:
10 is 2.89g/ml
11 in attachment
Explanation:
10. D=M/V
M=250G-120G=130G
V=75ML-30ML=45ML
D=130/45=2.89g/ml
