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aleksandr82 [10.1K]
3 years ago
9

One gram of an acid, HA of unknown molecular mass is dissolved in water and 22.72 cm3 of 0.60 M sodium hydroxide solution added

to it in a neutralization reaction. Exactly 7.84 cm3 of 0.530 M HCl was required to neutralize sodium hydroxide by back-titration. i) Write the reaction equations involved in the titration? ii) What is the molecular mass of HA?
Chemistry
1 answer:
FinnZ [79.3K]3 years ago
6 0
Chemistry


















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How many moles of iron(lll) sulfide, Fe2S3, would be produced from the complete reaction of 449 g iron(lll) bromide, FeBr3?
SpyIntel [72]

Answer:

158.004 Grams of Fe2S3

Explanation:

iron(lll) sulfide = Fe2S3 = Molar mass: 207.9 g/mol

iron(lll) bromide = FeBr3 = Molar mass: 295.56 g/mol

Since we don't have the chemical equation we will have to make one I guess.

2 FeBr3 + 3 Na2S -----> 6 NaBr + Fe2S3

449 g iron(lll) bromide = FeBr3

So we have 449/295.56 = moles of FeBr3 = 1.52 Moles

So ratio is 2:1 FeBr3 to Fe2S3

So we need 0.76 moles of Fe2S3

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7 0
3 years ago
+
aleksklad [387]

Answer:

<h3>The answer is 41.05 %</h3>

Explanation:

The percentage error of a certain measurement can be found by using the formula

P(\%) =  \frac{error}{actual \:  \: number}  \times 100\% \\

From the question

actual density = 0.95 g/mL

error = 0.95 - 0.56 = 0.39

So we have

p(\%) =  \frac{0.39}{0.95}  \times 100 \\  = 41.0526315...

We have the final answer as

<h3>41.05 %</h3>

Hope this helps you

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