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vitfil [10]
3 years ago
14

What's the mass in grams of 0.334 moles of sucrose (C12H22O11)?

Chemistry
1 answer:
Sonbull [250]3 years ago
3 0

Answer:

114.23g

Explanation:

Mass = xg

Mole = 0.334 mol

Molar mass of Sucrose

(C = 12, H = 1, O = 16)

C12H22O11

(12×12)+(1×22)+(16×11)

144 + 22 + 176

342g/mol

Mass = No of mole × Molar mass

Mass = 0.334 × 342

Mass = 114.228

Mass ≈ 114.23g

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Answer:

NO \longrightarrow N_2O_2 \longrightarrow N_2O \longrightarrow N2

Explanation:

The intermediates are the products of all the steps of the reaction pathway, with the exception of the last one. So the intermediates will be:

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The list from reactant to final product:

NO \longrightarrow N_2O_2 \longrightarrow N_2O \longrightarrow N2

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Lithium reacts with bromine (Br2) in a synthesis reaction to produce lithium bromide. Determine the limiting reactant if 25.0 gr
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Answer: Bromine is the limiting reactant

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First of all let's generate a balanced equation for the reaction

2Li + Br2 —> 2LiBr

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Number of mole of Li = 25/7 = 3.6moles

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To know which is the limiting reactant, we have to compare the ratio of the number of mole of experimental Li and Br2 to that of theoretical Li and Br2

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For the theoretical yield:

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From the above, we see clear that Br2 is the limiting reactant because according to the equation( which gives the theoretical yield), for every 2moles of Li, 1mole of Br2 is used up. But this is not so from the experiment conducted as 23moles required 1mole of Br2.

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Answer:

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Down a periodic group, metallicity increases.

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Across a period from left to right electropositivity or metallicity decreases.

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