Question

In: Chemistry

Question 1 The reaction: X (g) + e- → X- (g) + energy represents the first...

Question 1

The reaction: X (g) + e- → X- (g) + energy

represents the first ionization energy

represents the metallic character

represents electron affinity

represents the second ionization energy

represents electronegativity

Question 2

4.000 moles of Cl2 (g) react with excess P4 (s) to produce 0.5741 mol of phosphorus trichloride. What is the percent yield for this reaction?

P4 (s) + 6 Cl2 (g) → 4 PCl3 (l)

19.14 %

43.06 %

11.48 %

21.53 %

28.71 %

Question 3

If 239 grams of O2 are produced in the reaction below, how many grams of HgO have reacted?

2 HgO (s) → 2 Hg (l) + O2 (g)

3240 grams

1620 grams

0.0405 grams

853 grams

203 grams

Question 4

8.84 g of a compound is analyzed and found to contain 3.53 g of C, 0.295 g of H, 4.08 g of N, and 0.935 g of O. Which of the following could be the molecular formula for this compound?

C8H8N8O4

C10H10N10O2

C5H10N3O3

C4H8N6O2

C2H4N6O

Expert Solution

Question 1

The reaction: X (g) + e- → X- (g) + energy

represents electron affinity >>>>>>answer

Question 2

P4 (s) + 6 Cl2 (g) → 4 PCl3 (l)

6 moles of Cl2 react with P4 to gives 4 moles of PCl3

4 moles of Cl2 react with P4 to gives = 4*4/6 = 2.67 moles of PCl3

Theoritical yiled = 2.67moles of PCl3

percentage yield = actual yield*100/theoritical yield

= 0.5741*100/2.67 = 21.53% >>>>answer

question3

2 HgO (s) → 2 Hg (l) + O2 (g)

1 mole of O2 produced from 2 moles of HgO

32g of O2 produced from 2*216.58 g of HgO

239 g of O2 produced from = 2*216.58*239/32 = 3240g of HgO >>>answer

question 4

Element wt At.wt no of moles simple ratio

C 3.53 g 12 3.53/12 = 0.294 0.294/0.0584 = 5

H 0.295g 1 0.295/1 = 0.295 0.295/0.0584 = 5

N 4.08g 14 4.08/14 = 0.291 0.291/0.0584 = 5

O 0.935 16 0.935/16 = 0.0584 0.0584/0.0584 = 1

empirical formula = C5H5N5O

C10H10N10O2 >>>>>answer

queen_honey_blossom answered 3 years ago

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