Question

Hydrazine, N2H4, once used as a rocket propellant, reacts with oxygen in the following equation. N2H4(g) +O 2(g) →N2(g) + 2 H2O(g) The reaction, at 50% yield, produces 4.0 moles of N2. What was the mass of hydrazine used? Assume that O2 is in excess. A) 64 g B) 0.063 g C) 260 g D) 32.0 g 42)

43) Which of the following is the lowest temperature? A) 313 K B) 37°C C) 54°F D) All of these temperatures are all equal. 43)

44) A compound responsible for the odor of garlic has a molecular weight of 146 g/mol. A 0.650 g sample of the compound contains 0.321 g of carbon, 0.044 g of hydrogen, and 0.285 g of sulfur. What is the molecular formula of the compound? A) C3H5S B) C3H15S3 C) C6H10S2 D) CH5S 44)

45) When 12.00 g of calcium metal is reacted with water, 12.00 g of calcium hydroxide is produced. Using the following balanced equation, calculate the percent yield for the reaction? Ca(s) + 2 H2O(l) →Ca(OH)2(aq) + H2(g) A) 30.00 % B) 100.0% C) 27.02% D) 54.05% 45)

46) All of the following are fundamental SI units except the A) Kelvin. B) mole. C) year. D) kilogram. 46)

47) Convert 1 !m to meters. A) 1 × 10 -9 m B) 1 × 10 -6 m C) 1 × 10 -3 m D) 1 × 10 6 m 47)

48) Convert 4.300 × 10 -3 to ordinary notation. A) 430.0 B) 0.00 4300 C) 4300 D) 0.000 4300 48)

49) Which of the following is the greatest mass? A) 1.000 × 10 -3 cg B) 1.000 × 10 -3 kg C) 10,000 !g D) 1.000 × 10 -7 Mg 49)

50) Which of the following underlined items is not an extensive property? A) the diameter of a gold nugget B) the color of a cobalt compound C) the volume of a glucose solution D) the mass of a diamond

Answer 1

42)

N2H4(g) + O 2(g) →N2(g) + 2 H2O(g) The reaction, at 50% yield, produces 4.0 moles of N2.

thus theoretical yield of N2 = 4.0 * 100 / 50 = 8.0 mole.

mole of hydrazine required = 8.0 mole.

mass of hydrazine used = 8.0 mole * 32.0 g / mole = 256 g

C) 260 g (answer)

43) Which of the following is the lowest temperature?

A) 313 K (answer)

B) 37°C = 273 + 37 = 310 K

C) 54°F

C = 5 * (54 - 32) / 9 = 12.2 oC = 12.2 + 273 = 285.2 K

44)

C : H : S

0.321 / 12 : 0.044 / 1 : 0.285 / 32

or

0.02675 : 0.044 : 0.00891

or

3 : 5 : 1

empirical formula is C3H5S

[12 * 3 + 5 * 1+ 32]n = 146

n = 2

  C) C6H10S2 (answer)

45)

Ca(s) + 2 H2O(l) →Ca(OH)2(aq) + H2(g)

mole of Ca = 12.00 / 40 = 0.30 mole.

theoretical yield of Ca(OH)2 = 0.30 mole = 0.30 mole * 74.093 g/mole = 22.2 g

percent yield = 12.00 * 100 / 22.2 = 54.05 %

D) 54.05% (answer)