13# A 29.3 g of a new element was isolated from 660 kg of the ore molybdenite. T

Question

13# A 29.3 g of a new element was isolated from 660 kg of the ore molybdenite. The percent by mass of this element in the ore molybdenite was:

A) 44%

B)6.6%

C)29.3%

D) 0.0044%

E)19.3%

14# SO2 reacts with H2S as follows:

2H2S + SO2----3S + 2H2O

When 7.50g of H2S reacts with 12.75 g of SO2, which statement applies?

A) 6.38g of sulfur are formed

B) 10.6g of sulfur are formed

C)0.0216 moles of H2S remain

D) 1.13g of H2S remain

E) SO2 is the limiting reagent

15# Roundup,an herbicide manufactured by Monsanto, has the formula C3H8NO5P. How many moles are there in a 295.1-g sample of roundup?

18# You mix 55ml of 1.00 M silver nitrate with 25Ml of 0.84M sodium choride. What mass of silver choride should you form ?

Explanation / Answer

13# A 29.3 g of a new element was isolated from 660 kg of the ore molybdenite. The percent by mass of this element in the ore molybdenite was:

m = 29.3 g

m = 660 kg = 660000 g

% = mass of element / total mass * 100 = 29.3/660000 * 100 = 0.0044 %

then choose

D)

14# SO2 reacts with H2S as follows:

2H2S + SO2----3S + 2H2O

When 7.50g of H2S reacts with 12.75 g of SO2, which statement applies?

mol o H2S = masS/MW = 7.5/34.08 = 0.220 mol

mol of SO2 = mass/MW = 12.75/64.066 = 0.199 mol

ratio is 2:1 so H2S is in limint

0.220 mol of H2S form 3/2*0.220 = 0.33 mol of S

mas S= 0.33*32 = 10.56

answer is B

15)

C3H8NO5P = 169.0731 g/mol

mass = 295.1 g

mol = mass/MW = 169.0731/295.1 = 0.5729 mol

18)

V = 55 ml M = 1 AGNO3

V = 25 ml M = 0.74 NaCl

expec AgCl

mmol of Ag = MV = 55*1 = 55 mol of Ag+

mmol of Cl = MV = 25*0.74 = 18.5 mmol of Cl-

expect 18.5 mmol of AgCl

mass = mol*¨MW = (18.5*10^-3)(143.32 ) = 2.651 g

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