FALL CHEM 1711& 1811 FINAL EXAM scale factor; choices may be used once or not at

Question

FALL CHEM 1711& 1811 FINAL EXAM scale factor; choices may be used once or not at all. 12. Match each quantity with one most appropriate number of atoms in one mole f. energy in joules for one photon of visible light y number of picometers in one nanometer à. mass, in grams, of one iron atom A. 10 B. 103 C. 10 D. 101 E. 10 F. 10 G. 10 Part II. PRINT the letter for the one best answer (120 points) any compound of lead and oxygen is 9066 % lead by mass, what is its empirical formula? Pb,O, -1, A bi B. Pb,O D. PbO A. C. Po 2. 5.00cm-_ micrometers B. 5.00 x 10 D. 500 × 10" A. 5.00 x 10 C. 5.00x10" -3. What mass of oxygen is in 62.0 g of cuso,5H,0 ? B. 35.8 8 D. 19.98 A. 25.9 g C. 1798 For an ideal gas, which one of these quantities is inversely proportional to pressure? A. density Which one of these is a unit of volume? A. L3 -A B. volume C. mole amount D. temperature 5. B. ft C. mol D. g/cm 6. In the reaction 2 MnQ. (aq)+5Gof (aq) + 16 H. (aq) 2 Mn® (aq) + 10 co, (g) + 8 H00, the oxidizing agent (oxidizer) is A. H B. C. CO A fictional element (X) consists of only three naturally occurring isotopes: isotope 7. abundanoe 35.39 % 35.25 % 29.36 % measured mass 150.9377 amu 151.9791 amu 156.9332 amu What is the average atomic mass of element X? A. 151.5 amu C. 154.9 amu B. 153.9 amu D. 153.1 amu

Explanation / Answer

1. a) 1 mole of any substance contains 6.023*1023 atoms. Hence most appropriate is 1024.( C is correct)

b) wave length of visible light = 490 nm. E=hc/wave length , h is plack's constant= 6.627*10-34 J.s c= speed of light= 3*108 m/s, E= 6.627*10-34*3*108/490*10-9 m=4*10-19 J ( F is correct

c) 1 pico =10-12 m and 1 nano= 10-9 m

no of pico meter in 1 nano meter= 10-9/10-12 = 1000 ( E is correct)

d) atomic weight of iron =56 g/mole, 1 mole of any substance contains 6.023*1023 atoms

moles in 1 atom= 1/(6.023*1023)

mass of 1 atom of iron = moles* atomic weight = 56/(6.023*1023) =9.3*10-23 gm ( G is correct)

2. Basis : 100 gm of binary compound

Pb= 96 and O= 100-96=4 gm

atomic weights : Pb=207 and O= 16

moles = mass/atomic weight

moles :Pb= 96/207= 0.47 and O= 4/16=0.25

mole ratio of P:b:O= 0.47:0.25

dividing by the lowest no which is 0.25

the ratio becomes Pb: O2= 0.47/0.25:0.25/0.25= 2:1

the compound is Pb2O.

3.1cm= 10-2 m , 1m= 106 micrometer

1cm =106*10-2 micrometer=104 micrometers

5 cm= 5*104 micrometers ( C is correct)

4.atomic weights : Cu= 63.5, S= 32, O= 16 and H= 1

molar mass of CuSO4.5H2O= 63.5+32+64+5*18= 249.5 g/mole

there are 4 oxygen atoms whose contribution to molar mass is 4*16=64+5*16 =144 gm of oxygen

249.5 gm of CuSO4.5H2O contains 144 gm of O

62 gm of CuSO4.5H2O contain 62*144/249.5 =35.78 gm of Oxygen. ( B is correct)

5. From Boyles law, the pressure of gas is inversly proportional to volume at constant temperature for a given mass of gas. Hence volume is correct answer(B)

6. oxudation stateof Mnin MnO4- = x-8= -1 and x= 7 Manganese +7 each in the MnO4- is reduced, having its oxidation number reduced down to +2 each in Mn+2. So this is the oxidizing agent since an oxidizing agent gets rediced.

7.

D is correct.

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