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Multiple Choice Questions

1. Nitrogen may be combined with oxygen in proportion by mass of 1.75 g nitrogen to 1.00 g oxygen to form the gas nitrous oxide, N2O. The same elements may be combined in the ratio of 0.438g nitrogen to 1.00g oxygen. What is the atomic composition of the second compound?

a. NO(2)
b. NO
c. N(2)O
d. NO(3)

2. In an electrolysis experiment, a student plated silver onto an electrode, maintaining a current of 0.800 A for 46.0 min. What mass of silver was deposited?

a. 1.23 g
b. 2.47 g
c. 3.62 g
d. 3.10 g

3. The density of particles n is a suspension is found to go as n=n0e-lamdah, where lamda is a constant and h is the height. What is the value of lamda if the density falls to no/2 When h = 10 microm?

a. 69,300 /m
b. 34,650 /m
c. 23,000 /m
d. 11,000 /m

4. Silicon has a density of 2.23 g/cm3 and an atomic mass of 28.09 g/mol. Calculate the number of atoms in a cubic centimeter of silicon?

a. 6.02 x 10^22
b. 2.02 x 10^22
c. 4.78 x 10^22
d. 6.02 x 10^23

5. X rays of lamda= 0.709 X 10^-10 m are diffracted in second order ( n=2) from a silver bromide crystal. If the scattering angle is 2q = 28.50 degree , What is the spacing between the atomic planes?

a. .288 nm
b. .388 nm
c. .62 nm
d. 1.02 nm

6. X rays are Bragg-diffracted from a cesium chloride crystal with interplanar spacing of 4.11 X 10^-10m In first order, the scattering 2q is 7.59 degree, What is the wavelength of theX rays?

a. .0544 nm
b. 0.544 nm
c. 5.44 nm
d. 54.4 nm

7. Calculate the angular deflection of an electron beam traveling at 1.0 X 10^7 m/s through a region of magnetic field B=1.0 X 10^-3 T that is 1.5 cm long

a. 0.52 rad
b. 0.26 rad
c. .13 rad
d. .33 rad

Solution Preview

1. Nitrogen may be combined with oxygen in proportion by mass of 1.75 g nitrogen to 1.00 g oxygen to form the gas nitrous oxide, N2O. The same elements may be combined in the ratio of 0.438g nitrogen to 1.00g oxygen. What is the atomic composition of the second compound?

a. NO(2)
b. NO
c. N(2)O
d. NO(3)

Answer: a. NO(2)
Molecular weight of Nitrogen= 28 g/mol
Molecular weight of Oxygen= 32 g/mol

In 0.438 g of Nitrogen there are 0.0156 moles =0.438/28
In 1 g of Oxygen there are 0.0313 moles =1/32

Therfore ratio of Nitrogen to Oxygen = 0.0156 : 0.0313
or 1 : 2

2. In an electrolysis experiment, a student plated silver onto an electrode, maintaining a current of 0.800 A for 46.0 min. What mass of silver was deposited?

a. 1.23 g
b. 2.47 g
c. 3.62 g
d. 3.10 g

Answer: b. 2.47 g

Charge (in Coulombs)= Current (in Amperes ) x time ( in seconds)
Therefore charge= 2208 Coulomb =0.8*46*60

1 mole of electron ...

Solution Summary

Answers to Multiple Choice Questions

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