A covalent bond describes two atoms (most likely nonmetals) that share their valence electrons to satisfy the octet rule. Carbon and oxygen are both nonmetals, and they would share electrons with each other through a bond that is not polar enough to be considered ionic. The answer should be B
<u>Answer:</u> The density of gold in
is 
<u>Explanation:</u>
Density is defined as the ratio of mass of the object and volume of the object. Mathematically,

We are given:
Density of gold = 
Using conversion factors:
1 lb = 453.6 g
1 feet = 12 inches
1 inch = 2.54 cm
Converting given quantity into
, we get:

Hence, the density of gold in
is 
Answer:
d. its effective nuclear charge is lower than the other noble gases.
Explanation:
Xenon belongs to group O on the periodic table. Most of the elements here are unreactive.
Due to the large size of Xenon, the outermost electrons have very low effective nuclear charge. Effective nuclear charge is the effect of the positive charges of the nucleus on the electrons in orbits. This effect decreases outward as atomic shell increases.
Xenon has a very large atomic radius and there is weak a nuclear charge on the outermost electrons. The more electronegative elements would be able to attract some of its outermost electrons easily and form chemical bonds with xenon much more readily.
F = (mass)(acceleration) = ma
m = 55 kg
Vi = 20 m/s
t = 0.5 s
Vf = 0 m/s (since she was put to rest)
a=(Vf-Vi)/t
a=(0-20)/5
a = 40 m/s^2 (decelerating)
F = ma = (55 kg)(40 m/s^2)
F = 2200 N
For this item, we need to assume that air behaves like that of an ideal gas. Ideal gases follow the ideal gas law which can be written as follow,
PV = nRT
where P is the pressure,
V is the volume,
n is the number of mols,
R is the universal gas constant, and
T is temperature
In this item, we are to determine first the number of moles, n. We derive the equation,
n = PV /RT
Substitute the given values,
n = (1 atm)(5 x 10³ L) / (0.0821 L.atm/mol.K)(0 + 273.15)
n = 223.08 mols
From the given molar mass, we calculate for the mass of air.
m = (223.08 mols)(28.98 g/mol) = 6464.9 g
<em>ANSWER: 6464.9 g</em>