Answer:
Passive Transport
Explanation:
The three examples of passive transport are
Diffuison
Osmosis
facilated diffuison
So the answer can be A or 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 
The anion<span> is also </span>larger than<span> the </span>atom<span> because of </span>electron-electron repulsion<span>. As more </span>electrons are<span> added to the </span>outer shell<span>, and even to </span>higher<span> principle energy levels, the </span>repulsion<span> bewteen the negatively charged particles grows, pushing the </span>shells<span> farther from the nucleus.</span>
H2SO4 (1) H20 (g) + SO3 (g)