Answer:
I think the right answer is c/ number of atomic orbitals
Assume an original volume of blood of one deciliter (100 ml). if 5 ml of oxygen diffuses into the blood, 100 ml will be its final volume.
A tissue is made up of white blood cells, platelets, red blood cells, and other elements suspended in a liquid. Blood transports waste away and delivers nutrients and oxygen to the tissues. The entire amount of fluid moving through the heart's arteries, capillaries, veins, venules, and chambers is referred to as blood volume. Red blood cells (erythrocytes), white blood cells (leukocytes), platelets, and plasma are the elements that give blood volume.
The amount of water and sodium ingested, expelled by the kidneys into the urine, and lost through the digestive system, lungs, and skin determines blood volume. The amounts of salt and water that are consumed and excreted vary greatly.
To know more about blood volume refer to: brainly.com/question/7313563
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D = m / V
It even gives you the density of gold in the problem. Major hint. Once you know the volume (using V = m / D) then you can calculate the height (thickness) from the equation...
V = L x W x H
Volume = Length x Width x Height
start by converting 200.0 mg into grams
1000 mg = 1 g
200. mg x (1 g / 10^3 mg) = 0.200 g
V = m / D
V = 0.200 g / (19.32 g/cm^3)
V = 0.01035 cm^3
Convert 2.4 ft and 1 ft to cm
2.4 ft x (12 in / 1 ft) x (2.54 cm / 1 in) = 73.15 cm
1 ft = 30.48 cm
Compute the height (thickness)
V = LxWxH
H = V / LW = 0.01035 cm^3 / 73.15 cm / 30.48 cm
H = 4.64 x 10^-6 cm
Convert to nanometers
4.64 x 10^-6 cm x (1 m / 100 cm) x (10^9 nm / 1 m) = 46.4 nm
Knowing the atomic radius of gold, I might have asked my students for the minimum number of gold atoms in this thickness of gold. This would assume that the gold atoms are all in a row. This would give the minimum number of gold atoms.
Atomic radius gold = 174 pm
Diameter = 348 pm
46.4 nm x (1 m / 10^9 nm) x (10^12 pm / 1 m) x (1 Au atom / 248 pm) = 133 atoms of gold
2NaBr + Ca(OH)2 ➡️ CaBr2 + 2NaOH
