-Are the most used in daily life and basic human needs
(such as oxygen,nitrogen,carbon, etc)
Organism are made up of organic molecules consisting of carbon, hydrogen, oxygen, nitrogen, sulfur, and phosphorus.
The answer is B.
Answer:
0.4515 M
Explanation:
In case of titration , the following formula can be used -
M₁V₁ = M₂V₂
where ,
M₁ = concentration of acid ,
V₁ = volume of acid ,
M₂ = concentration of base,
V₂ = volume of base .
from , the question ,
M₁ = ? M
V₁ = 10.0 mL
M₂ = 0.100 M
V₂ = 45.15 mL
Using the above formula , the molarity of acid , can be calculated as ,
M₁V₁ = M₂V₂
M₁ * 10.0mL = 0.100 M * 45.15 mL
M₁ = 0.4515 M
Hence, the concentration of the acetic acid = 0.4515 M
Answer:
Compound B.
Explanation:
The freezing point depression is a colligative property. It depends on the number of particles (moles) present in the solution.

where b is the molal concentration

If m is constant (5 g), then

The compound with the greater molar mass has fewer moles and therefore fewer particles to depress the freezing point.
That must be Compound B, because Compound A has the lower freezing point.
Answer:
16.7 g of glucose
Explanation:
Convert grams of H₂O to moles. <em>0.555 mol</em>
We can use the chemical equation to determine the theoretical yield. Based on the equation, for every 6 moles of H₂O, 1 mole of glucose is produced. The ratio of glucose moles to H₂O moles is 1/6.
Multiply the moles of H₂O by the ratio to find moles of glucose. <em>0.0925 mol</em>
Convert moles of glucose to grams. <em>16.7 g</em>