Answer:
The amount of water converted from liquid to gas with 6,768 joules is approximately 3.035 g
Explanation:
The amount of heat required to convert a given amount of liquid to gas at its boiling point is known as the latent heat of evaporation of the liquid
The latent heat of evaporation of water, Δ
≈ 2,230 J/g
The relationship between the heat supplied, 'Q', and the amount of water in grams, 'm', evaporated is given as follows
Q = m × Δ
Therefore, the amount of water, 'm', converted from liquid to gas at the boiling point temperature (100°C), when Q = 6,768 Joules, is given as follows;
6,768 J = m × 2,230 J/g
∴ m = 6,768 J /(2,230 J/g) ≈ 3.035 g
The amount of water converted from liquid to gas with 6,768 joules = m ≈ 3.035 g.
Roughly 39 grams, give or take 1 gram
Answer:
Buffalo and Watertown.
Explanation:
This is correct as the Interior lowlands are located ib the US. These are correct as the others are in the US but aren't located anywhere near the Interior Lowlands.
Answer:
Explanation:
Given parameters :
Volume of solution = 100mL
Absorbance of solution = 0.30
Unknown:
Concentration of CuSO₄ in the solution = ?
Solution:
There is relationship between the absorbance and concentration of a solution. They are directly proportional to one another.
A graph of absorbance against concentration gives a value of 0.15M at an absorbance of 0.30.
The concentration is 0.15M
Also, we can use: Beer-Lambert's law;
A = ε mC l
where εm is the molar extinction coefficient
C is the concentration
l is the path length
Since the εm is not given and assuming path length is 1;
Then we solve for the concentration.
