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Tems11 [23]
3 years ago
5

Assume the solubility of benzoic acid in ice-cold water is 1.70 g/L and the solubility of benzoic acid in hot water is 68.0 g/L.

Calculate the minimum volume of water (in mL) needed to recrystallize 0.700 g of benzoic acid.
Chemistry
1 answer:
lana [24]3 years ago
4 0

Answer:

The volume of water needed to recrystallize 0.700 g of benzoic acid is 10.29 mL.

Explanation:

For the complete recrystalization,

the amount of hot water should be such that, the benzoic acid is completely soluble in it.

As we are given that the solubility of benzoic acid in hot water is 68.0 g/L. Now we have to determine the volume of water is needed to recrystallize 0.700 g of benzoic acid.

we conclude that,

As, 68.0 grams of benzoic acid soluble in 1 L of water.

So, 0.700 grams of benzoic acid soluble in  of water.

The volume of water needed = 0.01029 L = 10.29 mL

conversion used : (1 L = 1000 mL)

Therefore, the volume of water needed to recrystallize 0.700 g of benzoic acid is 10.29 mL.

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How much faster will ammonia vapor travel across a room then carbon dioxide
FromTheMoon [43]

Answer:

D) 1.61 times faster

Explanation:

vrms_{NH3} = √(3)RTM

R constant= 0.08206

T=constant, so in this problem we dont need a value for it

M=17.031 g/mol

√(3)(0.08206)(17.031)= 2.047

vrms_{CO2} = √(3)RTM

R constant= 0.08206

T=constant, so in this problem we dont need a value for it

M= 44.01 g/mol

√(3)(0.08206)(44.01)= 3.29

Since we are trying to measure how much faster NH3 will be, we have to realize that mass and speed have an inverse relationship.

So instead of doing (2.047)/(3.29) = 6.22

we have to flip the values to get (3.29)/(2.047)= 1.61

6 0
2 years ago
when 824 J of heat is added to 9 grams of oil at 30°c , the temperature increases to 87°C. what is the specific heat of the oil
sattari [20]

Answer:

c = 1.61 j/g.°C

Explanation:

Given data:

Mass of oil = 9 g

Heat added = 824 j

Initial temperature =  30°C

Final temperature = 87°C

Specific heat of oil = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature - final temperature

ΔT = 87°C - 30°C

ΔT = 57 °C

Q = m.c. ΔT

824 j = 9 g × c ×  57 °C

824 j = 513 g. °C × c

c = 1.61 j/g.°C

5 0
3 years ago
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