Explain why lime is added to lakes. Refer to the problem that exists and what lime does to the lakes.

Chemistry

1 answer:

Darya [45]4 years ago

8 0

Explanation:

Lime is an oxide of calcium(CaO). As a metallic oxide, it produces a basic solution when it reacts with water.

The main reason why lime is added to lakes is to reduce acidity in the lake water.

Bottom muds in lakes are usually acidic and makes the water hypertonic. When lime is added, it reduces the pH of the water and causes acidity to reduce.

Lime also helps to reduce the hardness of lake water to a good level.
Lime also helps to release nutrients useful for photosynthesis of plants in the lake.

Learn more:

Neutralization brainly.com/question/5544078

#learnwithBrainly

You might be interested in

Calculate the entropy change for the surroundings of the reaction below at 350K: N2(g) + 3H2(g) -> 2NH3(g) Entropy data: NH3

krek1111 [17]

Answer : The entropy change for the surroundings of the reaction is, -198.3 J/K

Explanation :

We have to calculate the entropy change of reaction $(\Delta S^o)$ .

$\Delta S^o=S_{product}-S_{reactant}$

$\Delta S^o=[n_{NH_3}\times \Delta S^0_{(NH_3)}]-[n_{N_2}\times \Delta S^0_{(N_2)}+n_{H_2}\times \Delta S^0_{(H_2)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S^0{(NH_3)}$ = standard entropy of $NH_3$

$\Delta S^0{(H_2)}$ = standard entropy of $H_2$

$\Delta S^0{(N_2)}$ = standard entropy of $N_2$

Now put all the given values in this expression, we get:

$\Delta S^o=[2mole\times (192.5J/K.mole)]-[1mole\times (191.5J/K.mole)+3mole\times (130.6J/K.mole)]$

$\Delta S^o=-198.3J/K$

Therefore, the entropy change for the surroundings of the reaction is, -198.3 J/K

4 0

3 years ago

23.5 L of h2 is stored at a pressure of 58.7 Kpa what volume would the gas take up at stp

stepan [7]

Answer:- 13.6 L

Solution:- Volume of hydrogen gas at 58.7 Kpa is given as 23.5 L. It asks to calculate the volume of hydrogen gas at STP that is standard temperature and pressure. Since the problem does not talk about the original temperature so we would assume the constant temperature. So, it is Boyle's law.

Standard pressure is 1 atm that is 101.325 Kpa.

Boyle's law equation is:

$P_1V_1=P_2V_2$