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Verizon [17]
3 years ago
8

How can acid rain affect organisms that live in the water?

Chemistry
1 answer:
krok68 [10]3 years ago
4 0
If the there is enough acid rain to spread in the water then it will become more acidic, making an impact on any<span> plants and animals that </span>live<span> in the water.</span>
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Can scientific laws be proved wrong? Why or why not?
astra-53 [7]

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The scientific laws have been well proven before they are published so it is difficult to prove mistakes

Explanation:

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according to the balanced chemical equation below,how many grams of h20 are produced if 3.98g of co2 were produced. 2C18H18 + 25
telo118 [61]

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Primero debes usar los gramos de co2 y luego buscar su peso molecular, luego de eso usar la relación de moles entre CO2 y H2O y por último buscar el pm del H2O pata ver cuantos gramos de produce.

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Write any two differences between unsaturated<br>and saturated solution​
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Read 2 more answers
If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of H2SO4(aq)? For your ans
Alik [6]

Question is incomplete, complete question is;

A 34.8 mL solution of H_2SO_4 (aq) of an unknown concentration was titrated with 0.15 M of NaOH(aq).

H_2SO_4(aq)+2NaOH(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)

If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of H_2SO_4(aq)? For your answer, only type in the numerical value with two significant figures. Do NOT include the unit.

Answer:

0.044 M is the molarity of H_2SO_4(aq).

Explanation:

The reaction taking place here is in between acid and base which means that it is a neutralization reaction .

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_1,M_2\text{ and }V_2  are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=?\\V_1=34.8 mL\\n_2=1\\M_2=0.15 M\\V_2=20.4 mL

Putting values in above equation, we get:

2\times M_1\times 34.8 mL=1\times 0.15 M\times 20.4 mL\\\\M_1=\frac{1\times 0.15 M\times 20.4 mL\times 10}{2\times 34.8 mL}=0.044 M

0.044 M is the molarity of H_2SO_4(aq).

4 0
3 years ago
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