Ask question

Ask question

All categories

2 years ago

12

Why do you think that tsunamis occur most commonly in the Pacific Ocean?

1 answer:

zhenek [66]2 years ago

5 0

Tsunamis are most common in the Pacific Ocean Basin. The entire Pacific Ocean is ringed by areas known as subduction zones, where the tectonic plates of the earth are moving relative to one another. Subduction zones are where two or more plates are colliding and one is going underneath another.

You might be interested in

What is the solubility in moles/liter for silver chloride at 25 oC given a Ksp value of 1.6 x 10-10. Write using scientific nota

trapecia [35]

Answer:

Explanation:

AgCl ⇄ Ag⁺ + Cl⁻

m m m

If x mole of AgCl be dissolved in one litre .

[ Ag⁺ ] [ Cl⁻ ] = 1.6 x 10⁻¹⁰

m² = 1.6 x 10⁻¹⁰

m = 1.26 x 10⁻⁵ moles

So solubility of AgCl is 1.26 x 10⁻⁵ moles / L

5 0

3 years ago

Plz give me short note on isotopes in agriculture industries and medicine

Mumz [18]

Uses of isotopes:

In agriculture,

C-14 is used to trace the path of photosynthesis.

In medicine,

Iodine -131 is used in the treatment of

goiter

8 0

3 years ago

Using stoichiometry, determine the mass of powdered drink mix needed to make 1.0 M solution of 100 mL

Rzqust [24]

Answer:

34.23 g.

M = (no. of moles of solute)/(V of the solution (L)).

6 0

3 years ago

You are given a solution containing a pair of enantiomers (a and b). careful measurements show that the solution contains 77% a

san4es73 [151]

The ee solution would be 55%

6 0

3 years ago

Read 2 more answers

Titration of a 20.0-mL sample of acid rain required 1.7 mL of 0.0811 M NaOH to reach the end point. If we assume that the acidit

Rasek [7]

Answer:

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

Explanation:

Volume of NaOH = 1.7 ml = 0.0017 L

Molarity of NaOH = 0.0811 M

Moles of NaOH = n

$0.0811 M=\frac{n}{0.0017 L}$

n = 0.0001378 mol

$H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O$

According to reaction, 2 mol of NaOH neutralize 1 mol of sulfuric acid.

Then 0.0001378 mol of NaOH will neutralize:

$\frac{1}{2}\times 0.0001378 mol=6.8935\times 10^{-5} mol$ of sulfuric acid.

Concentration of sulfuric acid in the acid rain sample: x

$x=\frac{6.8935\times 10^{-5}}{0.02 L}=0.0034467 mol/L$

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

7 0

3 years ago