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Dima020 [189]
3 years ago
14

Which probing question lies within the scope of physics? A. Are fish in the open ocean attracted by underwater sounds? B. Does i

ncreasing the saltiness of ocean water affect the speed of sound in the water? C. What effect does the release of industrial wastewater have on the acidity of oceans? D. What is the effect of rising sea temperatures on ocean currents?
Chemistry
2 answers:
Alex73 [517]3 years ago
5 0

the answer is B.Does increasing the saltiness of ocean water affect the speed of sound in the water?

konstantin123 [22]3 years ago
3 0

I think its B

Hope this helps!

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Bainliest. Mole Conversion
Flauer [41]

Answer:

125g × 1 mol/16 grams = 7.81 × avogadros #

3 0
2 years ago
Calculate [H3O+] and [OH−] for each of the following solutions at 25 ∘C given the pH. pH= 8.74, pH= 11.38, pH= 2.81
Gnom [1K]

Answer:

Explanation:

Given parameters;

pH  = 8.74

pH = 11.38

pH = 2.81

Unknown:

concentration of hydrogen ion and hydroxyl ion for each solution = ?

Solution

The pH of any solution is a convenient scale for measuring the hydrogen ion concentration of any solution.

It is graduated from 1 to 14

      pH = -log[H₃O⁺]

      pOH = -log[OH⁻]

 pH + pOH = 14

Now let us solve;

   pH = 8.74

             since  pH = -log[H₃O⁺]

                           8.74 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻^{8.74}

                             [H₃O⁺]  = 1.82 x 10⁻⁹mol dm³

       pH + pOH = 14

                 pOH = 14 - 8.74

                  pOH = 5.26

                  pOH = -log[OH⁻]

                     5.26  = -log[OH⁻]

                     [OH⁻] = 10^{-5.26}

                      [OH⁻] = 5.5 x 10⁻⁶mol dm³

2.  pH = 11.38

             since  pH = -log[H₃O⁺]

                           11.38 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻^{11.38}

                             [H₃O⁺]  = 4.17 x 10⁻¹² mol dm³

           pH + pOH = 14

                 pOH = 14 - 11.38

                  pOH = 2.62

                  pOH = -log[OH⁻]

                     2.62  = -log[OH⁻]

                     [OH⁻] = 10^{-2.62}

                      [OH⁻] =2.4 x 10⁻³mol dm³

3. pH = 2.81

             since  pH = -log[H₃O⁺]

                           2.81 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻^{2.81}

                             [H₃O⁺]  = 1.55 x 10⁻³ mol dm³

           pH + pOH = 14

                 pOH = 14 - 2.81

                  pOH = 11.19

                  pOH = -log[OH⁻]

                     11.19  = -log[OH⁻]

                     [OH⁻] = 10^{-11.19}

                      [OH⁻] =6.46 x 10⁻¹²mol dm³

5 0
3 years ago
If the half-life of a radioactive substances is 590 million years and you have 40 atoms of it, how many half-lives will have pas
Nady [450]

Answer:

3

Explanation:

Applying,

2^{n'} = R/R'............... Equation 1

Where n' = number of halflives that have passed, R = Original atom of the substance, R' = atom of the substance left after decay.

From the question,

Given: R = 40 atoms, R' = 5 atoms

Substitute these values into equation 1

2^{n'} = 40/5

2^{n'} = 8

2^{n'} = 2³

Equation the base,

n' = 3

3 0
3 years ago
Mention any two uses of oxygen gas.​
Elodia [21]

Answer:

steel, plastics

<h3>Explanation:</h3>

Hope it helps!

3 0
3 years ago
Suppose that in an ionic compound, "m" represents a metal that could form more than one type of ion. in the formula mf2 , the ch
geniusboy [140]
F (Fluorine) is in column (group/family) VIIA, or the "halogens". When you see the halogens (Fluorine, Chlorine, Bromine, and Iodine) in combination with a metal, each halogen atom present will carry a -1 charge. We can see that the atom has no charge, so the metal must cancel out the negative charges brought by the two fluorine atoms.
(Charge on m) + 2*(charge on fluorine) = 0
(Charge on m) + 2*(-1) = 0
(Charge on m) - 2 = 0
Charge on m ion = +2
3 0
3 years ago
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