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Svet_ta [14]
3 years ago
13

Can sunscreen damage other organisms?

Chemistry
1 answer:
miv72 [106K]3 years ago
5 0

Sunscreen can have negative effects on corals and other marine organisms under certain circumstances.

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Determine the [ OH − ] , pH, and pOH of a solution with a [ H + ] of 8.2 × 10 − 10 M at 25 °C. [ OH − ] = 1.22 ×10 −5 M pH = 9.0
nika2105 [10]

Answer:

See explanation below

Explanation:

At the beggining of the  exercise, you already giving the solution for the 1st two problems, so I just going to answer the remaining question which are:

<em>"Determine the [ H + ] , [ OH − ] , and pOH of a solution with a pH of 7.20 at 25 °C. [ H + ] = M [ OH − ] = M pOH = 6.8 Determine the [ H + ] , [ OH − ] , and pH of a solution with a pOH of 6.43 at 25 °C."</em>

Now, to get every data in the problem, we need to write the expressions for every part, which are the following:

pH = -log[H⁺]   (1)

pH = 14 - pOH  (2)

[H⁺] = 10^(-pH)  (3)

[H⁺] = Kw / [OH⁻]   (4)

pOH = -log[OH⁻]  (5)

pOH = 14 - pH  (6)

[OH⁻] = 10^(-pOH)   (7)

[OH⁻] = Kw / [H⁺]  (8)

Now, according to the given data, we will use any of these eight expressions.

a) "<u><em>Determine the [ H + ] , [ OH − ] , and pOH of a solution with a pH of 7.20"</em></u>

We have pH, so in this case, we will use expression (3), (6) and (8):

[H⁺] = 10^(-7.20)

[H⁺] = 6.31x10⁻⁹ M

pOH = 14 - 7.2

pOH = 6.8

[OH⁻] = 1x10⁻¹⁴ / 6.31x10⁻⁹

[OH⁻] = 1.58x10⁻⁴ M

b) <u>"</u><u><em>The [ H + ] , [ OH − ] , and pH of a solution with a pOH of 6.43 "</em></u>

In this case, we can use expressions (2), (4) and (7). Using those expressions we have:

pH = 14 - 6.43

pH = 7.57

[H⁺] = 10^(-7.57)

[H⁺] = 2.69x10⁻⁸ M

[OH⁻] = 1x10⁻¹⁴ / 2.69x10⁻⁸

[OH⁻] = 3.72x10⁻⁷ M

4 0
4 years ago
Explain the rule about diluting acids—what is it and why?
IceJOKER [234]

Explanation:

<em>Dur</em><em>ing</em><em> </em><em>dilu</em><em>tion</em><em>,</em><em> </em><em>we</em><em> </em><em>pour</em><em> </em><em>the</em><em> </em><em>aci</em><em>d</em><em> </em><em>into</em><em> </em><em>water</em><em> </em><em>not</em><em> </em><em>water</em><em> </em><em>into</em><em> </em><em>the</em><em> </em><em>acid</em><em>.</em><em> </em><em>This</em><em> </em><em>is</em><em> </em><em>beca</em><em>use</em><em> </em><em>when</em><em> </em><em>we</em><em> </em><em>pour</em><em> </em><em>the</em><em> </em><em>water</em><em> </em><em>into</em><em> </em><em>the</em><em> </em><em>aci</em><em>d</em><em> </em><em>it</em><em> </em><em>will</em><em> </em><em>resul</em><em>ts</em><em> </em><em>in</em><em> </em><em>explotion</em><em> </em><em>or</em><em> </em><em>combu</em><em>stion</em><em>.</em>

8 0
4 years ago
The nucleus or center area of the atom contains two kinds of subatomic particles. which two?
neonofarm [45]
Protons and neutrons
5 0
3 years ago
Use ideas about ELECTRON TRANSFER to explain why green precipitate, Iron(ii) hydroxide turns red-brown on standing.
azamat
Fe(OH)2 exhibits Fe2+, that is easily oxidized by oxigen present in the air.

Fe(OH)2 (green) + H2O - e- = Fe(OH)3 (brown) + H+
this half-reaction shows how Fe(OH)2 is oxidized to yield Fe(OH)3
As it is being oxidized, it loses an electron

O2 (in the air) + 2 H2O + 4 e- =  4 OH-
this half-reaction shows oxygen perfoming its oxidizing properties
You can see that it gains 4 electrons, which he receives from iron (iron loses electron, remember?) So overall electrons are being transferred from iron to oxygen. Iron is oxidized and oxygen is thus reduced.

Since the electrons lost in the first half-equation are the same as the ones gained in the second, lets multiply the first half-equation with the second:

4 Fe(OH)2 (green) + 4 H2O - 4e- = 4Fe(OH)3 (brown)<span> + 4 H+
</span>O2 (in the air) + 2 H2O + 4 e- <span>=  4 OH-
</span>------------------------------
4Fe(OH)2 (green) + O2 (in the air) + 6 H2O  + 4 e- - 4 e- = 4Fe(OH)3 (brown) + 4 H+ + <span>4 OH-

</span><span>4 e- - 4 e- = 0
</span>and
4 H+ + <span>4 OH- = 4 H2O
</span>thus
4Fe(OH)2 (green) + O2 (in the air) + 6 H2O<span>  = </span>4Fe(OH)3 (brown) + 4 H2O
4Fe(OH)2 (green) + O2 (in the air) + 2 H2O  = <span>4Fe(OH)3 (brown)</span>
3 0
4 years ago
Read 2 more answers
At certain temperature, kc for the reaction, PCl5 ------&gt; PCl3 + Cl2, is equal to 3.33. After .20 mole of PCL5 and 1.0 mole o
galben [10]

Answer : The equilibrium concentration of PCl_5 is, 0.16 M

Explanation :

First we have to calculate the concentration of PCl_5\text{ and }PCl_3

\text{Concentration of }PCl_5=\frac{\text{Moles of }PCl_5}{\text{Volume of solution}}=\frac{0.20mol}{2.00L}=0.4M

and,

\text{Concentration of }PCl_3=\frac{\text{Moles of }PCl_3}{\text{Volume of solution}}=\frac{1.0mol}{2.00L}=2.0M

The given chemical reaction is:

                     PCl_5(g)\rightarrow PCl_3(g)+Cl_2(g)

Initial conc.      0.4           2.0          0

At eqm.          (0.4-x)      (2.0+x)       x

The expression for equilibrium constant is:

K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}

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

3.33=\frac{(2.0+x)\times (x)}{(0.4-x)}

x = -5.57 and x = 0.24

We are neglecting the value of x = -5.57 because equilibrium concentration can not be more than initial concentration.

Thus, the value of x = 0.24

The equilibrium concentration of PCl_5 = (0.4-x) = (0.4-0.24) = 0.16 M

Therefore, the equilibrium concentration of PCl_5 is, 0.16 M

6 0
3 years ago
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