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VashaNatasha [74]
4 years ago
15

What percent of the ocean is euphotic zone

Chemistry
1 answer:
valina [46]4 years ago
7 0
Depth where light intensity falls to one percent<span> of that at the surface, called the euphotic depth</span>
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How many neutrons does beryllium have?
PilotLPTM [1.2K]
5.

i answerd this b4
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How is 1 hectoliter different from 1 kiloliter?
LUCKY_DIMON [66]
1 hectoliter is 26.4172

1 kiloliter is 264.172
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It takes 2 g of chlorine to sanitize 1,000,000g of water. How much chlorine will it take to sanitize a 40,000-gallon pool? ( 1 g
dimulka [17.4K]

Answer:

0.1g (Gallon) of chlorine

Explanation:

<u>Formula</u>

1 gallon = 3.7L; the density of water is 1.0g/ml

<u>Given</u>

2g (gallon) of chlorine to sanitize = 1,000,000g (gallon) of water

<u>Solve</u>

If 2g (gallon) chlorine = 1,000,000g (gallon)

∴, ? chlorine = 40,000

The First step; set up an equation

1000000/2 = 40000/?

The Next step; divide 1 million to 2

1000000 ÷ 2 = 500000

Then, divide the result by 40000

40000 ÷ 500000 = 0.08

In the nearest unit that is 0.1

Therefore, it will take 0.1g (gallon) of chlorine to sanitize a 40,000-gallon pool.

7 0
3 years ago
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How does activation energy affect a chemical reaction ?
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Enzymes affect the rate of the reaction in both the forward and reverse directions; the reaction proceeds faster because less energy is required for molecules to react when they collide. Thus, the rate constant (k) increases. Figure 3: Lowering the Activation Energy of a Reaction by a Catalyst.

hope this helps!

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3 years ago
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50.0 mL of 0.200 M HNO2 is titrated to its equivalence point with 1.00 M NaOH. What is the pH at the equivalence point?
yKpoI14uk [10]

Answer:

8.279

Explanation:

The pH can be determined by hydrolysis of a conjugate base of weak acid at the equivalence point.

At the equivalence point, we have

$n_{NaOH}=n_{HNO_2}$

           = 25.00 x 0.200

           = 5.00 m-mol

           = 0.005 mol

Volume of the base that is added to reach the equivalence point is

$\frac{0.005}{1.00} \times 1000= 5.00 \ mL$

Number of moles of $NO^-_{2}=n_{HNO_2}$

                                           = 0.005 mol

Volume at the equivalence point is 25 + 5 = 30.00 mL

Therefore, concentration of $NO^-_{2}= \frac{5}{30}$

                                                        = 0.167 M

Now the ICE table :

            $NO^-_2 + H_2O \rightarrow HNO_3 + OH^-$

I (M)       0.167                   0            0

C (M)         -x                      +x          +x

E (M)      0.167-x                  x           x

Now, the value of the base dissociation constant is ,

$K_w=K_a \times K_b$            $(K_w \text{ is the ionic product of water })$

$K_b =\frac{K_w}{K_a}$

$K_b =\frac{1 \times 10^{-14}}{4.6 \times 10^{-4}}$

    = $2.174 \times 10^{-11}$

Base ionization constant, $K_b = \frac{\left[HNO_2\right] \left[OH^- \right]}{\left[NO^-_2 \right]}$

$2.174 \times 10^{-11}=\frac{x^2}{0.167 -x}$

$x= 1.9054 \times 10^{-6}$

So, $[OH^-]=1.9054 \times 10^{-6 } \ M$

pOH =- $\log[OH^-]$

       = $- \log(1.9054 \times 10^{-6} \ M)$

        =5.72

Now, since pH + pOH = 14

           pH = 14.00 - 5.72

                = 8.279

Therefore the ph is 8.279 at the end of the titration.

8 0
3 years ago
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