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Elodia [21]
3 years ago
6

What is the average salinity of ocean water?

Chemistry
2 answers:
sergiy2304 [10]3 years ago
7 0
35 parts per thousandOn average, seawater in the world's oceans has a salinity of approximately 3.5%, or 35 parts per thousand. (For every 1 litre (1000 mL) of seawater there are 35 grams of salts.)

Therefore, your answer is 34.5 g/mL
Ratling [72]3 years ago
6 0

Answer:

<h2>34.5 g/mL</h2>

Explanation:

<h3>Actually sea water salinity is 35 grams per litre (average).</h3><h3>Which means: \frac{35g}{1L}</h3>
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The pOH of a solution is 10.75. What is the concentration of OH– ions in the solution?. A. –1.031 M. B. 1.778 x 10–11 M. C. 5.62
zloy xaker [14]
We are given the pOH of the solution of 10.75. pOH is the property of the solution that is related to the OH ion concentration of the solution. THe formula to be followed is pOH = -log (OH); OH- = 10^-pOH. In this case, OH-  = 10^-10.75 equal to B. 1.778 x 10^-11 M
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How is the value of an atomic number detirme?
serious [3.7K]

The number of protons in the nucleus determines the atomic number of an element.

Every element has its own characteristic atomic number.

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4 years ago
If a boy eats a meal and then runs
aalyn [17]

Answer: A

Explanation: A, because when you eat it is you use chemical energy afterward you run and use mechanical energy. Ex of Chemical energy can be batteries and you digesting food. Mechanical is movement.

7 0
3 years ago
Read 2 more answers
How many liters of 0.615 M NaOH will be needed to raise the pH of 0.385 L of 5.13 M sulfurous acid (H2SO3) to a pH of 6.247?
givi [52]

Answer:

Volume of NaOH required = 3.61 L

Explanation:

H2SO3 is a diprotic acid i.e. it will have two dissociation constants given as follows:

H2SO3\leftrightarrow H^{+}+HSO3^{-} --------(1)

where,  Ka1 = 1.5 x 10–2  or pKa1 = 1.824

HSO3^{-}\leftrightarrow H^{+}+SO3^{2-} --------(2)

where,  Ka2 = 1.0 x 10–7 or pKa2 = 7.000

The required pH = 6.247 which is beyond the first equivalence point but within the second equivalence point.

Step 1:

Based on equation(1), at the first eq point:

moles of H2SO3 = moles of NaOH

i.e. \ 5.13\  moles/L*0.385L = moles\  NaOH\\therefore, \ moles\  NaOH = 1.98\ moles\\V(NaOH)\ required = \frac{1.98\ moles}{0.615\ moles/L} =3.22L

Step 2:

For the second equivalence point setup an ICE table:

                  HSO3^{-}+OH^{-}\leftrightarrow H2O+SO3^{2-}

Initial           1.98                    ?                                       0

Change      -x                       -x                                       x

Equil           1.98-x                 ?-x                                    x

Here, ?-x =0 i.e. amount of OH- = x

Based on the Henderson buffer equation:

pH = pKa2 + log\frac{[SO3]^{2-} }{[HSO3]^{-} } \\6.247 = 7.00 + log\frac{x}{(1.98-x)} \\x=0.634 moles

Volume of NaOH required is:

\frac{0.634\ moles}{0.615 moles/L}=0.389L

Step 3:

Total volume of NaOH required = 3.22+0.389 =3.61 L

3 0
3 years ago
How many moles are in 38 grams of Argon? Round to 2 decimal places. Use "e" for exponents
yuradex [85]

Answer: 39.948

Explanation:

How many grams Argon in 1 mol? The answer is 39.948. We assume you are converting between grams Argon and mole. You can view more details on each measurement unit: molecular weight of Argon or mol The molecular formula for Argon is Ar.

3 0
3 years ago
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