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3 years ago

How does the energy provided by the Sun affect global ocean currents?

Chemistry

1 answer:

777dan777 [17]3 years ago

8 0

The answer is C. The sun heats waters near the equator, which then move toward the polls where the waters cool.

You might be interested in

Which is an example of potential energy?

Setler79 [48]

Probably C hope this helps

5 0

3 years ago

Read 2 more answers

How many grams of sodium fluoride should be added to 300. mL of 0.0310 M of hydrofluoric acid to produce a buffer solution with

Kisachek [45]

Answer : The mass of sodium fluoride added should be 0.105 grams.

Explanation : Given,

The dissociation constant for HF = $K_a=6.8\times 10^{-4}$

Concentration of HF (weak acid)= 0.0310 M

First we have to calculate the value of $pK_a$ .

The expression used for the calculation of $pK_a$ is,

$pK_a=-\log (K_a)$

Now put the value of $K_a$ in this expression, we get:

$pK_a=-\log (6.8\times 10^{-4})$

$pK_a=4-\log (6.8)$

$pK_a=3.17$

Now we have to calculate the concentration of NaF.

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[NaF]}{[HF]}$

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

$2.60=3.17+\log (\frac{[NaF]}{0.0310})$

$[NaF]=0.00834M$

Now we have to calculate the moles of NaF.

$\text{Moles of NaF}=\text{Concentration of NaF}\times \text{Volume of solution}=0.00834M\times 0.300L=0.0025mole$

Now we have to calculate the mass of NaF.

$\text{Mass of }NaF=\text{Moles of }NaF\times \text{Molar mass of }NaF=0.0025mole\times 42g/mole=0.105g$

Therefore, the mass of sodium fluoride added should be 0.105 grams.

4 0

3 years ago

The concentration of the KCN solution given in Part A corresponds to a mass percent of 0.473 %. What mass of a 0.473 % KCN solut

tiny-mole [99]

If the Ka of HCN = 5.0 x 10^-10
Since
(Ka) (Ka) - 1 x 10^ -14
then
the Kb of its conjugate base (CN-) = 2.0 X 10^-5

since
pH + pOH = 14
when the pH = 10.00
then
the pOH = 4.00
& the OH-
would then equal 1.0 X 10^-4

NaCN as a base does a hydrolysis in water:
CN- & water --> HCN & OH-
notice that equal amounts of OH- & HCN are formed

Kb = [HCN] [OH-] / [CN-]

2.0 X 10^-5 = [1.0 X 10^-4] [1.0 X 10^-4] / [CN-]

[CN-] =(1.0 X 10^-8) / (2.0 X 10^-5)

[CN-] = (5.0 X 10^-4)

that's 0.00050 Molar
which is 0.00050 moles in each liter of aqueous KCN solution
which is
0.00025 moles KCN in 500. mL of aqueous KCN solution

use molar mass of KCN, to find grams:
(0.00025 moles KCN) (65.12 grams KCN / mole) = 0.01628 grams of KCn

which is 16.3 mg of KCN
& rounded to the 2 sig figs which are showing in the Ka of HCN , "5.0" X 10^-10
your answer would be
16 mg of KCN

sorry even after making a correction in calcs , I don't get one of your answers.
the only way that I could get one of them is to pretend that yours was a 1 sig fig problem,
in which case your 16 mg would round off to 20 mg.
but you have 3 sig figs in "500. ml", & 2 sig figs in both the "pH of 10.00."
& The Ka of HCN = "5.0 x 10^-10."

it does however take 12 mg of NaCN, to make 500. mL of aqueous solution pH of 10.00. the molar mass of NaCN has the smaller molar mass of 49.00 grams per mole.
maybe they meant NaCN, but wrote KCN instead.

I hope i answered this correctly for you.

8 0

4 years ago

An increase in mass will

maks197457 [2]

Answer:

As an object increases in speed, so does the amount of energy that it has, this energy is what we refer to as 'the increase in mass'

Explanation:

3 0

4 years ago

In a titration, 25.0 mL of KHP is titrated to the equivalence point with NaOH. The final solution volume is 45.5 mL. What was th

alukav5142 [94]

I think it A because it 25.0 ml