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2 years ago

6

CORRECT ANSWER GETS BRAINLIST

1 answer:

pashok25 [27]2 years ago

6 0

Answer:

B. halocline

Explanation:

it is a zone in the oceanic water that changes depending on the depth

Hope This Helped Sorry If Wrong

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How many grams of nano3 would you add to 500g of h2o in order to prepare a solution that is 0.500 molal in nano3?

VARVARA [1.3K]

When the concentration is expressed in molality, it is expressed in moles of solute per kilogram of solvent. Since we are given the mass of the solvent, which is water, we can compute for the moles of solute NaNO3.

0.5 m = x mol NaNO3/0.5 kg water
x = 0.25 mol NaNO3

Since the molar mass of NaNO3 is 85 g/mol, the mass is

0.25 mol * 85 g/mol = 21.25 grams NaNO3 needed

4 0

3 years ago

Calculate the pH of the cathode compartment solution if the cell emf at 298 K is measured to be 0.660 V when (Zn^2+)=0.22 M and(

yaroslaw [1]

Answer:

pH = 2.059

Explanation:

At the Cathode:

The reduction reaction is:

$2H^+ + 2e^- \to H_2 \ \ \ \mathbf{E^0_{red}= 0.00 \ V}$

At the anode:

At oxidation reaction is:

$Zn \to Zn^{2+} +2e^- \ \ \ \mathbf{E^0_{ox} = 0.76 \ V}$

The overall equation for the reaction is:

$\mathbf{Zn + 2H^+ \to Zn^{2+} + H_2}$

The overall cell potential is:

$\mathbf{E^0_{cell}= E^0_{ox} + E^0_{red}}$

$\mathbf{E^0_{cell}= 0.76 \ V +0.00 \ V}$

$\mathbf{E^0_{cell}= 0.76\ V}$

Using the formula for the Nernst equation:

$E = E^0 - ( \dfrac{0.0591}{n})log (Q)\\$

where;

E = 0.66

(Zn^2+)=0.22 M

Then

$0.66 =0.76- ( \dfrac{0.0591}{2})log \bigg ( \dfrac{[Zn^{2+} ] PH_2}{[H^+]^2} \bigg )$

$0.66 =0.76- 0.02955 * log \bigg ( \dfrac{0.22*0.87}{[H^+]^2} \bigg )$

3.4 = log ( 0.1914) - 2 log [H⁺]

3.4 = -0.7180 - 2 log [H⁺]

3.4 + 0.7180 = - 2 log [H⁺]

4.118 = - 2 log [H⁺]

pH = log [H⁺] = 4.118/2

pH = 2.059

8 0

3 years ago

What value is closest to the mass of the atom?

slavikrds [6]

C. 10 amu i am 100 sure that it is 10 amu

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The second step of Experiment 5's multistep synthesis is an electrophilic aromatic substitution (EAS) reaction, similar to the r

mafiozo [28]

Answer:

C

Explanation:

C. Nucleophilic attack and carbocation rearrangement to form the most stable carbocation before the substitution reaction

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3 years ago

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