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Arlecino [84]
3 years ago
5

Hot water deep within earth can _______ minerals and transport them someplace else.

Chemistry
1 answer:
Vedmedyk [2.9K]3 years ago
5 0

Answer:

Dissolve

Hope this helps!

Explanation:

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What is the name of HCL?
Alex787 [66]

Answer:

Hydrochloric acid

Explanation:

Hydrochloric acid

4 0
3 years ago
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I have 250 mL of an acid. If a friend gives me another 75 mL of the same acid, how much will I have?
Zigmanuir [339]

Answer:325 mL of acid

Explanation:250 mL + 75 mL = 325 mL

3 0
2 years ago
A concentration cell is constructed using two Ni electrodes with Ni2+ concentrations of 1.0 M and 1.00 � 10�4 M in the two half-
Komok [63]

<u>Answer:</u> The cell potential of the cell is +0.118 V

<u>Explanation:</u>

The half reactions for the cell is:

<u>Oxidation half reaction (anode):</u>  Ni(s)\rightarrow Ni^{2+}+2e^-

<u>Reduction half reaction (cathode):</u>  Ni^{2+}+2e^-\rightarrow Ni(s)

In this case, the cathode and anode both are same. So, E^o_{cell} will be equal to zero.

To calculate cell potential of the cell, we use the equation given by Nernst, which is:

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Ni^{2+}_{diluted}]}{[Ni^{2+}_{concentrated}]}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = ?

[Ni^{2+}_{diluted}] = 1.00\times 10^{-4}M

[Ni^{2+}_{concentrated}] = 1.0 M

Putting values in above equation, we get:

E_{cell}=0-\frac{0.0592}{2}\log \frac{1.00\times 10^{-4}M}{1.0M}

E_{cell}=0.118V

Hence, the cell potential of the cell is +0.118 V

5 0
3 years ago
Enter your answer in the provided box. Find the pH of a buffer that consists of 0.34 M HBrO and 0.89 M KBrO (PK, of HBrO = 8.64)
11111nata11111 [884]

Answer : The  pH of buffer is 9.06.

Explanation : Given,

pK_a=8.64

Concentration of HBrO = 0.34 M

Concentration of KBrO = 0.89 M

Now we have to calculate the pH of buffer.

Using Henderson Hesselbach equation :

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

pH=pK_a+\log \frac{[KBrO]}{[HBrO]}

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

pH=8.64+\log (\frac{0.89}{0.34})

pH=9.06

Therefore, the pH of buffer is 9.06.

6 0
4 years ago
A testable question is one that _____.
cestrela7 [59]

Answer:

One that “Can be answered by conducting an experiment”

Explanation:

8 0
3 years ago
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