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kompoz [17]
3 years ago
11

What is the pH of a 3.5 x 10 to the negative 11th M H+ solution

Chemistry
2 answers:
Ghella [55]3 years ago
4 0

Answer:

Explanation:

-log(3.5 * 10^-11)

= 10.4559

Be careful how you put this into your calculator. I had to use Exp to get it to work properly.

-log

(3.5 * 10 exp -11)

=

Morgarella [4.7K]3 years ago
4 0

Answer : The pH of the solution is, 10.4

Explanation :

pH : It is defined as the negative logarithm of hydrogen ion concentration.

Mathematically,

pH=-\log [H^+]

Given:

[H^+]=3.5\times 10^{-11}M

Now put all the given values in the above expression, we get the value of pH.

pH=-\log [H^+]

pH=-\log (3.5\times 10^{-11}M)

pH=10.4

Thus, the pH of the solution is, 10.4

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Answer:

I'm pretty sure its C. limestone.

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What states can electrons exist in
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8 0
3 years ago
What is the resistance of a 150 W lightbulb connected to a 24 V voltage source?
Luda [366]

Answer:

3.84 Ω

Explanation:

From the question given above, the following data were obtained:

Electrical power (P) = 150 W

Voltage (V) = 24 V

Resistance (R) =?

P = IV

Recall:

V = IR

Divide both side by R

I = V/R

P = V/R × V

P = V² / R

Where:

P => Electrical power

V => Voltage

I => Current

R => Resistance

With the above formula (i.e P = V²/R), we can calculate resistance as illustrated below:

Electrical power (P) = 150 W

Voltage (V) = 24 V

Resistance (R) =?

P = V²/R

150 = 24² / R

150 = 576 / R

Cross multiply

150 × R = 576

Divide both side by 150

R = 576 / 150

R = 3.84 Ω

Thus, the resistance is 3.84 Ω

4 0
2 years ago
does the nitro group on the pyridine ring make the ring more electron rich or more electron deficient
Arte-miy333 [17]

Answer:

more electron deficient

Explanation:

The nitro group is an electron withdrawing group. It withdraws electrons from the pyridine ring by resonance.

This electron withdrawal by resonance makes the pyridine ring less electron rich or more electron deficient.

Hence, the nitro group makes the pyrinde ring more electron deficient

3 0
2 years ago
Determine the oxidation number of the element "J" in H3JO2-
igor_vitrenko [27]

Answer:

Oxidation number:

3*1+ oxidation number of J+2*-2= -1

Oxidation number of J = 0

5 0
2 years ago
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