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IgorC [24]
3 years ago
6

What is 2000lbs in 1 ton

Chemistry
1 answer:
liq [111]3 years ago
3 0
You've answered your question it is 1 ton
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An insect population increases and then decreases as the food supply changes.
Arada [10]

Answer:

self regulation

Explanation:

4 0
3 years ago
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Use the following half-reactions to construct a voltaic cell:
velikii [3]

<u>Answer:</u> The correct answer is 3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V

<u>Explanation:</u>

We are given:

Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o=-0.73V\\\\Ag^+(aq.)+e^-\rightarrow Ag(s);E^o=+0.80V

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction. Here, silver will always undergo reduction reaction will get reduced.

Chromium will undergo oxidation reaction and will get oxidized.

The half reactions for the above cell is:

Oxidation half reaction: Cr(s)\rightarrow Cr^{3+}+3e^-;E^o_{Cr^{3+}/Cr}=-0.73V

Reduction half reaction: Ag^{+}+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.80V       ( × 3)

Net equation:  3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=0.80-(-0.73)=1.53V

Hence, the correct answer is 3Ag^+(aq.)+Cr(s)\rightarrow 3Ag(s)+Cr^{3+}(aq.);E^o_{cell}=+1.53V

4 0
3 years ago
Calculate the pH of a solution that is 0.235M benzoic acid and 0.130M sodium benzoate, a salt whose anion is the conjugate base
lesantik [10]

Hello!

We have the following data:

ps: we apply Ka in benzoic acid to the solution.

[acid] = 0.235 M (mol/L)

[salt] = 0.130 M (mol/L)

pKa (acetic acid buffer) =?

pH of a buffer =?

Let us first find pKa of benzoic acid, knowing that Ka (benzoic acid) = 6.20*10^{-5}

So:

pKa = - log\:(Ka)

pKa = - log\:(6.20*10^{-5})

pKa = 5 - log\:6.20

pKa = 5 - 0.79

\boxed{pKa = 4.21}

Now, using the abovementioned data for the pH formula of a buffer solution or (Henderson-Hasselbalch equation), we have:

pH = pKa + log\:\dfrac{[salt]}{[acid]}

pH = 4.21 + log\:\dfrac{0.130}{0.235}

pH = 4.21 + log\:0.55

pH = 4.21 + (-0.26)

pH = 4.21 - 0.26

\boxed{\boxed{pH = 3.95}}\end{array}}\qquad\checkmark

Note:. The pH <7, then we have an acidic solution.

I Hope this helps, greetings ... DexteR! =)

8 0
2 years ago
Please help on this one?
quester [9]

Answer:

the answer is= NUCLEAR FISSION, NUCLEAR FUSSION, RADIOACTIVE DECAY.

7 0
3 years ago
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PLS ANSWER ME QUICK!!
Murrr4er [49]

Answer:

100.09 amu is the answer.

4 0
1 year ago
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