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3 years ago

What is the pH when the poH value is 2?

Chemistry

1 answer:

dimaraw [331]3 years ago

6 0

Answer:

i hate my life PERIODT

Explanation:

WHY AM I HERE :/

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Be sure to answer all parts. Consider the reaction N2(g) + 3H2(g) → 2NH3(g)ΔH o rxn = −92.6 kJ/mol If 3.0 moles of N2 react with

Alex787 [66]

Answer:

Work done in Joules = -14865.432J

ΔU = -262.94KJ

Explanation:

The steps are as shown in the attachment

6 0

3 years ago

Which substance do plants use to capture light energy from the sun? a) Carbon dioxide b) Chlorophyll c) Glucose d) Oxygen

Akimi4 [234]

Answer: chlorophyll

Explanation:

The substance that plants use to capture light energy from the sun is referred to as chlorophyll.

The chlorophyll is a colored pigment that is used by the plants for photosynthesis. The function of the chlorophyll is that it absorbs the energy from the sun which is then turned into a chemical energy.

6 0

3 years ago

Read 2 more answers

Please help me with this science

MAVERICK [17]

The answer is no. 5 grams are equal to 5 milligrams.

"Milli" means million so a milligram is one-millionth of a gram.

6 0

3 years ago

Which of the following is a characteristic of alcohols

liq [111]

You forgot to add pictures ..

3 0

3 years ago

The standard emf for the cell using the overall cell reaction below is +2.20 V:

ANEK [815]

Answer:

emf generated by cell is 2.32 V

Explanation:

Oxidation: $2Al-6e^{-}\rightarrow 2Al^{3+}$

Reduction: $3I_{2}+6e^{-}\rightarrow 6I^{-}$

---------------------------------------------------------------------------------

Overall: $2Al+3I_{2}\rightarrow 2Al^{3+}+6I^{-}$

Nernst equation for this cell reaction at $25^{0}\textrm{C}$ -

$E_{cell}=E_{cell}^{0}-\frac{0.059}{n}log{[Al^{3+}]^{2}[I^{-}]^{6}}$

where n is number of electrons exchanged during cell reaction, $E_{cell}^{0}$ is standard cell emf , $E_{cell}$ is cell emf , $[Al^{3+}]$ is concentration of $Al^{3+}$ and $[Cl^{-}]$ is concentration of $Cl^{-}$

Plug in all the given values in the above equation -

$E_{cell}=2.20-\frac{0.059}{6}log[(4.5\times 10^{-3})^{2}\times (0.15)^{6}]V$

So, $E_{cell}=2.32V$

3 0

3 years ago