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Hitman42 [59]
2 years ago
6

How are organising my time?

Chemistry
1 answer:
igomit [66]2 years ago
3 0

well for a start you need to learn how to type correctly then you have to plan out your schedule to what you can handle. :)  

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What specifically determines how much energy is available in a molecule
irina1246 [14]
Possibly the amount of electrons and energy levels in which the electrons reside

5 0
3 years ago
Which one of the following systems has the highest entropy?
Veronika [31]

Answer: a

Explanation:

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Is this chemical equation balanced?
cluponka [151]

Answer:

Yes

Explanation:

5 0
3 years ago
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3Fe+4H2O(yields) Fe3O4+4H2. What is the mole ratio of Fe3O4 to Fe?
WINSTONCH [101]

<u>Answer:</u> The correct answer is Option A.

<u>Explanation:</u>

Mole ratio is defined as the ratio between the stoichiometric coefficients of the molecules present in the chemical reaction.

For the given balanced chemical equation:

3Fe+4H_2O\rightarrow Fe_3O_4+4H_2

By Stoichiometry of the reaction:

3 moles of iron metal reacts with 4 moles of water to produce 1 mole of iron oxide and 4 moles of hydrogen gas.

The mole ratio of Fe_3O_4:Fe=1:3

Hence, the correct answer is Option A.

7 0
3 years ago
Read 2 more answers
In the laboratory, a general chemistry student measured the pH of a 0.529 M aqueous solution of phenol (a weak acid), C6H5OH to
Artyom0805 [142]

Answer:

The dissociation constant of phenol from given information is 9.34\times 10^{-11}.

Explanation:

The measured pH of the solution = 5.153

C_6H_5OH\rightarrow C_6H_5O^-+H^+

Initially      c

At eq'm   c-x       x  x

The expression of dissociation constant is given as:

K_a=\frac{[C_6H_5O^-][H^+]}{[C_6H_5OOH]}

Concentration of phenoxide ions and hydrogen ions are equal to x.

pH=-\log[x]

5.153=-\log[x]

x=7.03\times 10^{-6} M

K_a=\frac{x\times x}{(c-x)}=\frac{x^2}{(c-x)}=\frac{(7.03\times 10^{-6} M)^2}{ 0.529 M-7.03\times 10^{-6} M}

K_a=9.34\times 10^{-11}

The dissociation constant of phenol from given information is 9.34\times 10^{-11}.

4 0
3 years ago
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