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tatuchka [14]
3 years ago
14

Which equilibrium describes a reversible reaction a short time after starting the reaction with reactants only?

Chemistry
1 answer:
valkas [14]3 years ago
4 0

Answer:

Should be A

Explanation:

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49.Which of the following statements regarding chemical equilibrium is true?
crimeas [40]
A asystem at equilibrium stops
8 0
4 years ago
The chemical equation, Cr + Fe(NO3)2 → Fe + Cr(NO3)3, is an example of which type of reaction?
Angelina_Jolie [31]

Answer:

Redox type

Explanation:

The reaction is:

2Cr +  3Fe(NO₃)₂ → 2Fe + 2Cr(NO₃)₃

2 moles of chromium can react to 3 moles of iron (II) nitrate in order to produce 2 moles of iron and 2 moles of chromium nitrate.

If we see oxidation state, we see that chromium changes from 0 to +3

Iron changed the oxidation state from +2 to 0

Remember that elements at ground state has 0, as oxidation state.

Iron is being reduced while chromium is oxidized. Then, the half reactions are:

Fe²⁺  +  2e⁻ ⇄  Fe    (Reduction)

Cr ⇄ Cr³⁺  +  3e⁻    (Oxidation)

When an element is being  reduced, while another is being oxidized, we are in prescence of a redox reaction.

8 0
3 years ago
What concentrations of acetic acid (pKa = 4.76) and acetate would be required to prepare a 0.15 M buffer solution at pH 5.0? Not
Leni [432]

Answer:

Acetic acid 0,055M and acetate 0,095M.

Explanation:

It is possible to prepare a 0,15M buffer of acetic acid/acetate at pH 5,0 using Henderson-Hasselblach formula, thus:

pH = pka + log₁₀ [A⁻]/[HA] <em>-Where A⁻ is acetate ion and HA is acetic acid-</em>

Replacing:

5,0 = 4,76 + log₁₀ [A⁻]/[HA]

<em>1,7378 =  [A⁻]/[HA] </em><em>(1)</em>

As concentration of buffer is 0,15M, it is possible to write:

<em>[A⁻] + [HA] = 0,15M </em><em>(2)</em>

Replacing (1) in (2):

1,7378[HA] + [HA] = 0,15M

2,7378[HA] = 0,15M

[HA] = 0,055M

Thus, [A⁻] = 0,095M

That means you need <em>acetic acid 0,055M</em> and <em>acetate 0,095M</em> to obtain the buffer you need.

i hope it helps!

7 0
3 years ago
Which of these correctly lists the stages of development of the inner planets?
ZanzabumX [31]
The correct answer for the question that is being presented above is this one:  "C. planetesimals ® heavier elements ® inner planets ® protoplanets" The list of the stages of development of the inner planets is this <span>C. planetesimals ® heavier elements ® inner planets ® protoplanets</span>
5 0
3 years ago
Read 2 more answers
It takes 167 s for an unknown gas to effuse through a porous wall and 99 s for the same volume of n2 gas to effuse at the same t
irakobra [83]

Answer : The molar mass of the unknown gas will be 79.7 g/mol


Explanation : To solve this question we can use graham's law;


Now we can use nitrogen as the gas number 2, which travels faster than gas 1;

So, 167 / 99 = 1.687 So the nitrogen gas is 1.687 times faster that the unknown gas 1

We can compare the rates of both the gases;


So here, Rate of gas 2 / Rate of gas 1 = \sqrt{(molar mass 1 / molar mass 2)}

Now, 1.687 = square root [\sqrt{(molar mass 1) / (28.01 g/mol N_{2})} ]


When we square both the sides we get;


2.845 = (molar mass 1) / (28.01 g/mol N2)


On rearranging, we get,


2.845 X (28.01 g/mol N2) = Molar mass 1

So the molar mass of unknown gas will be = 79.7 g/mol

3 0
3 years ago
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