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Art [367]
3 years ago
8

5 x 10^25 atoms of nickel equal how many moles

Chemistry
1 answer:
Rom4ik [11]3 years ago
3 0
1 mole  Ni ------------- 6.02x10²³ atoms
?? moles Ni ----------- 5x10²⁵ atoms

1 x ( 5x10²⁵ ) / 6.02x10²³ =

=> 83.05 moles
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What reaction type is <br> 4+22⟶4+22
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Answer: CF_{4} + 2Br_{2} \rightarrow CBr_{4} + 2F_{2} is a single replacement reaction.

Explanation:

A chemical reaction in which one element of a compound is replaced by another element is called single replacement reaction.

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Here, fluorine element in the compound CF_{4} is replaced by the element bromine.

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3 years ago
Let us assume that cu(oh)2(s) is completely insoluble, which signifies that the precipitation reaction with naoh(aq) (presented
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4 years ago
Calculate the pH of a 1.00 L buffer of 0.97 M CH3COONa / 1.02 M CH3COOH before and after the addition of the following species.
ivanzaharov [21]

Answer:

a) 4.73

b) 4.78

c) 4.66 (further addition)  or 4.60 (starting from the original buffer solution)

Explanation:

<u>Step 1:</u> Data given

volume of the buffer = 1.00 L

Buffer = 0.97 M CH3COONa / 1.02 M CH3COOH

pKa CH3COOH = 4.75

<u>Step 2: </u>pH = pKa + log [CH3COONa]/[CH3COOH]

pH = 4.75 + log (0.97/1.02)

pH =<u> 4.73</u>

(b) pH after addition of 0.065 mol NaOH

Adding 0.065 mol NaOH will reduce the acid by that amount leaving 1.02 - 0.065 = 0.955 moles HA in 1 L so [HA] = 0.955; the neutralized acid produces A- in the same amount, increasing [A-] to 0.97 +0.065 = 1.035

pH = pKa + log[CH3COONa]/[CH3COOH]

pH = 4.75 + log(1.035/0.955)

pH = <u>4.78</u>

c) pH after<u> further</u> addition of 0.144 mol HCl

The reverse will happen after the addition of HCl:

[HA] = 0.955 + 0.144 = 1.099

[A-] = 1.035 - 0.144 = 0.891

pH = 4.75 + log(0.891/1.099)

pH = 4.66

If we add 0.144 mol of HCl to the original buffer we will get:

[HA] = 1.02 + 0.144 = 1.164

[A-] = 0.97 - 0.144 = 0.826

pH = 4.75 + log(0.826/1.164)

pH = 4.60

3 0
3 years ago
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