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soldi70 [24.7K]
3 years ago
15

What volume (in L) of a 1.25 M

Chemistry
1 answer:
sleet_krkn [62]3 years ago
4 0

Answer:

V1= 0.305L

Explanation:

To find the initial volume of 1.25M potassium fluoride needed to make tge dilution specified in the question, we can use: C1 × V1 = C2 × V2

Since the question wants the volume in litres, convert 455 mL to L

455/ 1000

= 0.455 L

Now make the substitution

1.25 × V1 = 0.838 × 0.455

Rearrange to make V1 the subject

V1=

\frac{0.838 \times 0.455}{1.25 }  = 0.305

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ΔHrxn = ΣδΗ(bond breaking) - ΣδΗ(bond making)

Bond enthalpies,
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According to the balanced equation,
ΣδΗ(bond breaking)  = N ≡ N x 1 + Cl - Cl x 3
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ΣδΗ(bond making)    = N - Cl x 3 x 2
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δHrxn = ΣδΗ(bond breaking) - ΣδΗ(bond making)
          = 1671 kJ mol⁻¹ - 1152 kJ mol⁻¹
          = 519 kJ mol⁻¹
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