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vitfil [10]
2 years ago
11

Perhatikan persamaan termokimia berikut.

Chemistry
1 answer:
worty [1.4K]2 years ago
7 0

the reaction is

HCl (aq) + NaOH (aq) -> NaCl (aq) + H2O (l).

the enthalpy of reaction is H=-54 kJ

Which means that one mole of HCl reacts with one mole of NaOH and gives 54 kJ of energy / heat

The reaction is exothermic and the graph will be like as shown in figure (1)

The question is

What is the enthalpy change if 10 mL HCl 1 M is reacted with 20 mL 1 M NaOH?

Moles of HCl  = molarity X volume = 1 M X 10 mL = 10 mmoles

moles of NaOH reacted = moles of NaOH

so total moles of acid base reacted = 10mmoles = water formed

when one mole of acid base reacted to give one mole of H2O

so when 10mmole of water are formed energy released is = 54 X 10 X 10^-3 = 0.54 kJ


reaksinya

HCl (aq) + NaOH (aq) -> NaCl (aq) + H2O (l).

entalpi reaksi adalah H = -54 kJ

Yang berarti satu mol HCl bereaksi dengan satu mol NaOH dan memberikan 54 kJ energi / panas

Reaksinya eksotermik dan grafiknya akan seperti yang ditunjukkan pada gambar (1)

Pertanyaannya adalah

Apa perubahan entalpi jika 10 mL HCl 1 M direaksikan dengan 20 mL 1 M NaOH?

Moles HCl = molaritas X volume = 1 M X 10 mL = 10 mmoles

Tahi lalat NaOH bereaksi = mol NaOH

sehingga jumlah mol asam basa bereaksi = 10mmol = air terbentuk

ketika satu mol asam basa bereaksi untuk memberikan satu mol H2O

jadi ketika 10mmole air terbentuk, energi yang dilepaskan adalah = 54 X 10 X 10 ^ -3 = 0,54 kJ


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Answer:

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E = 839 kJ/mol = 839,000 J/mol

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E'=\frac{hc}{\lambda } (Using planks equation)

\lambda =\frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{1.393\times 10^{-18} J}

\lambda =1.427\times 10^{-7} m =142.7 nm = 143 nm

(1 m = 10^9 nm)

The maximum wavelength of light for which a carbon-carbon triple bond could be broken by absorbing a single photon is 143 nm.

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A collection of coins contains 14 pennies, 16 dimes, and 7 quarters. What is the percentage of pennies in the collection?A colle
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The formula for the calculation of moles is shown below:

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2 years ago
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