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3 years ago

Does a set mousetrap that snaps shut potential energy to kinetic energy

Chemistry

1 answer:

Romashka [77]3 years ago

6 0

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Deuterium (D or 2H) is an isotope of hydrogen. The molecule D2 undergoes an exchange reaction with ordinary hydrogen, H2, that l

Paladinen [302]

Answer:

$Kp_2} =2.0$

Explanation:

$K_{p1}= 1.8$

$K_{p2}= ???$

$T_1= 298K$

$T_2 = 415 K$

$\delta H$ = 0.64 kJ/mol = 640 J/mol

R = 8.314 J/mol.K

Using Van't Hoff Equation:

$In (\frac{Kp_2}{Kp_1} )=(-\frac{DH}{R} *(\frac{1}{T_2}-\frac{1}{T_1} )$

$In (\frac{Kp_2}{1.8} )=(-\frac{640}{8.314} *(\frac{1}{415}-\frac{1}{298} )$

$In (\frac{Kp_2}{1.8} )=0.072827$

$(\frac{Kp_2}{1.8} )= e^{0.0728727}$

$\frac{Kp_2}{1.8} =1.075593605$

$Kp_2} =1.075593605*1.8$

$Kp_2} =1.936068489$

$Kp_2} =2.0$

3 0

4 years ago

Read 2 more answers

er solutions can be produced by mixing a weak acid with its conjugate base or by mixing a weak base with its conjugate acid. The

liubo4ka [24]

Answer:

You need to add 19,5 mmol of acetates

Explanation:

Using the Henderson-Hasselbalch equation:

pH = pKa + log₁₀ [base]/[acid]

For the buffer of acetates:

pH = pKa + log₁₀ [CH₃COO⁻]/[CH₃COOH]

As pH you want is 5,03, pka is 4,74 and milimoles of acetic acid are 10:

5,03 = 4,74 + log₁₀ [CH₃COO⁻]/[10]

1,95 = [CH₃COO⁻]/[10]

<em>[CH₃COO⁻] = 19,5 milimoles</em>

Thus, to produce an acetate buffer of 5,03 having 10 mmol of acetic acid, you need to add 19,5 mmol of acetates.

I hope it helps!

7 0

4 years ago

Is there any evidence that methanol forms ions either in the pure state or when dissolved in water?

worty [1.4K]

According to research, the answer is no. Although alcohol is dissolves in water solution but dissociation of methyl alcohol in into ions is barely not possible. Thus, methanol is considered as a non-electrolyte.

3 0

3 years ago

By referring to particles, explain why the white ring formed in the glass tube

Leya [2.2K]

Answer:

Ammonia and Hydrogen Chloride.

Explanation:

Assuming this is what you're referring to, Ammonia (NH3) and Hydrogen Chloride(NCI3). The concentrated ammonia (NH3) is placed on a pad in one end of a tube and hydrochloric acid (NCI3) on a pad at the other. Shortly after the gases will begin diffusing far enough to meet, a ring of solid ammonium chloride (NH4Cl) will be formed.

7 0

3 years ago

For the reaction: P4 + 6C12 – 4 PC13

bagirrra123 [75]

Answer:

332.3g

Explanation:

hope this helps loves x

5 0

3 years ago