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1 year ago

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Chemistry

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3241004551 [841]1 year ago

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No is the answer your welcome well I don’t.

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in a certain solvent is first order with respect to (CH3)3CBr and zero order with respect to OH2. In several experiments the rat

Inga [223]

Answer:

Ea= -175.45J

A= 3.5×10^14

k=3.64 ×10^14 s^2.

Explanation:

From

ln k= -(Ea/R) (1/T) + ln A

This is similar to the equation of a straight line:

y= mx + c

Where m= -(Ea/R)

c= ln A

y= ln k

Therefore

21.10 3 104= -(Ea/8.314)

Ea=-( 21.10 3 104×8.314)

Ea= -175.45J

b) ln A= 33.5

A= e^33.5

A= 3.5×10^14

k= Ae^-Ea/RT

k= 3.5×10^14 × e^ -(-175.45/8.314×531)

k = 3.64 ×10^14 s^2.

8 0

2 years ago

What causes interstellar dust and clouds of gas to condense into planets and stars?

aleksklad [387]

The gravitational pull

3 0

2 years ago

Be sure to answer all parts. Calculate the pH of the following two buffer solutions: (a) 1.4 M CH3COONa/1.6 M CH3COOH. (b) 0.1 M

aalyn [17]

Answer:

a) pH = 4.68 (more effective)

b) pH =4.44.

Explanation:

The pH of buffer solution is obtained by Henderson Hassalbalch's equation.

The equation is:

$pH =pKa +log\frac{[salt]}{[acid]}$

a) pKa of acetic acid = 4.74

[salt] = [CH₃COONa] = 1.4 M

[acid] = [CH₃COOH] = 1.6 M

$pH = 4.74 + log \frac{1.4}{1.6}= 4.68$

This is more effective as there is very less difference in the concentration of salt and acid.

b) pKa of acetic acid = 4.74

[salt] = [CH₃COONa] = 0.1 M

[acid] = [CH₃COOH] = 0.2 M

$pH = 4.74 + log \frac{0.1}{0.2}= 4.44$

3 0

2 years ago

When Luis burns wood in a campfire, the particles in the wood break down and energy is released. What type of energy does the wo

matrenka [14]

Answer:

chemical energy is stored in the wood before combustion.

8 0

2 years ago

The following question appears on a quiz: ""You fill a tank with gas at 60Â°C to 100 kPa and seal it. You decrease the temperatu

dusya [7]

Answer: The final pressure will decrease ad the value is 85 kPa

Explanation:

To calculate the final pressure of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1\text{ and }T_1$ are the initial pressure and temperature of the gas.

$P_2\text{ and }T_2$ are the final pressure and temperature of the gas.

We are given:

$P_1=100kPa\\T_1=60^0C=(60+273)K=333K\\P_2=?\\T_2=10^0C=(10+273)K=283K$

Putting values in above equation, we get:

$\frac{100kPa}{333K}=\frac{P_2}{283K}\\\\P_2=85kPa$

Hence, the final pressure will decrease ad the value is 85 kPa

8 0

2 years ago

Anyone want to talk ??​

Anyone want to talk ??