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kramer
3 years ago
5

A certain reaction with an activation energy of 205 kj/mol was run at 485 k and again at 505 k . what is the ratio of f at the h

igher temperature to f at the lower temperature?
Chemistry
1 answer:
lapo4ka [179]3 years ago
8 0
Arrhenius' Law relates activation energy, Ea, rate constant, K, and temperature, T as per this equation:

K (T) = A * e ^ (-Ea / RT), where R is the universal constant of gases and A is a constant which accounts for collision frequency..

Then you can find the ration between K's at two different temperatures as:

K1 = A * e ^ (-Ea / RT1)

K2 = A* e ^(-Ea / RT2)

=> K1 / K2 = e ^ { (-Ea / RT1) - Ea / RT2) }

=> K1 / K2 = e ^ {(-Ea/ R ) *( 1 / T1 - 1 T2) }

=> K1 / K2 = e^ { (-205,000 j/mol / 8.314 j/mol*k )* ( 1 / 505K - 1/ 485K) }

=> K1 / K2 = e ^ (2.0134494) ≈ 7.5

Answer: 7.5




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Balanced equation :

Cu(NO₃)₂(aq) + 2KOH(aq) → Cu(OH)₂(s)  + 2KNO₃(aq)

Balancing a chemical equation :

A chemical equation shows us the substances involved in a chemical reaction - the substances that react (reactants) and the substances that are produced (products). In general, a chemical equation looks like this:

                                 Reactant →Product

According to the law of conservation of mass, when a chemical reaction occurs, the mass of the products should be equal to the mass of the reactants. Therefore, the amount of the atoms in each element does not change in the chemical reaction. As a result, the chemical equation that shows the chemical reaction needs to be balanced. A balanced chemical equation occurs when the number of the atoms involved in the reactants side is equal to the number of atoms in the products side.

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3 0
2 years ago
A diver exhales a bubble with a volume of 250 mL at a pressure of 2.4 atm and a temperature of 15 °C. What is the volume of the
alexdok [17]

Answer : The volume of the bubble is, 625 mL

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 2.4 atm

P_2 = final pressure of gas = 1.0 atm

V_1 = initial volume of gas = 250 mL

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 15^oC=273+15=288K

T_2 = final temperature of gas = 27^oC=273+27=300K

Now put all the given values in the above equation, we get:

\frac{2.4atm\times 250mL}{288K}=\frac{1.0atm\times V_2}{300K}

V_2=625mL

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7 0
3 years ago
Explain how impurities are selected for doping in group 14 semiconductors
aleksklad [387]

Impurities selection for doping in group 14 semiconductors is: based on their ability to add more holes and fewer electrons or to add more electrons and reduce the holes.

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A semiconductor can be used as either a conductor or an insulator when worked upon.

In conclusion, Impurities selection for doping in group 14 semiconductors is: based on their ability to add more holes and fewer electrons or to add more electrons and reduce the holes.

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6 0
2 years ago
The Average Speed of the orbiting space shuttle is
maxonik [38]

Explanation:

It is given that, the Average Speed of the orbiting space shuttle is  17500 miles/hour.

We need to convert the speed in kilometers/ second

We know that,

1 mile = 1.609 km

or

1 km = 0.621 miles

1 hour = 3600 seconds

17500\ \dfrac{\text{miles}}{\text{hour}}=17500\ \dfrac{\text{miles}}{\text{h}}\times \dfrac{1\ h}{3600\ s}\\\\=17500\times \dfrac{\text{miles}}{3600\ s}

Now cancel the miles in numerator.

17500\times \dfrac{\text{miles}}{3600\ s}=17500\times \dfrac{\text{miles}}{3600\ s}\times \dfrac{1.609\ km}{1\ \text{miles}}\\\\=17500\times \dfrac{1.609}{3600}\ km/s\\\\=7.82\ km/s

So, 17500 miles/hour is equal to 7.82 km/s.

8 0
3 years ago
Acetic acid is a weak acid with a pKa of 4.76. What is the concentration of acetate in a buffer solution of 0.2M at pH 4.9. Give
Ghella [55]

Answer:

[base]=0.28M

Explanation:

Hello,

In this case, by using the Henderson-Hasselbach equation one can compute the concentration of acetate, which acts as the base, as shown below:

pH=pKa+log(\frac{[base]}{[acid]} )\\\\\frac{[base]}{[acid]}=10^{pH-pKa}\\\\\frac{[base]}{[acid]}=10^{4.9-4.76}\\\\\frac{[base]}{[acid]}=1.38\\\\

[base]=1.38[acid]=1.38*0.20M=0.28M

Regards.

6 0
3 years ago
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