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3 years ago

The rate constant k for a certain reaction is measured at two different temperatures:

Chemistry

1 answer:

bogdanovich [222]3 years ago

8 0

Answer:

Ea=5.5 Kcal/mole

Explanation:

Let rate constant are $K_1$ and $K_2$ at temperature $T_1$ and $T_2$

By using Arrhenius equation at two different two different temperature,

$Log K_1/K_2 =E_a/2.303R*(1/T_2 -1/T_2 );T_1=273+376=649K ;K_1=4.8*10^8;T_2=273+280=553K ;K_2=2.3*10^8;R=2 cal/(mole.K);Log (4.8*10^8)/(2.3*10^8 )=E_a/2.303R*(1/553K-1/649); Log 4.8/2.3=E_a/2.303R*96/358897 ;0.32=E_a/2.303R*96/358897;E_a=(0.32*2.303R*358897)/96;$

By putting value of R=2 cal/mole.K

$E_a=5510.265cal/mole;$

By rounding off upto 2 significant figure;

$E_a=5.5Kcal/mole;$

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Swift has a balloon of gas at 64psi. The volume of the balloon is 6.958L. If the volume of the balloon is reduced to 3L. what wi

aniked [119]

We could use the Boyle's law for gases.

This law states that:

$P_1V_1=P_2V_2$

Which means that at a constant temperature, as pressure increases, volume decreases. (P1 and V1 are the initial pressure and volume and P2 and V2 are the final pressure and volume respectively)

We could replace the values of the problem to get:

$\begin{gathered} P_1=64psi \\ V_1=6.958L \\ P_2=? \\ V_2=3L \end{gathered}$

Now, replacing in the equation:

$\begin{gathered} 64psi\cdot6.958L=P_2\cdot3L \\ P_2=\frac{64psi\cdot6.958L}{3L}=148.44psi \end{gathered}$

Therefore, the new pressure will be 148.44psi.

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1 year ago

Disease caused by virus are more dangerous.Why?

nlexa [21]

Answer:

Explanation:

1. Virus's are hard to detect because of their simple construction.

2. Some mutate very easily.

3. It is hard to isolate the virus and kill it without doing damage to the host.

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3 years ago

A voltaic cell has one half-cell with a Cu bar in a 1.00 M Cu²⁺ salt, and the other half-cell with a Cd bar in the same volume o

Verizon [17]

The $E^o_{cell}$ of the voltaic cell is 0.743 V. The ΔG°of the given cell is -143.377 kJ and K is 1.35 × 10²⁵

<h3>What is a voltaic cell?</h3>

A voltaic cell often called a galvanic cell, is an electrochemical device that produces electricity through spontaneous redox processes.

It is divided into two distinct half-cells. A half-cell is made up of an electrode (a metal strip, M) dissolved in a solution containing Mn⁺ ions. M can be any metal.

A wire from one electrode to the other connects the two half-cells. Additionally, a salt bridge links the two half-cells.

To solve the question, we need to write the equations of two half-cells

At the anode, oxidation occurs

$Cd(s) \rightarrow Cd^{2+}(aq) + 2e^-$

E° = -0.403 V

At the cathode, reduction occurs

$Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$

E° = 0.34 V

Overall reaction:

$Cd(s) + Cu^{2+}(aq) \rightarrow Cd^{2+}(aq) + Cu(s)$

We know

$E^o_{cell} = E^o_{cathode} - E^o_{anode}$

= 0.34 -(-0.403) = 0.743 V

Also,

ΔG° = -nF $E^o_{cell}$

Where, n = no of electrons gained or lost

F = Faraday constant

$E^o_{cell}$ = standard potential

ΔG° = -2×96485×0.743 = -143376.71 J = -143.377 kJ

Also,

ΔG° = -RTlnK

-143.377 = -8.314 × 10-3 × 298 × lnK

lnK = 57.87

K = 1.35 × 10²⁵

Hence, The $E^o_{cell}$ of the voltaic cell is 0.743 V. The ΔG°of the given cell is -143.377 kJ and K is 1.35 × 10²⁵

Learn more about Voltaic cell:

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6 0

1 year ago

Please help me, please i really need it :)

Mekhanik [1.2K]

GGCCATAGGTCCCTTTAGCG

I believe this is correct (I used the complementary base)

4 0

3 years ago

For each of the following balanced chemical equations, calculate how many moles and how many grams of each product would be prod

natima [27]

Answer:

(a)

Moles of ammonium chloride = 0.5 mole

Mass of ammonium chloride formed = 26.7455 g

(b)

Mole of $CS_2$ = 0.125 mole

Mass = Moles * Molar mass = 0.125 * 76.139 g = 9.52 g

Mole of $H_2S$ = 0.25 mole

Mass = Moles * Molar mass = 0.25 * 34.1 g = 8.525 g

Explanation:

(a)

For the first reaction:-

$NH_3_{(g)}+HCl_{(g)}\rightarrow NH_4Cl_{(s)}$

The mole ratio of the reactants = 1 : 1

0.5 moles of ammonia react with 0.5 moles of hydrochloric gas to give 0.5 moles of ammonium chloride

So, Moles of ammonium chloride formed = 0.5 moles

Molar mass of ammonium chloride = 53.491 g/mol

Mass = Moles * Molar mass = 0.5 * 53.491 g = 26.7455 g

(b)

For the first reaction:-

$CH_4_{(g)}+4S_{(s)}\rightarrow CS_2_{(l)}+2H_2S_{(g)}$

The mole ratio of the reactants = 1 : 4

It means

0.5 moles of methane react with 2.0 moles of sulfur to give 0.5 moles of Carbon disulfide and 1.0 moles of hydrogen sulfide gas.

But available moles of S = 0.5 moles

Limiting reagent is the one which is present in small amount. Thus, S is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

4 moles of S produces 1 mole of $CS_2$

Thus,

0.5 moles of S produces $\frac{1}{4}\times 0.5$ mole of $CS_2$

Mole of $CS_2$ = 0.125 mole

Molar mass of $CS_2$ = 76.139 g/mol

Mass = Moles * Molar mass = 0.125 * 76.139 g = 9.52 g

4 moles of S produces 2 moles of $H_2S$

Thus,

0.5 moles of S produces $\frac{2}{4}\times 0.5$ mole of $H_2S$

Mole of $H_2S$ = 0.25 mole

Molar mass of $H_2S$ = 34.1 g/mol

Mass = Moles * Molar mass = 0.25 * 34.1 g = 8.525 g

7 0

3 years ago