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3 years ago

What does an increase in temperature do to the reaction rate?

Chemistry

1 answer:

Goshia [24]3 years ago

8 0

Answer:

An increase in temperature typically increases the rate of reaction because an increase in temperature will raise the average kinetic energy of the reactant molecules. Therefore, a greater proportion of molecules will have the minimum energy necessary for an effective collision.Increasing the temperature increases reaction rates because of the disproportionately large increase in the number of high energy collisions. It is only these collisions (possessing at least the activation energy for the reaction) which result in a reaction.

Explanation:

Can you please give this answer a brainliest answer please

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What is the maximum mass of ethyl alcohol you could boil with 1000 j of heat, starting from 22 ∘c?

kramer

solution:

1000 = m*2400*(78-22) + m*8.79*10^5

1000= 134400m + 879000m

1000= 1030200m

m = 1000/1013400

m= 1013.4 grams

the final answer is 0.9706 grams

5 0

3 years ago

A reaction A(aq)+B(aq)↽−−⇀C(aq) has a standard free‑energy change of −4.20 kJ/mol at 25 °C. What are the concentrations of A, B,

Dafna1 [17]

Answer : The concentration of $A,B\text{ and }C$ at equilibrium are 0.132 M, 0.232 M and 0.168 M respectively.

Explanation :

The given chemical reaction is,

$A(aq)+B(aq)\rightleftharpoons C(aq)$

First we have to calculate the equilibrium constant for the reaction.

The relation between the equilibrium constant and standard free‑energy is:

$\Delta G^o=-RT \ln k$

where,

$\Delta G^o$ = standard free‑energy change = -4.20 kJ/mole

R = universal gas constant = 8.314 J/mole.K

k = equilibrium constant = ?

T = temperature = $25^oC=273+25=298K$

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

$-4.20kJ/mole=-(8.314J/mole.K)\times (298K) \ln k$

$k=5.45$

Now we have to calculate the concentrations of A, B, and C at equilibrium.

The given equilibrium reaction is,

$A(aq)+B(aq)\rightleftharpoons C(aq)$

Initially 0.30 0.40 0

At equilibrium (0.30-x) (0.40-x) x

The expression of equilibrium constant will be,

$k=\frac{[C]}{[A][B]}$

$5.45=\frac{x}{(0.30-x)\times (0.40-x)}$

By solving the term x, we get

$x=0.168\text{ and }0.716$

From the values of 'x' we conclude that, x = 0.716 can not more than initial concentration. So, the value of 'x' which is equal to 0.716 is not consider.

The value of x will be, 0.168 M

The concentration of $A$ at equilibrium = (0.30-x) = 0.30 - 0.168 = 0.132 M

The concentration of $B$ at equilibrium = (0.40-x) = 0.40 - 0.168 = 0.232 M

The concentration of $C$ at equilibrium = x = 0.168 M

3 0

4 years ago

If an element is more reactive, is it more likely to be found as an element or a compound?

laila [671]

It is more likely to be found as a COMPOUND, as it is more reactive, by the time we found them, they're already reacted with other elements or compounds to form new compounds.
Example is oxygen, it is very reactive, therefore we often found oxygen in water, which is H2O, in earth, instead of just pure oxygen.

5 0

3 years ago

What is the mass in grams of 1.00*10^12 lead atoms

pav-90 [236]

34.5 X 10^-11 grams of lead

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3 years ago

How many energy levels does sulfur have?<br> A. 2<br> B. 3<br> C. 4<br> D. 14

Karolina [17]

Its electronic confuguration is 2.8.6 so it occupies 3 shells.

7 0

4 years ago