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3 years ago

At higher temperatures,

Chemistry

1 answer:

valentinak56 [21]3 years ago

3 0

Answer:

reaction rates increase because reactants move faster, collide more often, and produce more collisions with the required energy of activation.

Explanation:

At higher temperatures, the reactant atoms gain high kinetic energy which decreases the intermolecular forces of attraction between these atoms/molecules.

Hence the rate of collision is very high which decreases the activation energy and products are formed with in a short period of time.

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A student prepares a solution of Potassium Nitrate (KNO3) containing 95g at 40 C. This solution is -

Sholpan [36]

Answer:

unsaturated

Explanation:

4 0

2 years ago

The acid dissociation constant Ka of boric acid (H3BO3) is 5.8 times 10^-10. Calculate the pH of a 4.4 M solution of boric acid.

madam [21]

Answer: The pH of a 4.4 M solution of boric acid is 4.3

Explanation:

$H_3BO_3\rightarrow H^+H_2BO_3^-$

at t=0 cM 0 0

at eqm $c-c\alpha$ $c\alpha$ $c\alpha$

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Give c= 4.4 M and $\alpha$ = ?

$K_a=5.8\times 10^{-10}$

Putting in the values we get:

$5.8\times 10^{-10}=\frac{(4.4\times \alpha)^2}{(4.4-4.4\times \alpha)}$

$(\alpha)=0.000011$

$[H^+]=c\times \alpha$

$[H^+]=4.4\times 0.000011=4.8\times 10^{-5}M$

Also $pH=-log[H^+]$

$pH=-log[4.8\times 10^{-5}]=4.3$

Thus pH of a 4.4 M $H_3BO_3$ solution is 4.3

3 0

4 years ago

Four springs are stretched to the same distance from the equilibrium position. The spring constants are listed in the table. A 2

svp [43]

Answer:

The correct option is;

X, W, Y, Z

Explanation:

The parameters given are;

Spring (S), Spring Constant (N/m)

W, 24

X, 35

Y, 22

Z, 15

The equation for elastic potential energy, $E_e$ , is $E_e = 0.5 \times k \times x^2$

The above equation can also be written as $E_e =\dfrac{1}{2} \times k \times x^2$

Where:

k = The spring constant in (N/m)

x = The spring extension

Therefore, since the elastic potential energy, $E_e$ , of the spring is directly proportional to the spring constant, k, we have the springs with higher spring constant will have higher elastic potential energy, $E_e$ , therefore the correct order is as follows;

X > W > Y > Z

9 0

3 years ago

4Na + O2 2Na2O

beks73 [17]

4
N
a
+
O
2
→
2
N
a
2
O
.
By the stoichiometry of this reaction if 5 mol natrium react, then 2.5 mol
N
a
2
O
should result.
Explanation:
The molecular mass of natrium oxide is
61.98

g
⋅
m
o
l
−
1
. If
5

m
o
l
natrium react, then
5
2

m
o
l
×
61.98

g
⋅
m
o
l
−
1

=

154.95

g
natrium oxide should result.
So what have I done here? First, I had a balanced chemical equation (this is the important step; is it balanced?). Then I used the stoichiometry to get the molar quantity of product, and converted this molar quantity to mass. If this is not clear, I am willing to have another go

5 0

3 years ago

Ba(OH)2 Ba+2 + 2 OH- (dissolved in solution). Which will NOT happen to the equilibrium of this solution as H+ ions are added? H+

prohojiy [21]

Answer:

The reaction will move to the left.

Explanation:

For the reaction:

Ba(OH)₂ = Ba²⁺ + 2OH⁻,

Ba(OH)₂ is dissociated to Ba²⁺ and 2OH⁻.

If H⁺ ions are added to the equilibrium:

H⁺ will combine with OH⁻ to form water.

So, the concentration of OH⁻ will decrease and the equilibrium is disturbed.

According to Le Châtelier's principle: when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.

So, the reaction will move to the right to suppress the effect of decreasing OH⁻ concentration.

The base will dissociate to form more OH⁻ and thus, the quantity of Ba(OH)₂ will decrease.

So, the right choice is: the reaction will move to the left, is the choice that will not happen to the equilibrium.

5 0

3 years ago