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2 years ago

Calculate the Ka for 0.220 M solution of H3AsO4, pH = 1.50

1 answer:

6 0

Remember that the formula to calculare KA is the following
Ka = [H+ ion]*[conj. base] / [acid]
Now that you have the data you can do as the following:
Ka = [H+]*[base] / [acid] 
Ka = [x]*[x] / [acid - x] 
Ka = x^2 / [acid - x]
then we go
Ka = [.032]^2 / [.220 - .032]
Ka = 5.4 x 10^-3.
Or what is the same
Ka = .0054

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In acidic solution, the breakdown of sucrose into glucose and fructose has this rate law: rate = k[H+][sucrose].

Karo-lina-s [1.5K]

Answer:

a)If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

b)If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

c)If concentration of $[H^+]$ is changed to 0.0001 M than rate will be increased by the factor of 0.01.

d) If concentration when [sucrose] and $[H^+]$ both are changed to 0.1 M than rate will be increased by the factor of 1.

Explanation:

Sucrose + $H^+\rightarrow$ fructose+ glucose

The rate law of the reaction is given as:

$R=k[H^+][sucrose]$

$[H^+]=0.01M$

[sucrose]= 1.0 M

$R=k[0.01M][1.0 M]$ ..[1]

The rate of the reaction when [Sucrose] is changed to 2.5 M = R'

$R'=[0.01 M][2.5 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.01 M][2.5 M]}{k[0.01M][1.0 M]}$

$R'=2.5\times R$

If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

The rate of the reaction when [Sucrose] is changed to 0.5 M = R'

$R'=[0.01 M][0.5 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.01 M][0.5 M]}{k[0.01M][1.0 M]}$

$R'=2.5\times R$

If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

The rate of the reaction when $[H^+]$ is changed to 0.001 M = R'

$R'=[0.0001 M][1.0 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.0001 M][1.0M]}{k[0.01M][1.0 M]}$

$R'=0.01\times R$

If concentration of $[H^+]$ is changed to 0.0001 M than rate will be increased by the factor of 0.01.

The rate of the reaction when [sucrose] and $[H^+]$ both are changed to 0.1 M = R'

$R'=[0.1M][0.1M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.1M][0.1M]}{k[0.01M][1.0 M]}$

$R'=1\times R$

If concentration when [sucrose] and $[H^+]$ both are changed to 0.1 M than rate will be increased by the factor of 1.

5 0

2 years ago

Consider the two facts below:

OLEGan [10]

Answer:

A. There is more dissolved oxygen in colder waters than in warm water.

D. If ocean temperature rise, then the risk to the fish population increases.

Explanation:

Conclusion that can be drawn from the two facts stated above:

*Dissolved oxygen is essential nutrient for fish survival in their aquatic habitat.

*Dissolved oxygen would decrease as the temperature of aquatic habit rises, and vice versa.

*Fishes, therefore, would thrive best in colder waters than warmer waters.

The following are scenarios that can be explained by the facts given and conclusions arrived:

A. There is more dissolved oxygen in colder waters than in warm water (solubility of gases decreases with increase in temperature)

D. If ocean temperature rise, then the risk to the fish population increases (fishes will thrive best in colder waters where dissolved oxygen is readily available).

4 0

2 years ago

Freon-12, CF2Cl2, which has been widely used in air conditioning systems, is considered a threat to the ozone layer in the strat

Vilka [71]

Answer:

The root mean squared velocity for CF2Cl2 is $v_{rms}= 207.06 m/s$