How much energy would be required to vaporize 45.0 grams of water?

Suppose the half-life is 9.0 s for a first order reaction and the reactant concentration is 0.0741 M 50.7 s after the reaction s

bazaltina [42]

<u>Answer:</u> The time taken by the reaction is 84.5 seconds

<u>Explanation:</u>

The equation used to calculate half life for first order kinetics:

$k=\frac{0.693}{t_{1/2}}$

where,

$t_{1/2}$ = half-life of the reaction = 9.0 s

k = rate constant = ?

Putting values in above equation, we get:

$k=\frac{0.693}{9}=0.077s^{-1}$

Rate law expression for first order kinetics is given by the equation:

$k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}$ ......(1)

where,

k = rate constant = $0.077s^{-1}$

t = time taken for decay process = 50.7 sec

$[A_o]$ = initial amount of the reactant = ?

[A] = amount left after decay process = 0.0741 M

Putting values in equation 1, we get:

$0.077=\frac{2.303}{50.7}\log\frac{[A_o]}{0.0741}$

$[A_o]=3.67M$

Now, calculating the time taken by using equation 1:

$[A]=0.0055M$

$k=0.077s^{-1}$

$[A_o]=3.67M$

Putting values in equation 1, we get:

$0.077=\frac{2.303}{t}\log\frac{3.67}{0.0055}\\\\t=84.5s$

Hence, the time taken by the reaction is 84.5 seconds

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maria [59]

Answer:

A change in form or appearance of matter

Explanation:

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What kind of energy is required for photosynthesis to occur

Makovka662 [10]

Answer:

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How many grams of sulfuric acid is needed to neutralize 380 ml of solution with pH = 8.94

erma4kov [3.2K]

Answer : The mass of sulfuric acid needed is $16.23\times 10^{-5}g$ .

Solution : Given,

pH = 8.94

Volume of solution = 380 ml = $380\times 10^{-3}$ $(1ml=10^{-3}L)$

Molar mass of sulfuric acid = 98.079 g/mole

As we know,

$pH+pOH=14\\pOH=14-8.94=5.06$

$pOH=-log[OH^-]$

$5.06=-log[OH^-]$

$[OH^-]=0.00000871=8.71\times 10^{-6}mole/L$

Now we have to calculate the moles of $OH^-$ .

Formula used : $Moles=Concentration\times Volume$

$\text{ Moles of }[OH^-]=\text{ Concentration of }[OH^-]\times Volume\\\text{ Moles of }[OH^-]=(8.71\times 10^{-6}mole/L)\times (380\times 10^{-3}L)=3309.8\times 10^{-9}moles$