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4 years ago

9

1 example of o neutral

1 answer:

katrin [286]4 years ago

5 0

Answer:

NaCl and CO are two neutral compounds as well as these are popular and these Neutral substances are those which have a pH of 7 even they are neither acidic nor basic. The substances which are neither acidic, nor basic and these are called neutral substance. For example distilled water, sugar solution etc. 3.9. 17 votes.

Explanation:

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How many half-lives are required for the concentration of reactant to decrease to 1.56% of its original value?4247.56.56

neonofarm [45]

Answer:

6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.

Explanation:

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given:

Concentration is decreased to 1.56 % which means that 0.0156 of $[A_0]$ is decomposed. So,

$\frac {[A_t]}{[A_0]}$ = 0.0156

Thus,

$\frac {[A_t]}{[A_0]}=e^{-k\times t}$

$0.0156=e^{-k\times t}$

kt = 4.1604

The expression for the half life is:-

Half life = 15.0 hours

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

Where, k is rate constant

So,

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

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<u>6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.</u>

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4 years ago

A sample of a gas at 25°C has a volume of 150 mL when its pressure is 0.947 atm. What will the temperature of the gas be at a pr

sertanlavr [38]

Answer: The temperature of the gas at a pressure of 0.987 atm and volume of 144mL is $25.16^0C$

Explanation:

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 0.947 atm

$P_2$ = final pressure of gas = 0.987 atm

$V_1$ = initial volume of gas = 150 ml

$V_2$ = final volume of gas = 144 ml

$T_1$ = initial temperature of gas = $25^0C=(25+273.15)K=298.15K$

$T_2$ = final temperature of gas = ?

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

$\frac{0.947\times 150}{298.15}=\frac{0.987\times 144}{T_2}$

$T_2=298.31K=(298.31-273.15)^0C=25.16^0C$

The temperature of the gas at a pressure of 0.987 atm and volume of 144mL is $25.16^0C$

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