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forsale [732]
4 years ago
5

What is the concentration of 10.00mL of HBr if it takes 16.73 mL of a 0.253M LiOH solution to neutralize it?

Chemistry
1 answer:
Fudgin [204]4 years ago
3 0
The chemical reaction would be expressed as follows:

HBr + LiOH = LiBr + H2O

We are given the volumes and corresponding concentration to be used for the reaction. We use these values to solve for the concentration of the other reactant. We do as follows:

0.253 mol LiOH / L solution ( 0.01673 L ) ( 1 mol HBr / 1 mol LiOH ) = 0.00423 HBr needed

Concentration of HBr =0.00423mol / .010 L = 0.423 M HBr 
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<u>Answer:</u> The sample of Carbon-14 isotope will take 2377.9 years to decay it to 25 %

<u>Explanation:</u>

The equation used to calculate rate constant from given half life for first order kinetics:

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

where,

t_{1/2} = half life of the reaction = 5730 years

Putting values in above equation, we get:

k=\frac{0.693}{5730yrs}=1.21\times 10^{-4}yrs^{-1}

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

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}

where,

k = rate constant  = 1.21\times 10^{-4}yr^{-1}

t = time taken for decay process = ? yr

[A_o] = initial amount of the sample = 100 grams

[A] = amount left after decay process =  (100 - 25) = 75 grams

Putting values in above equation, we get:

1.21\times 10^{-4}=\frac{2.303}{t}\log\frac{100}{75}\\\\t=2377.9yrs

Hence, the sample of Carbon-14 isotope will take 2377.9 years to decay it to 25 %

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3 years ago
What happens to the atoms in a scoop of ice cream as it melts?
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If a system performs 147 kJ of work while receiving 47 kJ of heat, what is change in its internal energy?
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Answer:

-100 kJ

Explanation:

We can solve this problem by applying the first law of thermodynamics, which states that:

\Delta U = Q-W

where:

\Delta U is the change in internal energy of a system

Q is the heat absorbed/released by the system (it is positive if absorbed by the system, negative if released by the system)

W is the work done by the system (it is positive if done by the system, negative if done on the system)

For the system in this problem we have:

W = +147 kJ is the work done by the system

Q = +47 kJ is the heat absorbed by the system

So , its change in internal energy is:

\Delta U = +47 - (+147) =-100 kJ

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