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horsena [70]
4 years ago
11

A heterogeneous mixture is a

Chemistry
2 answers:
11111nata11111 [884]4 years ago
7 0
A heterogeneous mixture is a mixture in which its components retain their identity. The correct answer is B. 
vladimir2022 [97]4 years ago
6 0
Heterogenous mixtures are those in which solute and solvents are in different states so option B is right answer.
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If the number of moles of a gas initially contained in a 2.10 l vessel is doubled, what is the final volume of the gas in liters
Anon25 [30]
Avagadro's law gives the relationship between volume of gas and number of moles of the gas. When the temperature and pressure are constant, volume is directly proportional to number of moles of gas.
\frac{V1}{n1} =  \frac{V2}{n2}
V -volume and n - number of moles of gas.
parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation 
initial number of moles n1 - N
n2 - 2N as the number of moles are doubled 
V1 - initial volume - 2.10 L
V2 - final volume - V
substituting the values in the equation 
\frac{2.10L}{N}  \frac{V}{2N}
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3 years ago
g The decomposition reaction of A to B is a first-order reaction with a half-life of 2.42×103 seconds: A → 2B If the initial con
Tanzania [10]

Answer:

In 23.49 minutes the concentration of A to be 66.8% of the initial concentration.

Explanation:

The equation used to calculate the constant for first order kinetics:

t_{1/2}=\frac{0.693}{k}} .....(1)

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

t=\frac{2.303}{k}\log\frac{[A_o]}{[A]} ......(2)

where,  

k = rate constant

t_{1/2} =Half life of the reaction = 2.42\times 10^3 s

t = time taken for decay process = ?

[A_o] = initial amount of the reactant = 0.163 M

[A] = amount left after time t =  66.8% of [A_o]

[A]=\frac{66.8}{100}\times 0.163 M=0.108884 M

k=\frac{0.693}{2.42\times 10^3 s}

t=\frac{2.303}{\frac{0.693}{2.42\times 10^3 s}}\log\frac{0.163 M}{0.108884 M}

t = 1,409.19 s

1 minute = 60 sec

t=\frac{1,409.19 }{60} min=23.49 min

In 23.49 minutes the concentration of A to be 66.8% of the initial concentration.

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