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2 years ago

9 + (1/2)^4x48 I couldn't find a correct answer for this.

Mathematics

1 answer:

Rufina [12.5K]2 years ago

3 0

$9+ (\dfrac{1}{2} )^4 \times 48 = 9 + \dfrac{1}{16} \times 48 = 9 + 3 = 12$

Answer: 12

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2 years ago

-34x+6=-28 I need to learn how to solve for x

Dimas [21]

Answer:

x=1

Step-by-step explanation:

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2 years ago

Read 2 more answers

Radioactive Decay:

Vadim26 [7]

The question is incomplete, here is the complete question:

The half-life of a certain radioactive substance is 46 days. There are 12.6 g present initially.

When will there be less than 1 g remaining?

Answer: The time required for a radioactive substance to remain less than 1 gram is 168.27 days.

Step-by-step explanation:

All radioactive decay processes follow first order reaction.

To calculate the rate constant by given half life of the reaction, we use the equation:

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

where,

$t_{1/2}$ = half life period of the reaction = 46 days

k = rate constant = ?

Putting values in above equation, we get:

$k=\frac{0.693}{46days}\\\\k=0.01506days^{-1}$

The formula used to calculate the time period for a first order reaction follows:

$t=\frac{2.303}{k}\log \frac{a}{(a-x)}$

where,

k = rate constant = $0.01506days^{-1}$

t = time period = ? days

a = initial concentration of the reactant = 12.6 g

a - x = concentration of reactant left after time 't' = 1 g

Putting values in above equation, we get:

$t=\frac{2.303}{0.01506days^{-1}}\log \frac{12.6g}{1g}\\\\t=168.27days$

Hence, the time required for a radioactive substance to remain less than 1 gram is 168.27 days.

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2 years ago