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2 years ago

HELP!! PLEASE!!!!!

Mathematics

2 answers:

aleksandr82 [10.1K]2 years ago

5 0

Answer:

Isolate the variable by dividing each side by factors that don't contain the variable.

n=5, -5

8090 [49]2 years ago

4 0

Answer:

$n=\frac{3i\sqrt{22}-2}{-8}$

Step-by-step explanation:

$(2-8n)^2=-198$

$\sqrt{(2-8n)^2} =\sqrt{-198}$

$2-8n=i\sqrt{198}$

$2-8n=3i\sqrt{22}$

$-8n=3i\sqrt{22} -2$

$n=\frac{3i\sqrt{22}-2}{-8}$

I hope this helps!

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Modeling Radioactive Decay In Exercise, complete the table for each radioactive isotope.

Travka [436]

Answer :

The amount after 1000 years will be, 5.19 grams.

The amount after 10000 years will be, 0.105 grams.

Step-by-step explanation :

Half-life = 1599 years

First we have to calculate the rate constant, we use the formula :

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

$k=\frac{0.693}{1599\text{ years}}$

$k=4.33\times 10^{-4}\text{ years}^{-1}$

Now we have to calculate the amount after 1000 years.

Expression for rate law for first order kinetics is given by:

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

where,

k = rate constant = $4.33\times 10^{-4}\text{ years}^{-1}$

t = time passed by the sample = 1000 years

a = initial amount of the reactant = 8 g

a - x = amount left after decay process = ?

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

$1000=\frac{2.303}{4.33\times 10^{-4}}\log\frac{8}{a-x}$

$a-x=5.19g$

Thus, the amount after 1000 years will be, 5.19 grams.

Now we have to calculate the amount after 10000 years.

Expression for rate law for first order kinetics is given by:

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

where,

k = rate constant = $4.33\times 10^{-4}\text{ years}^{-1}$

t = time passed by the sample = 10000 years

a = initial amount of the reactant = 8 g

a - x = amount left after decay process = ?

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

$10000=\frac{2.303}{4.33\times 10^{-4}}\log\frac{8}{a-x}$

$a-x=0.105g$

Thus, the amount after 10000 years will be, 0.105 grams.

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