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nordsb [41]
3 years ago
5

The half–life of rubidium–89 is 15 minutes. If the initial mass of the isotope is 250 grams, how many grams will be left after 1

00 minutes?
Mathematics
2 answers:
nikitadnepr [17]3 years ago
8 0

Answer:

32.5 grams  will be left after 100 minutes.                        

Step-by-step explanation:

Given : The half–life of rubidium–89 is 15 minutes. If the initial mass of the isotope is 250 grams.

To find : How many grams will be left after 100 minutes?          

Solution :

Let the exponential equation of rubidium is Q=Q_oe^{rt}

Where, Q_o=250 is the initial value

t is the time taken i.e. t=15 minutes

The half–life of rubidium–89 is 15 minutes.

i.e. Q=\frac{Q_o}{2}

Substitute in the formula,

\frac{Q_o}{2}=Q_oe^{r\times 15}

\frac{1}{2}=e^{r\times 15}

Taking log both side,

\log (\frac{1}{2})=r\times 15

-0.301=r\times 15

\frac{-0.301}{15}=r

r=-0.02

Now, we have to find Q in 100 minutes,

Q=Q_oe^{rt}

Substitute in the formula,

Q=250e^{-0.02\times 100}

Q=250e^{-2}

Q=250\times 0.13

Q=32.5

Therefore, 32.5 grams  will be left after 100 minutes.

irina1246 [14]3 years ago
5 0
M(100) = 2.46 grams    . 
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Step-by-step explanation:

1.

Simplify the expression by combining like terms. Remember, like terms have the same variable part, to simplify these terms, one performs operations between the coefficients. Please note that a variable with an exponent is not the same as a variable without the exponent. A term with no variable part is referred to as a constant, constants are like terms.

2h^2-7h+2h^2-h+6+4h-9h

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Use a very similar method to solve this problem as used in the first. Please note that all of the rules mentioned in the first problem also apply to this problem; for that matter, the rules mentioned in the first problem generally apply to any pre-algebra problem.

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Use the same rules as applied in the first problem. Also, keep the distributive property in mind. In simple terms, the distributive property states the following (a(b+c)=(a)(b)+(a)(c)=ab+ac). Also note, a term raised to an exponent is equal to the term times itself the number of times the exponent indicates. In the event that the term raised to an exponent is a constant, one can simplify it. Apply these properties here,

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The same method used to solve problem (3) can be applied to this problem.

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