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Olegator [25]
4 years ago
9

Nuclear power plants produce a waste product of cesium-137, which has a half-life of 30 years. how long would it take for the ce

sium to decay to 1/8 of its original amount?
Chemistry
1 answer:
artcher [175]4 years ago
7 0
m(final)=m(initial)*( \frac{1}{2} )^{ \frac{time}{half-life} 

m(final)= (1/8) *m(initial)



\frac{1}{8} m(initial)=m(initial)*( \frac{1}{2})^{ \frac{time}{halg-life} } 


\frac{1}{8} = \left( \frac{1}{2} \right)^{\frac{time}{30} }



( \frac{1}{2})^{3}= (\frac{1}{2} )^{ \frac{time}{30}}
\\ \\ 3= \frac{time}{30} 
\\ \\ time = 90 (years)
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A sample of gas is held at 1000C at a volume of 20 L. If the volume is increased to 40 L, what is the new temperature of the gas
kaheart [24]

Answer:

The new temperature will be 2546 K or 2273 °C

Explanation:

Step 1: Data given

The initial temperature = 1000 °C =1273 K

The volume = 20L

The volume increases to 40 L

Step 2: Calculate the new temperature

V1/T1 = V2/T2

⇒with V1 = the initial volume = 20L

⇒with T1 = the initial temperature = 1273 K

⇒with V2 = the increased volume = 40L

⇒with T2 = the new temperature = TO BE DETERMINED

20L/ 1273 K = 40L / T2

T2 = 40L / (20L/1273K)

T2 = 2546 K

The new temperature will be 2546 K

This is 2546-273 = 2273 °C

Since the volume is doubled, the temperature is doubled as well

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Two commonly used sulfonic acids are methanesulfonic acid (CH3SO3H) and trifluoromethanesulfonic acid (CF3SO3H). Which acid's co
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