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I am Lyosha [343]
3 years ago
5

Which statements best describe half-lives of radioactive isotopes? Check all that apply.

Chemistry
2 answers:
AURORKA [14]3 years ago
5 0

Answer: The half-life varies depending on the isotope.  Half-lives range from fractions of a second to billions of years.   The half-life of a particular isotope is constant.  

Explanation: On Edgenuity!!!!

Igoryamba3 years ago
3 0

Answer:

The half-life varies depending on the isotope.

Half-lives range from fractions of a second to billions of years.

The half-life of a particular isotope is constant.

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How many oxygen atoms are in 3.70 g of quartz?
Ad libitum [116K]

Answer:

The formula of Quartz is SiO2. So, 1 mole of Quartz will have 2 moles of oxygen atoms, i.e. 2 x 6.022 x 10^23 atoms of oxygen. The molar mass of quartz is 60 g per mol. So, 60 g quartz means 1 mole quartz.

Explanation:

3 0
3 years ago
Five million gallons per day (MGD) of wastewater, with a concentration of 10.0 mg/L of a conservative pollutant, is released int
hjlf

Answer:

a) The concentration in ppm (mg/L) is 5.3 downstream the release point.

b) Per day pass 137.6 pounds of pollutant.  

Explanation:

The first step is to convert Million Gallons per Day (MGD) to Liters per day (L/d). In that sense, it is possible to calculate with data given previously in the problem.  

Million Gallons per day 1 MGD = 3785411.8 litre/day = 3785411.8 L/d

F_1 = 5 MGD (\frac{3785411.8 L/d}{1MGD} ) = 18927059 L/d\\F_2 =10 MGD (\frac{3785411.8 L/d}{1MGD} )= 37854118 L/d

We have one flow of wastewater released into a stream.  

First flow is F1 =5 MGD with a concentration of C1 =10.0 mg/L.

Second flow is F2 =10 MGD with a concentration of C2 =3.0 mg/L.  

After both of them are mixed, the final concentration will be between 3.0 and 10.0 mg/L. To calculate the final concentration, we can calculate the mass of pollutant in total, adding first and Second flow pollutant, and dividing in total flow. Total flow is the sum of first and second flow. It is shown in the following expression:  

C_f = \frac{F1*C1 +F2*C2}{F1 +F2}

Replacing every value in L/d and mg/L

C_f = \frac{18927059 L/d*10.0 mg/L +37854118 L/d*10.0 mg/L}{18927059 L/d +37854118 L/d}\\C_f = \frac{302832944 mg/d}{56781177 L/d} \\C_f = 5.3 mg/L

a) So, the concentration just downstream of the release point will be 5.3 mg/L it means 5.3 ppm.

Finally, we have to calculate the pounds of substance per day (Mp).  

We have the total flow F3 = F1 + F2 and the final concentration C_f. It is required to calculate per day, let's take a time of t = 1 day.  

F3 = F2 +F1 = 56781177 L/d \\M_p = F3 * t * C_f\\M_p = 56781177 \frac{L}{d} * 1 d * 5.3 \frac{mg}{L}\\M_p = 302832944 mg

After that, mg are converted to pounds.  

M_p = 302832944 mg (\frac{1g}{1000 mg} ) (\frac{1Kg}{1000 g} ) (\frac{2.2 lb}{1 Kg} )\\M_p = 137.6 lb

b) A total of 137.6 pounds pass a given spot downstream per day.

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3 years ago
Which two characteristics describe all animals
Maru [420]

All animals can be dangerous and they would fight for their family. (This might be wrong)

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3 years ago
How do I do this and what are the answers?
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I cannot see your question to help you... sorry
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