All categories

2 years ago

What are the units for the measurements of radioactive element activity?

Chemistry

1 answer:

torisob [31]2 years ago

7 0

Answer:

Explanation:

The number of decays per second, or activity, from a sample of radioactive nuclei is measured in becquerel (Bq), after Henri Becquerel. One decay per second equals one becquerel.

You might be interested in

Please help me to do this assignment

kirill [66]

Answer:

1. Objective

2. Objective

3. Opinion

4. Objective

5. Opinion

6. Opinion

7. Opinion (I think.)

8. Opinion (I think.)

Explanation:

7 0

2 years ago

3. If the percent by volume is 2.0% and the volume of solution is 250 mL, what is the volume of solute in solution? (1 point) 0.

igor_vitrenko [27]

Volume percent = Volume of solute
----------------------------------
Volume of the solution

   2 Volume of the solute
------- =   ------------------------------
100 250

Volume of the solute = 2 x 250
------------
100

= 5 mL.

Hope this helps!

4 0

2 years ago

Which descriptions apply to emitted radiation? check all that apply?

Murrr4er [49]

<span>measurement in Ci/Bq

the amount of radioactive materials released into the environment.

number of disintegrations of radioactive atoms in a radioactive material over a period of time</span>

7 0

2 years ago

The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process. In the first step, manganese(I

ale4655 [162]

Answer:

The answer is "6.52 kg and 13.1 kg"

Explanation:

For point a:

$Actual\ yield = 6.52 \ kg\\\\Percent \ yield= 66\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical\ yield} \times 100 \%\\\\Theoretical\ yield \ of \ MnO_2 = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\=\frac{6.52 \ kg}{66 \%} \times 100\% =9.88 \ kg\\\\$

Equation:

$3MnCO_3 +O_2 \longrightarrow 2MnO_2 + 2CO_2\\\\$

Calculating the amount of $MnCO_3$

$= 9.88 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ MnO_2}{86.94 \ g} \times \frac{2 \ Mol \ MnCO_3}{2 \ mol \ MnO_2} \times \frac{114.95\ g}{1 \ mol \ MnCO_3 }\times \frac{1\ kg}{1000\ g}\\\\= 13.1 \ kg$

For point b:

$Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield} \times 100 \% \\\\Theoretical \ yield\ of\ Mn = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\$

$=\frac{4.0 \ kg}{97.0\%} \times 100\% =4.12 \ kg$

Equation:

$3MnO_2 +4AL \longrightarrow 3Mn + 2AL_2O_3\\\\$

Calculating the amount of $MnO_2:$

$= 4.12 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ Mn}{54.94 \ g} \times \frac{3 \ Mol \ MnO_2}{3 \ mol \ Mn} \times \frac{86.94 \ g}{1 \ mol \ MnO_2 }\\\\= 6516 \ g \\\\=6.52 \ kg\\\\$

7 0

2 years ago

17.0 g of Cl_2 has a volume of 9.22 L at 17°C. What is its pressure?

TiliK225 [7]

Answer:

17.0 g of hask2 lahhwle 11c u2b8ss

7 0

2 years ago