Answer:

Explanation:
Hello,
In this case, we need to remember that for the required time for a radioactive nuclide as radium-226 to decrease to one half its initial amount we are talking about its half-life. Furthermore, the amount of remaining radioactive material as a function of the half-lives is computed as follows:

Therefore, for an initial amount of 100 mg with a half-life of 1590 years, after 1000 years, we have:

Best regards.
You must add 7.5 pt of the 30 % sugar to the 5 % sugar to get a 20 % solution.
You can use a modified dilution formula to calculate the volume of 30 % sugar.
<em>V</em>_1×<em>C</em>_1 + <em>V</em>_2×<em>C</em>_2 = <em>V</em>_3×<em>C</em>_3
Let the volume of 30 % sugar = <em>x</em> pt. Then the volume of the final 20 % sugar = (5 + <em>x</em> ) pt
(<em>x</em> pt×30 % sugar) + (5 pt×5 % sugar) = (<em>x</em> + 5) pt × 20 % sugar
30<em>x</em> + 25 = 20x + 100
10<em>x</em> = 75
<em>x</em> = 75/10 = 7.5
Answer:
2 moles
Explanation:
Hi there !!
Given volume = 44.8 litres
Molar volume = 22.4 litres
No: of moles = Given volume / Molar volume
= 44.8 / 22.4 = 2 moles
Answer:
0.35 milli moles of ethanol can be theoretically be produced under these conditions.
Explanation:

Moles of glucose =
milli mole
Moles of ADP = 0.35 milli mole
Moles of Pi = 0.35 milli mole
Moles of ATP = 0.70 milli mole
As we can see that ADP and Pi are in limiting amount which means tat they are limiting reagent. So, the moles of ethanol produced will depend upon the moles of ADP and Pi.
According to reaction, 2 moles of ADP gives 2 moles of glucose.
Then 0.35 milli moles of ADp will give :
of ethanol
0.35 milli moles of ethanol can be theoretically be produced under these conditions.