Answer:
18.018 seconds.
Explanation:
Given that the half life of Manganese, Mn = 3 seconds. The initial sample mass = 90.0 gram, the final sample mass = 1.40 gram.
The general idea to the question is to look for the time it will take to decay from the initial mass that is 90 gram to 1.40 gram.
Therefore, we will be making use of the formula below;
J(t) = J(o) × (1/2)^t/t(hL).
Where t(hL) is the half life, t is the time taken, J(t)= mass after time,t and J(o) is the initial mass. So, let us slot in the values into the equation above.
1.4 = 90 × (1/2)^ t/3.
1.4/90 = (1/2)^t/3.
t/3 = log(0.5) (1.4/90).
+Please note that the 0.5 of the log is at the subscript).
That is the base 0.5 logarithm of (1.4/90) 0.01556 is 6.0060141295.
t = 3 × 6.0060141295.
t = 18.018 seconds.
Answer:
It is 20. g HF
Explanation:
H2 + F2 ==> 2HF ... balanced equation
Since the question is asking us to find the mass of product formed, we will want to first convert the molecules of H2 into moles of H2 (we could do this at the end of the calculations, but it's just as easy to do it now).
moles of H2 present (using Avogadro's number):
3.0x1023 molecules H2 x 1 mole H2/6.02x1023 molecules = 0.498 moles H2
From the balanced equation, we see that 1 mole H2 produces 2 moles HF. Therefore, we can now find the theoretical mass of HF produced from 0.498 moles H2:
0.498 moles H2 x 2 moles HF/1 mol H2 = 0.996 moles HF formed.
The molar mass of HF = 20.01 g/mole, thus...
0.996 moles HF x 20.01 g/mole = 19.93 g HF = 20. g HF formed (to 2 significant figures)
It can be done. Normally the boiling point of water is 100°C. It will boil at temperature greater than 100°C more quickly. Water can be boiled at 95°C but for that the atmospheric pressure of the water should be decreased which will decrease the boiling point of water.
<h3>
Concept :</h3>
To boil water at 95°C, decrease the atmospheric pressure.
At 105°C, the water will be boiling quickly than normal at 100°C.
Answer:
one mole of atom of any element contains6.022×1033 atoms regardless of the type of elements the mass of one mole of an element depend on what that element is and is equal to atom mass of that element in gram
Answer:
B
Explanation:
If the student needs one gram but so far only has 0.37 grams, then the amount they need is the difference between what they need and how much they already have. 1-0.37=0.63 grams.
...which isn't actually an option because none of them have decimal points but I would say it is B anyway because it is the equivalent ratio and maybe there was a typo.
Hope this helped!
