They didnt have a powerfull enough microscope to see them an when they could they were moving to fast unless they were cold
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Answer:</h3>
0.248 g
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Explanation:</h3>
<u>We are given;</u>
- Time as 40 minutes
- Original amount of N-13
We are required to calculate the amount left after decay.
- We need to know the half life of N-13 first;
- Half life of N-13 is 9 minutes 58 seconds (9.965 minutes0
We are going to use the formula;
- Remaining amount = Original amount × 0.5^n
That is; N = N₀ × 0.5^n , where n is the number of half lives.
n = 40 minutes ÷ 9.965 minutes
= 4.014
Therefore;
N = 4 g × 0.5^4.014
= 0.248 g
Therefore, the mass of Nitrogen-13 that will be left is 0.248 g
The partial pressure of nitrogen in the problem is 725 - 231 = 494 mmHg.
494 / 760 = 0.650 atm. To continue, we will use the ideal gas equation to find the mole of N2 and from that, we can find the mass of N2.
PV = nRT ----> n= PV/RT
Find the mole of Nitrogen
Given P=0.650 atm
n= (0.650)x(255/1000) / (0.082 x 338)
= (0.650) (0.255) / 27.716
= 0.16575 / 27.716
= 0.0060 moles
Find the mass of nitrogen
G= n x Mr (N2)
= 0.0060 x 28
= 0.168 grams
So the mass of nitrogen is 0.168
When a warm air mass and a cold air mass meet and cant move its called "stationary front"
