A(n) ______ is the interaction that allows for atoms to join together.
2 answers:
You might be interested in
1. <em>Balancing nuclear equations </em>
Answer:
Explanation:
The main point to remember in balancing nuclear equations is that the <em>sums of the superscripts</em> (the mass numbers) and the <em>subscripts</em> (the nuclear charges) <em>must balance</em>.
Mass numbers: 37 + 0 = 37; balanced.
Charges: 18 + 1 = 17; balanced
B is <em>wrong</em>. Mass numbers not balanced. 6 +2(1) ≠ 4 + 3.
C is <em>wron</em>g. Mass numbers not balanced. 254 + 4 ≠ 258 + 2(1).
D is <em>wron</em>g. Mass numbers not balanced. 14 + 4 ≠ 17 + 2.
===============
2. <em>Amount remaining </em>
Answer:
D. 5.25 g
Explanation:
The half-life of Th-234 (24 da) is the time it takes for half the Th to decay.
After one half-life, half (50 %) of the original amount will remain.
After a second half-life, half of that amount (25 %) will remain, and so on.
We can construct a table as follows:
No. of Fraction Amount
<u>half-lives</u> <u>t/(da</u>) <u>remaining</u> <u>remaining/g</u>
1 24 ½ 21.0
2 48 ¼ 10.5
3 72 ⅛ 5.25
4 96 ¹/₁₆ 2.62
We see that 72 da is three half-lives, and the amount of Th-234 remaining is 5.25 g.
===============
3.<em> Calculating the half-life </em>
Answer:
a. 2.6 min
Explanation:
The fraction of the original mass remaining is 1.0 g/4.0 g ≈ ¼.
We saw from the previous table that it takes two half-lives to decay to ¼ of the original amount.
2 half-lives = 5.2 min Divide both sides by 2
1 half-life = 5.2 min/2 = 2.6 min
Answer:
(a) The proportion of dry air bypassing the unit is 14.3%.
(b) The mass of water removed is 1.2 kg per 100 kg of dry air.
Explanation:
We can express the proportion of air that goes trough the air conditioning unit as and the proportion of air that is by-passed as
, being
.
The amount of water that goes into the drier inlet has to be 0.004 kg/kg, and can be expressed as:
Replacing the first equation in the second one we have
(b) Of every kg of dry air feed, 85.7% goes in to the air conditioning unit.
It takes (0.016-0.002)=0.014 kg water per kg dry air feeded.
The water removed of every 100 kg of dry air is
It can also be calculated as the difference in humiditiy between the inlet and the outlet: (0.016-0.004=0.012 kgW/kDA) and multypling by the total amount of feed (100 kgDA).
100 * 0.012 = 1.2 kgW