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
> How many valence electrons does a helium atom have?
The electron configuration of Helium is simply 1s2. We see
that its outermost shell is the s shell and it contains 2 electrons, therefore
the number of valence electrons is also 2.
> What is the formula of the ion formed when potassium
achieves noble-gas electron configuration?
Potassium has a electron configuration of [Ar] 4s1. To
have an electron configuration of only [Ar] which is a noble gas, the one electron
from 4s1 should be removed, hence:
<span>K+</span>
Answer:
4–ethyl–2,3–dimethylheptane
Explanation:
To name the compound given above, do the following:
1. Locate the longest continuous carbon chain. This gives the parent name of the compound. In this case, the longest continuous carbon chain is 7. Thus the parent name is Heptane.
2. Identify the substituent group attached to the compound. In this case, the substituent group attached are:
a. Methyl (–CH₃). There are two methyl group attached.
b. Ethyl (–CH₂CH₃)
3. Locate the position of the substituent group attached to the compound by naming alphabetically.
a. The two Methyl (–CH₃) groups are located at carbon 2 and 3
b. The Ethyl (–CH₂CH₃) is located at carbon 4.
NOTE: The position of the Ethyl (–CH₂CH₃) group is the same from both side so we consider the lowest count for the methyl group.
4. Combine the above to obtain the name of the compound.
The name of the compound is:
4–ethyl–2,3–dimethylheptane
Density is given by mass divided by the volume of a substance.
In this case the mass is 0.794 g and the volume is 0.82 cm ×1.45 cm ×1.25 cm = 1.48625 cm³.
Therefore, density = 0.794/ 1.48625
= 0.5342 g/cm³
Thus, the density of lithium metal is 0.5342 g/cm³