Answer:
scientists often communicate their research results in three general ways:
1) One is to publish their results in peer-reviewed journals that can be ready by other scientists.
2) Two is to present their results at national and international conferences where other scientists can listen to presentations
Explanation:
Answer:
0.005 mol
Explanation:
Moles is denoted by given mass divided by the molecular mass ,
Hence ,
n = w / m
n = moles ,
w = given mass ,
m = molecular mass .
From the question ,
w = given mass of Gold = 1.05 g ,
m = molecular mass of Gold = 197 g/mol
<u>Hence , moles can be calculated as -</u>
n = w / m
= 1.05 g / 197 g/mol = 0.005 mol
Since
potassium and phosphate is what we are to find for and they are both found in
the potassium phosphate solution, therefore we solve for this one first on the
basis of the phosphate.
The formula
for finding the volume given the concentration and number of moles is:
Volume =
number of moles / concentration in Molarity
Volume
potassium phosphate required = 30 mmol phosphate / (3 mmol / mL)
<u>Volume
potassium phosphate required = 10 mL</u>
This would
also contain potassium in amounts of:
Amount of
potassium in potassium phosphate = 10 mL (4.4 meg / mL)
Amount of
potassium in potassium phosphate = 44 meg
Therefore
the potassium chloride required is:
Volume of
potassium chloride = (80 meg – 44 meg) / (2 meg / mL)
<span><u>Volume of
potassium chloride = 72 mL</u></span>
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
KE = 1/2 * m * v^2
KE = 1/2 * 130 * 23^2
KE = 34385J
