Answer:
Protons are positive, electrons are negative, neutrons have to charge. Most of an Atoms mass comes from its protons and neutrons
Explanation:
I took some notes in my notebook about Atoms.
Answer:
A. 0.90 L.
Explanation:
- NaOH solution will react with H₂SO₄ according to the balanced reaction:
<em>H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O.</em>
<em>1.0 mole of H₂SO₄ reacts with 2.0 moles of NaOH.</em>
- For NaOH to react completely with H₂SO₄, the no. of millimoles should be equal.
<em>∴ (MV) NaOH = (xMV) H₂SO₄.</em>
x for H₂SO₄ = 2, due to having to reproducible H⁺ ions.
<em>∴ V of NaOH = (xMV) H₂SO₄/ M of NaOH</em> = 2(0.6 L)(3.0 M)/(4.0 M) = <em>0.90 L.</em>
A is obviously out because it leads to a volume of 125.0 milliliters of the new solution and gives you a lower concentration than you were aiming for.
D is out because you are adding 75 milliliters of the stock solution, so your concentration would be too high. You only need 25.0 milometers of stock solution per 100 milliliters of the new solution.
C is also out because it leads to 50.0 milliliters stock solution per 100 milliliters of the new solution and hence the wrong concentration.
B is by default the correct answer. It also details the correct technique. First you add the stock solution (This you know from your calculations to be 25 milliliters.) then you add the water up to the volume you needed. (Because the calculations only tell you the total volume of water not what you need to add) You also add the water last so you can rinse the neck of the flask to make sure you also get all the stock solution residue into the stock solution.
I would add the final step of stirring, but B is the only answer that can be correct.
Answer:
fluorine plus chlorides think
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
