Answer:
1) 10.0 moles of NO
2) 25 moles of NaCl
3) 1200 moles of CO2
1. How many moles of nitrogen monoxide can be made using 5.0 moles of oxygen in the following composition reaction?
N2 + O2 → 2NO
For 1 mol N2 we need 1 mol O2 to produce 2 moles of NO
For 5.0 moles of N2 we need 5.0 moles of O2 to produce 10.0 moles of NO
2. The neutralization of an acid with a base is a double replacement reaction in which a salt and water are formed. If you start with 25 moles of HCl and neutralize it with NaOH how many moles of NaCl will be formed?
HCl + NaOH → NaCl + H2O
For 1 mol HCl we need 1 mol NaOH to produce 1 mol of NaCl and 1 mol H2O
For 25 moles of HCl we need 25 moles of NaOH to produce 25 moles of NaCl and 25 moles of H2O
3. A car burns gasoline (octane – C8H18) with oxygen. If you drive to Salt Lake and burn 150 moles of octane how many moles of carbon dioxide are you producing?
2C8H18 + 25O2 → 16CO2 + 18H2O
For 2 moles of octane we need 25 moles of O2 to produce 16 moles of CO2 and 18 moles of H2O
For 150 moles of octane we need 25*75 = 1875 moles of O2
To produce 16*75 = 1200 moles of CO2 and 18*75= 1350 moles
Explanation:
Answer: The "Forchhammer" principle
I hope that this helps you !
From the reaction of ammonia and sulfuric acid in aqueous solution
2nh3(aq)+h2so4(aq)<span>→</span> (NH4)2SO4 + H2O
will be formed.Sulfuric acid is diprotic so is able to give up 2 H+ ions.
It is an acid-base neutralisation reaction forming ammonium sulphate as the salt.
2NH3 with H2SO4 reacts in a neutralization reaction to form salt water, with ammonium sulphate left behind to crystallize after evaporation.
The researcher may first weight the beaker with water and then start to heat the water to a constant temperature, for example 30 °C and then start adding salt and stirring. He should add salt slowly until solid salt starts to become visible and the solution starts becoming cloudy. When this happens, he should quickly weigh the beaker. The increase in mass is the mass of salt dissolved at that temperature.
The procedure is then repeated but at an increased temperature until 5-6 temperatures have been tested.
Answer:
year 1 is 5.5%
year 2 is 7.5%
year 3 is 10.2%
Explanation:
since,
length of the transect covered in seaweed / total lenth of transect x 100
then,
0.55 / 10.0 x 100 = 5.5
and
0.75 / 10.0 x 100 = 7.5
and
1.02 / 10.0 x 100 = 10.2
you could also just move the decimal to the right once
:)
