Haber process is the large scale manufacture of a ammonia by reacting nitrogen and hydrogen at a ratio of 1:3
Change in hydrogen concentration is 0.45 - 0.16 = 0.29 moles/l
Therefore, the average rate of reaction of hydrogen = 0.29 / 30 = 0.0096
= 0.0096 moles/liter/sec
Answer:
A is the closest thing. You change the composition of the steak. You don't in any of the others.
Explanation:
Usually when you cook something, you are doing something to the composition of the object being cooked. A steak might not be obvious, but boiling an egg should be.
Chopping a tree is something physical. You are removing mass in such a way that the tree will fall. There's nothing chemical about that.
Heating a cup of tea looks like it might be chemical. After all steam is sometimes given off which looks like it is chemical. It's not. The water in the tea is just changing phase.
Drying clothes in a dryer. Again, this looks like something might have changed. After all the mass of the clothes just became less. But all you are doing is separating two masses (leaving one of them behind).
Answer:
C. 0.4.
Explanation:
<em>∵ mole fraction of acetic acid (X acetic acid) = (no. of moles acetic acid)/(total no. of moles) = (no. of moles acetic acid)/(no. of moles of acetic acid + no. of moles of water).</em>
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- no. of moles of acetic acid = 2, no. of moles of water = 3.
- Total no. of moles = no. of moles of acetic acid + no. of moles of water = 2 + 3 = 5.
<em>∴ mole fraction of acetic acid (X acetic acid) = (no. of moles acetic acid)/(total no. of moles) =</em> (2)/(5)<em> = 0.4.</em>
Answer:
pH = 12.33
Explanation:
Lets call HA = butanoic acid and A⁻ butanoic acid and its conjugate base butanoate respectively.
The titration reaction is
HA + KOH ---------------------------- A⁻ + H₂O + K⁺
number of moles of HA : 118.3 ml/1000ml/L x 0.3500 mol/L = 0.041 mol HA
number of moles of OH : 115.4 mL/1000ml/L x 0.400 mol/L = 0.046 mol A⁻
therefore the weak acid will be completely consumed and what we have is the unreacted strong base KOH which will drive the pH of the solution since the contribution of the conjugate base is negligible.
n unreacted KOH = 0.046 - 0.041 = 0.005 mol KOH
pOH = - log (KOH)
M KOH = 0.005 mol / (0.118.3 +0.1154)L = 0.0021 M
pOH = - log (0.0021) = 1.66
pH = 14 - 1.96 = 12.33
Note: It is a mistake to ask for the pH of the <u>acid solutio</u>n since as the above calculation shows we have a basic solution the moment all the acid has been consumed.
Answer:
Lithosphere: Volcanos..
Hydrosphere: Rain falls..
biosphere: Plants..
Explanation:
