The volume of CO= 2.912 L
<h3>Further explanation
</h3>
Given
Dehydration of 6 g of methanoic acid with H2SO4
Required
The volume of CO
Solution
Reaction
HCOOH ⇒ CO+H₂O
MW = 1.Ar C+2.Ar H+2.Ar O
MW = 1.12+2.1+2.16
MW= 46 g/mol
mol = mass : MW
mol = 6 g : 46 g/mol
mol = 0.13
From equation, mol ratio HCOOH : CO, so mol CO = 0.13
Vm = 22.4 L
so for 0.13 mol :
= 0.13 x 22.4 L
= 2.912 L
You can use the mole ratios given by the coefficients in the balanced chemical equation. The coefficients are the smallest whole-number number of moles.
<span>Aluminum hydroxide is insoluble in water, while calcium sulfate is only marginally soluble. </span>
<span>Al2(SO4)3(aq) + 3Ca(OH)2(aq) --> 2Al(OH)3(s) + 3CaSO4(s) </span>
<span>1 mol ................. 3 moles ............... 2 moles ......... 3 moles </span>
<span>0.725 mol </span>
<span>0.725 mol ......... 3(0.725) mol ........ 2(0.725) mol ..3(0.725) mol </span>
<span>Al2(SO4)3(aq) + 3Ca(OH)2(aq) --> 2Al(OH)3(s) + 3CaSO4(s) </span>
<span>1 mol ................. 3 moles ............... 2 moles ......... 3 moles </span>
<span>........ .................. 0.774 mol </span>
<span>0.774 / 3 mol ..... 0.774 mol .......... 2/3(0.774)mol.. 0.774 mol</span>
I would say “ends with -ic” , “starts with hydro”,
Answer:
See below
Step-by-step explanation:
"Pseudo" means "false" or "fake". Pseudoscience is fake science.
Some of the differences are:
1S — based on observation and experiments to confirm or reject a hypothesis
1P — starts with hypothesis, looks only for supporting evidence, and ignores conflicting evidence
2S — based on reproducibility of results
2P — focuses on alleged exceptions to disprove results
3S — convinces by using factual evidence
3P — convinces by appealing to faith and emotions
4S — rejects anecdotal evidence
4P — relies on anecdotal evidence
1. In first reaction reactant a is the electron donor, while b is the electron acceptor,
the oxidized product is c while the reduced product is d
2. in the second equation e is the electron donor, f is the electron acceptor
g is the oxidized product while h is the reduced product.
3. In the third reaction i is the electron donor, j is the electron acceptor , k is the oxidized product while l is the reduced product.