Knowing the ratio between atoms we can write an empirical formula:
<span>C4H6O </span>
<span>we compute the molar mass of this single formula: </span>
<span>4x12 + 6 x 1 + 16 x1 = 70 g / mol </span>
<span>Now, as we know the actual molar mas being 280 g/mol, we divide this number by 70 and we get the ratio between empirical formula and molecular actual formula: </span>
<span>280 / 70 = 4 </span>
<span>This means that actual molecular formula is: </span>
<span>(C4H6O)4 or </span>
<span>C16H24O4 </span>
Explanation:
Moles of metal,
=
4.86
⋅
g
24.305
⋅
g
⋅
m
o
l
−
1
=
0.200
m
o
l
.
Moles of
H
C
l
=
100
⋅
c
m
−
3
×
2.00
⋅
m
o
l
⋅
d
m
−
3
=
0.200
m
o
l
Clearly, the acid is in deficiency ; i.e. it is the limiting reagent, because the equation above specifies that that 2 equiv of HCl are required for each equiv of metal.
So if
0.200
m
o
l
acid react, then (by the stoichiometry), 1/2 this quantity, i.e.
0.100
m
o
l
of dihydrogen will evolve.
So,
0.100
m
o
l
dihydrogen are evolved; this has a mass of
0.100
⋅
m
o
l
×
2.00
⋅
g
⋅
m
o
l
−
1
=
?
?
g
.
If 1 mol dihydrogen gas occupies
24.5
d
m
3
at room temperature and pressure, what will be the VOLUME of gas evolved?
We know that domestic dogs are descended from wolves. Exactly when and how this occurred is still being studied; many experts indicate that dogs were first domesticated about 12,000 years ago, but a few DNA studies indicate a much earlier date for the split between the wolf and the dog, perhaps as long ago as 130,000 years.
Regardless of the exact timing of the divergence between wolves and dogs, one thing is true: both species display quite a bit of genetic diversity, particularly in size and coloration.In order to breed a dog of a specific size, all one has to do is choose parents with the appropriate variations. For instance, if you want to create a small dog breed, you would just breed the smallest dogs you have, then you would choose their smallest offspring to breed the next generation, and so on until you have reached the desired size.
Answer:
1.06 V
Explanation:
The standard reduction potentials are:
Ag^+/Ag E° = 0.7996 V
Ni^2+/Ni E° = -0.257 V
The half-cell and cell reactions for Ni | Ni^2+ || Ag^+ | Ag are
Ni → Ni^2+ + 2e- E° = 0.257 V
<u>2Ag^+ 2e- → 2Ag </u> <u>E° = 0.7996 V
</u>
Ni + 2Ag^+ → Ni^2+ + 2Ag E° = 1.0566 V
To three significant figures, the standard potential for the cell is 1.06 V
.