Answer:
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Winds blowing across the ocean surface often push water away from an area. When this occurs, water rises up from beneath the surface to replace the diverging surface water. This process is known as upwelling.
Explanation:
Answer:
stable isotopes have stable nuclei and do not show radioactivity, but for unstable isotopes it is the opposite
Explanation:
hope this helps, ask more questions if needed.
In nature there are many more variations amino acids than the simple 20 found in humans. However, when analyzing the human genome sequence, there is a code for all 64 permutations (4^3), only some of them share amino acids. This is a safe-guard against mutations of one or two nucleotides. For example, the amino acid Alanine is coded with four different nucleotide sequences: GCA, GCC, GCG, GCU. Also some amino acids code the same like UUU &UUC
This is a tricky question. a mole of any compound contains the same number of molecules of that certain compound. so, one mole of chlorine gas has the same number of molecules as one mole of glucose, which is 6.02 x 10^23.
this is avogadro's number and it applies for any mole of molecules.
the question is tricky because it is like asking. " what weighs more, a pound of feathers or a pound of rocks?" the both weigh the same, a pound. when ewe talking about moles, same as pounds, it is a quantity unit. one mole will aways be equal to 6.02 x 10^23 molecules.
Answer:
(a) ΔSº = 216.10 J/K
(b) ΔSº = - 56.4 J/K
(c) ΔSº = 273.8 J/K
Explanation:
We know the standard entropy change for a given reaction is given by the sum of the entropies of the products minus the entropies of reactants.
First we need to find in an appropiate reference table the standard molar entropies entropies, and then do the calculations.
(a) C2H5OH(l) + 3 O2(g) ⇒ 2 CO2(g) + 3 H2O(g)
Sº 159.9 205.2 213.8 188.8
(J/Kmol)
ΔSº = [ 2(213.8) + 3(188.8) ] - [ 159.9 + 3(205.) ] J/K
ΔSº = 216.10 J/K
(b) CS2(l) + 3 O2(g) ⇒ CO2(g) + 2 SO2(g)
Sº 151.0 205.2 213.8 248.2
(J/Kmol)
ΔSº = [ 213.8 + 2(248.2) ] - [ 151.0 + 3(205.2) ] J/K = - 56.4 J/K
(c) 2 C6H6(l) + 15 O2(g) 12 CO2(g) + 6 H2O(g)
Sº 173.3 205.2 213.8 188.8
(J/Kmol)
ΔSº = [ 12(213.8) + 6(188.8) ] - [ 2(173.3) + 15( 205.2) ] = 273.8 J/K
Whenever possible we should always verify if our answer makes sense. Note that the signs for the entropy change agree with the change in mol gas. For example in reaction (b) we are going from 4 total mol gas reactants to 3, so the entropy change will be negative.
Note we need to multiply the entropies of each substance by its coefficient in the balanced chemical equation.
