Answer:
The standard entropy or reaction ΔS° =+324.6J/mol.K
Explanation:
The reaction is
2H2O(l) → 2H2(g) + O2(g)
Standard Molar Entropies of Selected Substances at 298 K are
Substance S∘ (J/mol⋅K)
H2(g) 130.6
O2(g) 205.0
H2O(g) 188.8
H2O(l) 69.9
The standard entropy of reaction will be obtained from standard entropy of formation as given below
ΔH°= 2(130.6)+(205.0)-2(69.9) (Jol/mol.K)
ΔH°=261.2+205.0-139.8
ΔH°=+324.6Jol/mol.K
Answer:0.000000001
Explanation:you will move the decimal point 9 times to your left hand side
if O is -2 and hydrogen +1 then phosphorus is +5
What this tells you is that the oxidation number of P in phosphorus acid must be +3 or +4, while the value for phosphoric acid must be +5 or +6. Since phosphorus is a member of group 5A, it has 5 electrons in its valence shell. As a result, the most common oxidation states it can have are +3 (s2p0) or +5 (s0p0).
So far we have determined the oxidation state at P. What needs to be done now is to add as many oxygens and hydroxyl (OH) to make the molecule neutral. The correct combination will have the correct Lewis-dot structure. For phosphorous acid we need a combination that will add up to +3. This can be done by adding 3 OH- to the central atom to yield the structure H3PO3. There is a little caveat though. Because this is not a hydro___acid, it is implied that there must be at least one oxo ligand (O^2-) bonded to P. Therefore, the actual bonding structure is not P(OH)3 but rather H-P(=O)(OH)2, where one H is bonded directly to the phosphorus atom and is the least acidic of the protons. The great thing is that the oxidation charge of P is still +3 because P is slightly more electronegative than H (some theories will say otherwise); thus the hydrogen is regarded as H+ for this example.
With phosphoric acid, charge of +5, you can have 3 OH- and 1 O^2- to make a neutral molecule:
O=P(OH)3.
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Answer: d :The blue and orange soccer balls; they have more mass than the black soccer ball, but changed speed by the same amount.