I can’t see the question :/
Answer:
No, it's a physical reaction.
Explanation:
A chemical change produces new chemical compounds, but combining water and powder is just mixing the powder with the water. It's not a new compound.
I don't know how to really explain, sorry :)
Gasoline is predominantly octane, C8H18. Something like soap would be a great homogenizer. Soap is composed of a long hydrocarbon chain with a tiny, highly polar tip on one end. Usually, the soap is the anion of a salt, NaX. This allows the polar end of the soap to stick to water, while the nonpolar end sticks to the oil.
Answer:
- What is the AGⓇ of this reaction? 0.
- Which will be favoured - the forward reaction, the reverse reaction, or neither? Neither.
- What effect does the presence of the enzyme aspartate transaminase have on the Key value when compared with its value in the absence of enzyme? It does not affect the value of Keq.
- If one of the products of reaction 1, oxaloacetate, is removed by converting it to citrate as follows: Reaction 2: oxaloacetate + acetyl-CoA citrate + COASH will the key for Reaction l be changed? No, the Keq does not change.
Explanation:
1. To calculate the delta G of a reaction given the K, we use the following equation:
ΔG°= -RT ln K.
Which gives us 0 when K is 1.
2.None of the reactions is favoured. Given that the K equals 1, the system will try to keep the concentration of both products and reagents the same.
3. A catalyst is a substance that, when added, provides a different and faster mechanism through which a reaction takes place. This only means that the speed at which the equilibrium is attained is reduced, but the enzyme does nothing to alter the difference in energy (ΔG°) of the start and end points of the reaction, which ultimately gives us the value of Keq.
4. The addition of a side reaction does not change the value of Keq for the main reaction. They are both separate ways of making oxaloacetate disappear. While the Keq does not change, keep in mind that the end concentrations will not be the same, for any set of starting concentrations of your substances.
Answer is: molarity of hydrofluoric solution is 0.09 M.
Chemical reaction: HF(aq) + KOH(aq) → KF(aq) + H₂O(l).
V(HF) = 30.0 mL.
c(KOH) = 0.122 M.
V(KOH) = 22.15 mL:
c(HF) = ?.
From chemical reaction: n(HF) : n(KOH) = 1 : 1.
n(HF) = n(KOH).
c(HF) · V(HF) = c(KOH) · V(KOH).
c(HF) = c(KOH) · V(KOH) ÷ V(HF).
c(HF) = 0.122 M · 22.15 mL ÷ 30 mL:
c(HF) = 0.09 M.
