The answer to this question would be: <span>NaOH solution
Limiting reagent is the reagent that totally consumed in the reaction. To answer this question you need to know the equation for the chemical reaction. The equation would be:
</span><span>HCl + NAOH = H2O + NACl
</span>
All of the coefficient in the reaction is 1, which mean 1 HCl will react to 1 NaOH. In this question, the concentration of HCl and NaOH is same so the amount of the molecule can be reflected by the volume. NaOH has lower volume compared to HCl so it is clear that NaOH will be totally consumed in this reaction.
Answer:
the study of the movements and relative positions of celestial bodies interpreted as having an influence on human affairs and the natural world
Explanation:
Since valence shells in nonmetal atoms are almost full, the atoms attract electrons and hold them tightly to fill their valence shells.
Answer:
Diphosphorus pentoxide
Carbon dichloride
BCl3
N2H4
Explanation:
These are all covalent compounds. To name covalent compounds, you add prefixes to the beginning of their names depending on what the subscript is of each element. The prefixes are:
1: Mono
2: Di
3: Tri
4: Tetra
5: Penta
6: Hexa
7: Hepta
8: Octa
9: Nona
10: Deca
For example, since the first one is Phopsphorus with a 2 next to it, you add the prefix Di to it.
If the first element in the compound only has one, meaning no number next to it, you do not say mono. This is why we just say "Carbon" for the second one instead of "Monocarbon."
Finally, you always have to end the second element in the compound with "ide." So, "chlorine" becomes "chloride," "oxygen" becomes "oxide," and so on.
Consider the halogenation of ethene is as follows:
CH₂=CH₂(g) + X₂(g) → H₂CX-CH₂X(g)
We can expect that this reaction occurring by breaking of a C=C bond and forming of two C-X bonds.
When bond break it is endothermic and when bond is formed it is exothermic.
So we can calculate the overall enthalpy change as a sum of the required bonds in the products:
Part a)
C=C break = +611 kJ
2 C-F formed = (2 * - 552) = -1104 kJ
Δ H = + 611 - 1104 = - 493 kJ
2C-Cl formed = (2 * -339) = - 678 kJ
ΔH = + 611 - 678 = -67 kJ
2 C-Br formed = (2 * -280) = -560 kJ
ΔH = + 611 - 560 = + 51 kJ
2 C-I Formed = (2 * -209) = -418 kJ
ΔH = + 611 - 418 = + 193 kJ
Part b)
As we can see that the highest exothermic bond formed is C-F bond so from bond energies we can found that addition of fluoride is the most exothermic reaction
