Answer:
In order of decreasing miscibility
C₉H₂₀ (nonane)→C₂H₅F (fluoroethane)→C₂H₅Cl (chloroethane)→H₂O (water)
Explanation:
The solubility of a solid is a measure of its ability to dissolve in a liquid while for liquids, the miscibility is a measure of thhe liquid to mix with anoyjer liquid resulting in a soltion which can hold any amount of either liquids. Immiscible liquids are those that are not soluble or have very limited solibility with each other.
C₉H₂₀ (nonane)→C₂H₅F (fluoroethane)→C₂H₅Cl (chloroethane)→H₂O (water)
In the order of decreasing miscibility as like dissolve like, ability to dissociate and polar and organic characteristics are considered
It is called exothermic reaction because it releases heat and light and it is called combustion reaction because it is reacting and is being oxidised by O2 to MgO.
It can also be called as oxidation reaction since Mg is oxidised to MgO.
Answer:
B. calcium has electrons in more energy levels than magnesium
Explanation:
atoms are made of three types of subatomic particles - electrons, protons and neutrons.
Neutrons and protons reside in the nucleus of the atom and electrons reside in energy shells
electronic configuration for both Ca and Mg are as follows
Mg - 2,8,2
Ca - 2,8,8,2
outermost energy shell of Mg with electrons is the third energy level
whereas outermost energy shell of Ca with electrons is the fourth energy shell
therefore Ca has a larger atomic radius than Mg as it has one more energy shell than Mg in which electrons reside
Answer:
amount of charge
Explanation:
Oxygen and sulfur are both in Group 16, which means they have a -2 charge. They have two more electrons than protons, making the charge of the ion negative.
Hope that helps.
Answer:
A
Explanation:
Recall that Δ<em>H</em> is the sum of the heats of formation of the products minus the heat of formation of the reactants multiplied by their respective coefficients. That is:
Therefore, from the chemical equation, we have that:
![\displaystyle \begin{aligned} (-317\text{ kJ/mol}) = \left[\Delta H^\circ_f \text{ N$_2$H$_4$} + \Delta H^\circ_f \text{ H$_2$O} \right] -\left[3 \Delta H^\circ_f \text{ H$_2$}+\Delta H^\circ_f \text{ N$_2$O}\right] \end{aligned}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cbegin%7Baligned%7D%20%28-317%5Ctext%7B%20kJ%2Fmol%7D%29%20%3D%20%5Cleft%5B%5CDelta%20H%5E%5Ccirc_f%20%5Ctext%7B%20N%24_2%24H%24_4%24%7D%20%2B%20%20%5CDelta%20H%5E%5Ccirc_f%20%5Ctext%7B%20H%24_2%24O%7D%20%20%5Cright%5D%20%20%20-%5Cleft%5B3%20%5CDelta%20H%5E%5Ccirc_f%20%5Ctext%7B%20H%24_2%24%7D%2B%5CDelta%20H%5E%5Ccirc_f%20%5Ctext%7B%20N%24_2%24O%7D%5Cright%5D%20%5Cend%7Baligned%7D)
Remember that the heat of formation of pure elements (e.g. H₂) are zero. Substitute in known values and solve for hydrazine:
![\displaystyle \begin{aligned} (-317\text{ kJ/mol}) & = \left[ \Delta H^\circ _f \text{ N$_2$H$_4$} + (-285.8\text{ kJ/mol})\right] -\left[ 3(0) + (82.1\text{ kJ/mol})\right] \\ \\ \Delta H^\circ _f \text{ N$_2$H$_4$} & = (-317 + 285.8 + 82.1)\text{ kJ/mol} \\ \\ & = 50.9\text{ kJ/mol} \end{aligned}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cbegin%7Baligned%7D%20%28-317%5Ctext%7B%20kJ%2Fmol%7D%29%20%26%20%3D%20%5Cleft%5B%20%5CDelta%20H%5E%5Ccirc%20_f%20%5Ctext%7B%20N%24_2%24H%24_4%24%7D%20%2B%20%28-285.8%5Ctext%7B%20kJ%2Fmol%7D%29%5Cright%5D%20-%5Cleft%5B%203%280%29%20%2B%20%2882.1%5Ctext%7B%20kJ%2Fmol%7D%29%5Cright%5D%20%5C%5C%20%5C%5C%20%5CDelta%20H%5E%5Ccirc%20_f%20%5Ctext%7B%20N%24_2%24H%24_4%24%7D%20%26%20%3D%20%28-317%20%2B%20285.8%20%2B%2082.1%29%5Ctext%7B%20kJ%2Fmol%7D%20%5C%5C%20%5C%5C%20%26%20%3D%2050.9%5Ctext%7B%20kJ%2Fmol%7D%20%5Cend%7Baligned%7D)
In conclusion, our answer is A.
