Answer:
B. The electronegativities of the two atoms are equal.
Answer: The concentration of KOH for the final solution is 0.275 M
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.
![Molarity=\frac{n\times 1000}{V_s}](https://tex.z-dn.net/?f=Molarity%3D%5Cfrac%7Bn%5Ctimes%201000%7D%7BV_s%7D)
where,
n = moles of solute
= volume of solution in ml = 150 ml
moles of solute =![\frac{\text {given mass}}{\text {molar mass}}=\frac{10.0g}{56g/mol}=0.178moles](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%20%7Bgiven%20mass%7D%7D%7B%5Ctext%20%7Bmolar%20mass%7D%7D%3D%5Cfrac%7B10.0g%7D%7B56g%2Fmol%7D%3D0.178moles)
Now put all the given values in the formula of molality, we get
![Molality=\frac{0.178\times 1000}{150}=1.19M](https://tex.z-dn.net/?f=Molality%3D%5Cfrac%7B0.178%5Ctimes%201000%7D%7B150%7D%3D1.19M)
According to the dilution law,
![C_1V_1=C_2V_2](https://tex.z-dn.net/?f=C_1V_1%3DC_2V_2)
where,
= molarity of stock solution = 1.19 M
= volume of stock solution = 15.0 ml
= molarity of diluted solution = ?
= volume of diluted solution = 65.0 ml
Putting in the values we get:
![1.19\times 15.0=M_2\times 65.0](https://tex.z-dn.net/?f=1.19%5Ctimes%2015.0%3DM_2%5Ctimes%2065.0)
![M_1=0.275M](https://tex.z-dn.net/?f=M_1%3D0.275M)
Therefore, the concentration of KOH for the final solution is 0.275 M
![{\qquad\qquad\huge\underline{{\sf Answer}}}](https://tex.z-dn.net/?f=%20%7B%5Cqquad%5Cqquad%5Chuge%5Cunderline%7B%7B%5Csf%20Answer%7D%7D%7D%20)
Here we go ~
Energy difference btween the two electronic states can be expressed as :
![{ \qquad \sf \dashrightarrow \: \Delta E = h\nu}](https://tex.z-dn.net/?f=%7B%20%5Cqquad%20%5Csf%20%20%5Cdashrightarrow%20%5C%3A%20%5CDelta%20E%20%3D%20h%5Cnu%7D%20)
[ h = planks constant,
= frequency ]
![\qquad \sf \dashrightarrow \:214.68 = 39.79 \times 10 {}^{ - 14} \times \nu](https://tex.z-dn.net/?f=%20%5Cqquad%20%5Csf%20%20%5Cdashrightarrow%20%5C%3A214.68%20%3D%2039.79%20%5Ctimes%2010%20%7B%7D%5E%7B%20-%2014%7D%20%20%5Ctimes%20%20%5Cnu)
![\qquad \sf \dashrightarrow \: \nu = \cfrac{214.68}{39.79 \times 10 {}^{ - 4} }](https://tex.z-dn.net/?f=%5Cqquad%20%5Csf%20%20%5Cdashrightarrow%20%5C%3A%20%5Cnu%20%3D%20%20%5Ccfrac%7B214.68%7D%7B39.79%20%5Ctimes%2010%20%7B%7D%5E%7B%20-%204%7D%20%7D%20)
![\qquad \sf \dashrightarrow \: \nu = \cfrac{214.68}{39.79 } \times 10 {}^{14}](https://tex.z-dn.net/?f=%5Cqquad%20%5Csf%20%20%5Cdashrightarrow%20%5C%3A%20%5Cnu%20%3D%20%20%5Ccfrac%7B214.68%7D%7B39.79%20%7D%20%20%5Ctimes%2010%20%7B%7D%5E%7B14%7D%20)
![\qquad \sf \dashrightarrow \: \nu \approx 5.395 \times10 {}^{14} \:\:hertz](https://tex.z-dn.net/?f=%5Cqquad%20%5Csf%20%20%5Cdashrightarrow%20%5C%3A%20%5Cnu%20%20%5Capprox%20%205.395%20%5Ctimes10%20%7B%7D%5E%7B14%7D%20%20%5C%3A%5C%3Ahertz)
<span>Bodies of magma rise in the crust until they reach a point of neutral buoyancy. The expansion of gases brings the magma closer to the surface and drives eruptions. When more gases are dissolve into the magma, there would more chances of explosion. Hope this answers the question.</span>