Answer:
0.808 M
Explanation:
Using Raoult's Law
![\frac{P_s}{Pi}= x_i](https://tex.z-dn.net/?f=%5Cfrac%7BP_s%7D%7BPi%7D%3D%20x_i)
where:
= vapor pressure of sea water( solution) = 23.09 mmHg
= vapor pressure of pure water (solute) = 23.76 mmHg
= mole fraction of water
∴
![\frac{23.09}{23.76}= x_i](https://tex.z-dn.net/?f=%5Cfrac%7B23.09%7D%7B23.76%7D%3D%20x_i)
![x_i = 0.9718](https://tex.z-dn.net/?f=x_i%20%3D%200.9718)
![x_i+x_2=1](https://tex.z-dn.net/?f=x_i%2Bx_2%3D1)
![x_2 = 1- x_i](https://tex.z-dn.net/?f=x_2%20%3D%201-%20x_i)
![x_2 = 1- 0.9718](https://tex.z-dn.net/?f=x_2%20%3D%201-%200.9718)
![x_2 = 0.0282](https://tex.z-dn.net/?f=x_2%20%3D%200.0282)
------ equation (1)
------ equation (2)
where;
number of moles of sea water
number of moles of pure water
equating above equation 1 and 2; we have :
![\frac{n_2}{n_i}](https://tex.z-dn.net/?f=%5Cfrac%7Bn_2%7D%7Bn_i%7D)
![= \frac{0.0282}{0.9178}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B0.0282%7D%7B0.9178%7D)
![= 0.02901](https://tex.z-dn.net/?f=%3D%200.02901)
NOW, Molarity = ![\frac{moles of sea water}{mass of pure water }*1000](https://tex.z-dn.net/?f=%5Cfrac%7Bmoles%20of%20sea%20water%7D%7Bmass%20of%20pure%20water%20%7D%2A1000)
![= \frac{0.02901}{18}*1000](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B0.02901%7D%7B18%7D%2A1000)
![= 0.001616*1000](https://tex.z-dn.net/?f=%3D%200.001616%2A1000)
![= 1.616 M](https://tex.z-dn.net/?f=%3D%201.616%20M)
As we assume that the sea water contains only NaCl, if NaCl dissociates to Na⁺ and Cl⁻; we have ![\frac{1.616}{2} =0.808 M](https://tex.z-dn.net/?f=%5Cfrac%7B1.616%7D%7B2%7D%20%3D0.808%20M)
Answer: 2 moles
Explanation:
STP is Standard Temperature and Pressure. That means the pressure is 1.00 atm and the temperature is 273K. Since the oxygen is placed in the same container, we can use Ideal Gas Law to figure out what container the CO₂ used.
Ideal Gas Law: PV=nRT
P=1.00 atm
n=moles
R=0.08206 Latm/Kmol
T=273K
CO₂
![V=\frac{nRT}{P}](https://tex.z-dn.net/?f=V%3D%5Cfrac%7BnRT%7D%7BP%7D)
![V=\frac{(2.0 mol)(0.08206Latm/Kmol)(273K)}{1.00atm}](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B%282.0%20mol%29%280.08206Latm%2FKmol%29%28273K%29%7D%7B1.00atm%7D)
![V=44.8L](https://tex.z-dn.net/?f=V%3D44.8L)
Since we know that CO₂ has a 44.8 L container, we can use that to find the moles of oxygen.
![n=\frac{PV}{RT}](https://tex.z-dn.net/?f=n%3D%5Cfrac%7BPV%7D%7BRT%7D)
![n=\frac{(1.00atm)(44.8L)}{(0.08206Latm/Kmol)(273K)}](https://tex.z-dn.net/?f=n%3D%5Cfrac%7B%281.00atm%29%2844.8L%29%7D%7B%280.08206Latm%2FKmol%29%28273K%29%7D)
![n=1.99=2mol](https://tex.z-dn.net/?f=n%3D1.99%3D2mol)
There are 2 mol of oxygen.
₁₇Cl 1s²2s²2p⁶3s²3p⁵, 1 unpaired electron in 3pz orbital.
The answer is A
Mark Brainliest ☺☻