Answer:
Answer: a) 20g of H2O (18.02 g/mol) molecules=6.68x10^23
Explanation:
In order to find the amount of molecules of each of the options, we need to follow the following equation.
![molecules=\frac{mass(g)x6.022x10^{23}(molecules/mol) }{atomic weight(g/mol)}](https://tex.z-dn.net/?f=molecules%3D%5Cfrac%7Bmass%28g%29x6.022x10%5E%7B23%7D%28molecules%2Fmol%29%20%7D%7Batomic%20weight%28g%2Fmol%29%7D)
So, let´s get the number of molecules for each of the options.
![a) molecules=\frac{20(g)x6.022x10^{23}(molecules/mol) }{18.02(g/mol)}=6.68x10^{23}molecules](https://tex.z-dn.net/?f=a%29%20molecules%3D%5Cfrac%7B20%28g%29x6.022x10%5E%7B23%7D%28molecules%2Fmol%29%20%7D%7B18.02%28g%2Fmol%29%7D%3D6.68x10%5E%7B23%7Dmolecules)
![b) molecules=\frac{77(g)x6.022x10^{23}(molecules/mol) }{16.06(g/mol)}=2.89x10^{24}molecules](https://tex.z-dn.net/?f=b%29%20molecules%3D%5Cfrac%7B77%28g%29x6.022x10%5E%7B23%7D%28molecules%2Fmol%29%20%7D%7B16.06%28g%2Fmol%29%7D%3D2.89x10%5E%7B24%7Dmolecules)
![c) molecules=\frac{68(g)x6.022x10^{23}(molecules/mol) }{42.09(g/mol)}=9.73x10^{23}molecules](https://tex.z-dn.net/?f=c%29%20molecules%3D%5Cfrac%7B68%28g%29x6.022x10%5E%7B23%7D%28molecules%2Fmol%29%20%7D%7B42.09%28g%2Fmol%29%7D%3D9.73x10%5E%7B23%7Dmolecules)
![d) molecules=\frac{100(g)x6.022x10^{23}(molecules/mol) }{44.02(g/mol)}=1.37x10^{24}molecules](https://tex.z-dn.net/?f=d%29%20molecules%3D%5Cfrac%7B100%28g%29x6.022x10%5E%7B23%7D%28molecules%2Fmol%29%20%7D%7B44.02%28g%2Fmol%29%7D%3D1.37x10%5E%7B24%7Dmolecules)
![d) molecules=\frac{84(g)x6.022x10^{23}(molecules/mol) }{20.01(g/mol)}=2.53x10^{24}molecules](https://tex.z-dn.net/?f=d%29%20molecules%3D%5Cfrac%7B84%28g%29x6.022x10%5E%7B23%7D%28molecules%2Fmol%29%20%7D%7B20.01%28g%2Fmol%29%7D%3D2.53x10%5E%7B24%7Dmolecules)
the smalest number is in option a)
Best of luck.
The correct answers are A and C.
Answer:
648.5 mL
Explanation:
Here we will assume that the pressure of the gas is constant, since it is not given or specified.
Therefore, we can use Charle's law, which states that:
"For an ideal gas kept at constant pressure, the volume of the gas is proportional to its absolute temperature"
Mathematically:
![\frac{V}{T}=const.](https://tex.z-dn.net/?f=%5Cfrac%7BV%7D%7BT%7D%3Dconst.)
where
V is the volume of the gas
T is its absolute temperature
The equation can be rewritten as
![\frac{V_1}{T_1}=\frac{V_2}{T_2}](https://tex.z-dn.net/?f=%5Cfrac%7BV_1%7D%7BT_1%7D%3D%5Cfrac%7BV_2%7D%7BT_2%7D)
where in this problem we have:
is the initial volume of the gas
is the initial temperature
is the final temperature
Solving for V2, we find the final volume of the gas:
![V_2=\frac{V_1 T_2}{T_1}=\frac{(490)(315)}{238}=648.5 mL](https://tex.z-dn.net/?f=V_2%3D%5Cfrac%7BV_1%20T_2%7D%7BT_1%7D%3D%5Cfrac%7B%28490%29%28315%29%7D%7B238%7D%3D648.5%20mL)
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