Answer:
Na - 1 away from neon
Li - 1 away from helium
K - 1 away from argon
Cs - 1 away from xenon
Explanation:
We see this trend as group 1A elements are 1 away from noble gases and hence, have a valency of 1
Answer:
Total concentration of ions present in the final solution is 0.0726 M.
Explanation:
Mass of silver nitrate = 54.0 g
Moles of silver nitrate = ![\frac{54.0 g}{170 g/mol}=0.318 mol](https://tex.z-dn.net/?f=%5Cfrac%7B54.0%20g%7D%7B170%20g%2Fmol%7D%3D0.318%20mol)
Volume of the solution made = V = 350.0 mL = 0.350 L
![Molarity=\frac{Moles}{Volume(L)}](https://tex.z-dn.net/?f=Molarity%3D%5Cfrac%7BMoles%7D%7BVolume%28L%29%7D)
![M=\frac{0.318 mol}{0.35 L} =0.9086 M](https://tex.z-dn.net/?f=M%3D%5Cfrac%7B0.318%20mol%7D%7B0.35%20L%7D%20%3D0.9086%20M)
After dilution of 10.00 mL 0.9227 M solution.
![M_1=0.9086 M](https://tex.z-dn.net/?f=M_1%3D0.9086%20M)
![V_1=10.00 mL](https://tex.z-dn.net/?f=V_1%3D10.00%20mL)
![M_2=?](https://tex.z-dn.net/?f=M_2%3D%3F)
![V_2=250.0 mL](https://tex.z-dn.net/?f=V_2%3D250.0%20mL)
![M_1V_1=M_2V_2](https://tex.z-dn.net/?f=M_1V_1%3DM_2V_2)
![M_2=\frac{M_1V_1}{V_2}=\frac{0.9086 M\times 10.00 mL}{250.0 mL}](https://tex.z-dn.net/?f=M_2%3D%5Cfrac%7BM_1V_1%7D%7BV_2%7D%3D%5Cfrac%7B0.9086%20M%5Ctimes%2010.00%20mL%7D%7B250.0%20mL%7D)
![M_2=0.0363 M](https://tex.z-dn.net/?f=M_2%3D0.0363%20M)
Concentration of silver nitrate after dilution = 0.0363 M
![AgNO_3\rightarrow Ag^++NO_3^{-}](https://tex.z-dn.net/?f=AgNO_3%5Crightarrow%20Ag%5E%2B%2BNO_3%5E%7B-%7D)
![[AgNO_3]=[Ag^+]=[NO_3^{-}]=0.0363 M](https://tex.z-dn.net/?f=%5BAgNO_3%5D%3D%5BAg%5E%2B%5D%3D%5BNO_3%5E%7B-%7D%5D%3D0.0363%20M)
Total concentration of ions present in the final solution:
![[Ag^+]+[NO_3^{-}]=0.0363 M+0.0363 M=0.0726 M](https://tex.z-dn.net/?f=%5BAg%5E%2B%5D%2B%5BNO_3%5E%7B-%7D%5D%3D0.0363%20M%2B0.0363%20M%3D0.0726%20M)
The answer to this question Is sorenson
Answer:
1.82 L
Explanation:
We are given the following information;
- Initial volume as 2.0 L
- Initial temperature as 60.0°C
- New volume as 30.0 °C
We are required to determine the new volume;
From Charles's law;
![\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,
are initial and new volume respectively, while
are initial and new temperatures respectively;
![T_1= 333 K](https://tex.z-dn.net/?f=T_1%3D%20333%20K)
![T_2=303K](https://tex.z-dn.net/?f=T_2%3D303K)
![V_1 =2.0L](https://tex.z-dn.net/?f=V_1%20%3D2.0L)
Rearranging the formula;
![V_2=\frac{V_1T_2}{T_1}](https://tex.z-dn.net/?f=V_2%3D%5Cfrac%7BV_1T_2%7D%7BT_1%7D)
![= \frac{(2.0L)(303K)}{333K} \\=1.820 L](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B%282.0L%29%28303K%29%7D%7B333K%7D%20%5C%5C%3D1.820%20L)
Therefore, the new volume that would be occupied by the gas is 1.82 L
I believe it is a compound, seeing as compounds are basically different types of elements, only combined into one single compound. Please tell me if i'm wrong?