<u>Answer:</u> The value of <em>i</em> is 1.4 and 40% dissociation of 100 particles of zinc sulfate will yield 60 undissociated particles.
<u>Explanation:</u>
The equation used to calculate the Vant' Hoff factor in dissociation follows:
where,
= degree of dissociation = 40% = 0.40
i = Vant' Hoff factor
n = number of ions dissociated = 2
Putting values in above equation, we get:
The equation used to calculate the degee of dissociation follows:
Total number of particles taken = 100
Degree of dissociation = 40% = 0.40
Putting values in above equation, we get:
This means that 40 particles are dissociated and 60 particles remain undissociated in the solution.
Hence, 40% dissociation of 100 particles of zinc sulfate will yield 60 undissociated particles.
Answer:
3. crystal habit and cleavage.
Explanation:
Crystal habit is a distinctive characteristic which is shown in its general shape, crystallographic forms, how developed each form is.
Cleavage is the ability of a mineral to break in smooth planes parallel to zones of weak bonding. Cleavage in three directions at right angles (90o). Cubic cleavage. Cleavage in three directions not at right angles (120o and 60o).
Answer:
pH 
Explanation:
For every mole of hydrochloric acid, one mole of hydronium ion is required. Thus, in order to neutralize 0.014 moles of HCL, 0.014 moles of hydronium is required.
![[H_3O^+] = [HCl] = 0.014](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%20%3D%20%5BHCl%5D%20%3D%200.014)
pH ![= -log [H^+] = -log [H_3O^+]](https://tex.z-dn.net/?f=%3D%20-log%20%5BH%5E%2B%5D%20%3D%20-log%20%5BH_3O%5E%2B%5D)
Substituting the available values in above equation, we can say that the pH of the solution is equal to
pH
pH of a 
M HCL solution
Answer:
a. 1810mL
Explanation:
When conditions for a gas change under constant pressure (and the number of molecules doesn't change), it follows Charles' Law:
where the temperatures must be measured in Kelvin
To convert from Celsius to Kelvin, add 273, or use the equation: 
For this problem, one must also recall that standard temperature is 0°C (or 273K).
So, ![T_1 = 273[K]](https://tex.z-dn.net/?f=T_1%20%3D%20273%5BK%5D)
, and ![T_2 = (49.4+273)[K]=322.4[K]](https://tex.z-dn.net/?f=T_2%20%3D%20%2849.4%2B273%29%5BK%5D%3D322.4%5BK%5D)
.
![\dfrac{(1532.7[mL])}{(273[K])}=\dfrac{V_2}{(322.4[K])}](https://tex.z-dn.net/?f=%5Cdfrac%7B%281532.7%5BmL%5D%29%7D%7B%28273%5BK%5D%29%7D%3D%5Cdfrac%7BV_2%7D%7B%28322.4%5BK%5D%29%7D)
![\dfrac{(1532.7[mL])}{(273[K\!\!\!\!\!{-}])}(322.4[K\!\!\!\!\!{-}] )=\dfrac{V_2}{(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})}(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})](https://tex.z-dn.net/?f=%5Cdfrac%7B%281532.7%5BmL%5D%29%7D%7B%28273%5BK%5C%21%5C%21%5C%21%5C%21%5C%21%7B-%7D%5D%29%7D%28322.4%5BK%5C%21%5C%21%5C%21%5C%21%5C%21%7B-%7D%5D%20%29%3D%5Cdfrac%7BV_2%7D%7B%28322.4%5BK%5D%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%7B----%7D%29%7D%28322.4%5BK%5D%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%5C%21%7B----%7D%29)
![1810.04571428[mL]=V_2](https://tex.z-dn.net/?f=1810.04571428%5BmL%5D%3DV_2)
Adjusting for significant figures, this gives ![V_2=1810[mL]](https://tex.z-dn.net/?f=V_2%3D1810%5BmL%5D)
