Answer:
B. 
Explanation:
The unit for rate is M/s while the unit for each molecule should be M. You can find the unit for k by putting the units for rate and the molecules into the equation
rate= k{X][Y]
M/s= k *
* 
k= (M/s) / (
)
k= 
You can also use this predetermined formula to solve this problem faster: k= 
Where n is the number of molecule. There are 3 molecule(2X and 1Y) so n=3, so
k= 
k=
=
= 
Calcium fluoride.
Ca is metal, F is non-metal, so they form ionic bond.
Ca as metal can form only positive ion. Ca in the second group, so the charge of Ca ion is 2+. Ca²⁺
F is in the 17th group, so it has 7 electrons on the last level. It is non-metal, non-metal, so it has negative charge -(8-7)=-1. "8" because on the last level cannot be more than 8 electrons. F-ion is F¹⁻.
Ca²⁺ F¹⁻
Number of positive charges should be equal to number of negative charges,
Formula of calcium fluoride
CaF2.
2 atoms Fluorine bond with Calcium.
<u>Answer:</u> The concentration of
required will be 0.285 M.
<u>Explanation:</u>
To calculate the molarity of
, we use the equation:

Moles of
= 0.016 moles
Volume of solution = 1 L
Putting values in above equation, we get:

For the given chemical equations:

![Ni^{2+}(aq.)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K_f=1.2\times 10^9](https://tex.z-dn.net/?f=Ni%5E%7B2%2B%7D%28aq.%29%2B6NH_3%28aq.%29%5Crightleftharpoons%20%5BNi%28NH_3%29_6%5D%5E%7B2%2B%7D%2BC_2O_4%5E%7B2-%7D%28aq.%29%3BK_f%3D1.2%5Ctimes%2010%5E9)
Net equation: ![NiC_2O_4(s)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K=?](https://tex.z-dn.net/?f=NiC_2O_4%28s%29%2B6NH_3%28aq.%29%5Crightleftharpoons%20%5BNi%28NH_3%29_6%5D%5E%7B2%2B%7D%2BC_2O_4%5E%7B2-%7D%28aq.%29%3BK%3D%3F)
To calculate the equilibrium constant, K for above equation, we get:

The expression for equilibrium constant of above equation is:
![K=\frac{[C_2O_4^{2-}][[Ni(NH_3)_6]^{2+}]}{[NiC_2O_4][NH_3]^6}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BC_2O_4%5E%7B2-%7D%5D%5B%5BNi%28NH_3%29_6%5D%5E%7B2%2B%7D%5D%7D%7B%5BNiC_2O_4%5D%5BNH_3%5D%5E6%7D)
As,
is a solid, so its activity is taken as 1 and so for 
We are given:
![[[Ni(NH_3)_6]^{2+}]=0.016M](https://tex.z-dn.net/?f=%5B%5BNi%28NH_3%29_6%5D%5E%7B2%2B%7D%5D%3D0.016M)
Putting values in above equations, we get:
![0.48=\frac{0.016}{[NH_3]^6}}](https://tex.z-dn.net/?f=0.48%3D%5Cfrac%7B0.016%7D%7B%5BNH_3%5D%5E6%7D%7D)
![[NH_3]=0.285M](https://tex.z-dn.net/?f=%5BNH_3%5D%3D0.285M)
Hence, the concentration of
required will be 0.285 M.
Answer: Molar concentration of the tree sap have to be 0.783 M
Explanation:
To calculate the concentration of solute, we use the equation for osmotic pressure, which is:
where,
= osmotic pressure of the solution = 19.6 atm
i = Van't hoff factor = 1 (for non-electrolytes)
R = Gas constant =
T = temperature of the solution =
Putting values in above equation, we get:


Thus the molar concentration of the tree sap have to be 0.783 M to achieve this pressure on a day when the temperature is 32°C
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