<u>Answer:</u>
<u>For a:</u> The equilibrium mixture contains primarily reactants.
<u>For b:</u> The equilibrium mixture contains primarily products.
<u>Explanation:</u>
There are 3 conditions:
- When
; the reaction is product favored. - When
; the reaction is reactant favored. - When
; the reaction is in equilibrium.
For the given chemical reactions:
The chemical equation follows:

The expression of
for above reaction follows:
![K_{eq}=\frac{[CN^-][H_3O^+]}{[HCN][H_2O]}=6.2\times 10^{-10}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BCN%5E-%5D%5BH_3O%5E%2B%5D%7D%7B%5BHCN%5D%5BH_2O%5D%7D%3D6.2%5Ctimes%2010%5E%7B-10%7D)
As,
, the reaction will be favored on the reactant side.
Hence, the equilibrium mixture contains primarily reactants.
The chemical equation follows:

The expression of
for above reaction follows:
![K_{eq}=\frac{[HCl]^2}{[H_2][Cl_2]}=2.51\times 10^{4}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BHCl%5D%5E2%7D%7B%5BH_2%5D%5BCl_2%5D%7D%3D2.51%5Ctimes%2010%5E%7B4%7D)
As,
, the reaction will be favored on the product side.
Hence, the equilibrium mixture contains primarily products.
A volumetric flask is used to contain a predetermined volume of substance and only measures that volume, for example 250 ml.
Conical flasks can be used to measure the volume of substances but the accuracy they provide is usually up to 10ml. Conical flasks are used in titrations, reactions where the liquid may boil, and reactions which involve stirring.
Pippettes are of two types, volumetric and graduated. Pippettes are used where high accuracy is required and volumetric pippettes come in as little as 1 ml. Pippettes are usually used in titrations.
Graduated cylinders come in a wide variety of sizes and their accuracy can be down to as much as 1 ml. They are used to contain liquids.
Answer:
Explanation:
The following equation relates to Born-Haber cycle

Where
is enthalpy of formation
S is enthalpy of sublimation
B is bond enthalpy
is ionisation enthalpy of metal
is electron affinity of non metal atom
is lattice energy
Substituting the given values we have
-435.7 = 79.2 + 1/2 x 242.8 + 418.7 - 348 +U_L
= - 707 KJ / mol
<span>336*280 i believe... i hope this helps
</span>
Answer:
Amount of Ca(NO3)2 produced = 14.02 g
Explanation:
The given reaction can be depicted as follows:
Ca(OH)2 + 2HNO3 → Ca(NO3)2 + 2H2O
Since it is given that HNO3 is in excess, the limiting reactant is Ca(OH)2
Now, Mass of Ca(OH)2 = 6.33 g
Molar mass of Ca(OH)2 = 74 g/mol

Based on the reaction stoichiometry:
1 mole of Ca(OH)2 forms 1 mole of Ca(NO3)2
Therefore, moles of Ca(NO3)2 produced from the moles of Ca(OH)2 reacted = 0.0855 moles
Molar mass of Ca(NO3)2 = 164 g/mol
