Answer:
The correct answer is B.
The
is samller than
of the reaction . So,the reaction will shift towards the left i.e. towards the reactant side.
Explanation:
Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as 
K is the constant of a certain reaction when it is in equilibrium, while Q is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.
For the given chemical reaction:

The expression for
is written as:
![Q=\frac{[PCl_3][Cl_2]}{[[PCl_5]^1}](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B%5BPCl_3%5D%5BCl_2%5D%7D%7B%5B%5BPCl_5%5D%5E1%7D)


Given :
= 0.0454
Thus as
, the reaction will shift towards the left i.e. towards the reactant side.
Answer:
A . 6.3 In a healthy pond, the temperature is 16°C (61°F). What is the most likely pH of this pond
Here we have to calculate the amount of
ion present in the sample.
In the sample solution 0.122g of
ion is present.
The reaction happens on addition of excess BaCl₂ in a sample solution of potassium sulfate (K₂SO₄) and sodium sulfate [(Na)₂SO₄] can be written as-
K₂SO₄ = 2K⁺ + 
(Na)₂SO₄=2Na⁺ + 
Thus, BaCl₂+
= BaSO₄↓ + 2Cl⁻ .
(Na)₂SO₄ and K₂SO₄ is highly soluble in water and the precipitation or the filtrate is due to the BaSO₄ only. As a precipitation appears due to addition of excess BaCl₂ thus the total amount of
ion is precipitated in this reaction.
The precipitate i.e. barium sulfate (BaSO₄)is formed in the reaction which have the mass 0.298g.
Now the molecular weight of BaSO₄ is 233.3 g/mol.
We know the molecular weight of sulfate ion (
) is 96.06 g/mol. Thus in 1 mole of BaSO₄ 96.06 g of
ion is present.
Or. we may write in 233.3 g of BaSO₄ 96.06 g of
ion is present. So in 1 g of BaSO₄
g of
ion is present.
Or, in 0.298 g of the filtered mass (0.298×0.411)=0.122g of
ion is present.
2.47 moles are produce if 85g of N2 react with 180g of Mg
Answer:
symbol= F
proton= 9
neutron= 19 I understand this is it ok for you