The answer is A if the temperature increases, then it’s volume will also increase
Answer:
AB + CD → AC +BD
Explanation:
Double replacement:
It is the reaction in which two compound exchange their ions and form new compounds.
AB + CD → AC +BD
Example:
The following reaction is double replacement reaction.
CaCO₃ + 2HCl → CaCl₂ + H₂CO₃
While,
A + B → AB
Synthesis reaction:
It is synthesis reaction. The reaction in which two or more simple substance react to give one or more complex product.
AB → A + B
Decomposition:
It is the decomposition reaction. The reaction in which one reactant is break down into two or more product.
AB + C → AC + B
Single replacement:
It is the reaction in which one elements replace the other element in compound.
a) when Kc = concentration of products / concentration of reactants
So according to the reaction equation:
Br2(g) + Cl2(g) → 2BrCl(g)
∴ Kc =[BrCl] ^2 / [Br2][Cl2]
b) when q = [BrCl]^2 / [Br2][Cl2]
and we have [BrCl] = 3 m
[Br2] = 1 m
[Cl2] = 1 m
So by substitution:
q= 3^2 / 1*1 = 9
- and we can see that q > Kc
the reaction is not at equilibrium that means there are more products and the reaction shifts to the left to increase the reactants and decrease the products to achieve equilibrium.
C) by using ICE table:
Br2(g) + Cl2(g) → 2BrCl (g)
initial 1 1 3
change -X -X +X
Equ (1-X) (1-X) (3+X)
when Kc = [Brcl]^2/[Cl2][Br2]
by substitution:
7 = (3+X)^2 / (1+X) (1+X) by solving this equation for X
∴X = 0.215
so at equilibrium:
∴ [Br2] = [Cl2] = 1-0.215 = 0.785 m
[BrCl] = 3+0.215 = 3.215 m
It should be 8 O atoms. 3O atoms in Na2S2O3 and 5O atom in 5H2O. The reason there are 5 O atoms are because the 5 in front of H2O means you multiply each atom in the compound by that number (like the distributive property). The H2 molecule becomes 10 Hydrogen atoms (5*2) and the Oxygen becomes 5 Oxygen atoms (5*1). Then you add the 5O atoms to the 3O atoms which equals 8
<em><u>The Rutherford model shows that an atom is mostly empty space, with electrons orbiting a fixed, positively charged nucleus in set, predictable paths.</u></em>
