Answer: The reaction is not at equilibrium and will proceed to make more products to reach equilibrium.
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 reaction 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:
Given :
= 54.8
Thus as
, the reaction will shift towards the right i.e. towards the product side.
Answer:

Explanation:
HCl + NaOH ⟶ NaCl + H₂O
There are two energy flows in this reaction.
Heat of reaction + heat to warm water = 0
q₁ + q₂ = 0
q₁ + mCΔT = 0
Data:
m(HCl) = 50 g
m(NaOH) = 50 g
T₁ = 22 °C
T₂ = 28.87 °C
C = 4.18 J·°C⁻¹g⁻¹
Calculations:
m = 50 + 50 = 100 g
ΔT = 28.87 – 22 = 6.9 °C
q₂ = 100 × 4.18 × 6.9 = 2900 J
q₁ + 2900 = 0
q₁ = -2900 J
The negative sign tells us that the reaction produced heat.
The reaction produced
.
One mole any substance contains 6.022 ₓ 10²³ particles called Avogadro's Number.
The relation between moles and number of particles is given as,
# of particles = moles ₓ Avogadro's number
In our case the particles are formula units of MgCO₃. So, 1 mole of MgCO₃ contain 6.022 ₓ 10²³ formula units, then the number of formula units in 1.72 moles are calculated as,
# of formula units = 1.72 mol ₓ 6.022 ₓ 10²³ formula units / mol
# of formula units = 1.035 ₓ 10²⁴ Formula Units
Answer:
28.52 L
Explanation:
First, let's calculate the density of the ocean, which is the mass divided by the volume:
d = m/V
d = 35.06/1
d = 35.06 g/L
So, for a mass of 1.00 kg = 1000.00 g
d = m/V
35.06 = 1000.00/V
V = 1000.00/35.06
V = 28.52 L
How all the data are expressed with two significant figures, the volume must also be expressed with two.
A 0.200 M of K2SO4 solution is produced by diluting 20.0 mL of 5.00 M K2SO4 solution to 500.0 mL.
<u>Explanation</u>:
- When dealing with dilution we will use the following equation:
M1 V1 = M2 V2
where,
M1 = initial concentration
V1 = initial volume
M2 = final concentration
V2 = final volume
- By diluting 20.0 mL of 5.00 M K2SO4 solution to 500.0 mL, we get
M1 V1 = M2 V2
20.0 mL
5.00 M = M2
500.0 mL
M2 = (20.0 mL
5.00 M) / 500.0 mL
M2 = 0.200 M.
Hence A 0.200 M of K2SO4 solution is produced by diluting 20.0 mL of 5.00 M K2SO4 solution to 500.0 mL.