They're both are very destructed and have risk for the peiple
<span>First we can calculate the area of the rectangular lawn using the formula:
Area = Width x Length = 21 ft x 20 ft = 420 square feet
And the total number of snow flakes per minute on the entire lawn is:
(1350 snowflakes per minute per square foot) x (420 square feet) = 567,000 snowflakes per minute
In one hour (or 60 minutes) we get a total of:
(567,000 snowflakes per minute) x (60 minutes / 1 hour) = 34,020,000 snowflakes
The total mass of which would be:
34,020,000 snowflakes x 1.60 mg = 54,432,000 mg = 54.432 kg (as 1 kg = 1,000,000 mg).
So 54.432 kg of snow accumulates every hour on the lawn.</span>
Answer : The equilibrium will shift in the left direction.
Explanation :
Le-Chatelier's principle : This principle states that if any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.
The given reaction is:
As per question, when we are adding 
then the concentration of is increased on product side then the equilibrium will shift in the direction where decrease of concentration of takes place. Therefore, the equilibrium will shift in the left direction.
Thus, the equilibrium will shift in the left direction.
Explanation:
What will the question be ?
Answer:
131.5 kJ
Explanation:
Let's consider the following reaction.
CaCO₃(s) → CaO(s) + CO₂(g)
First, we will calculate the standard enthalpy of the reaction (ΔH°).
ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g)
) - 1 mol × ΔH°f(CaCO₃(s)
)
ΔH° = 1 mol × (-634.9 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1207.6 kJ/mol)
ΔH° = 179.2 kJ
Then, we calculate the standard entropy of the reaction (ΔS°).
ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g)
) - 1 mol × S°(CaCO₃(s)
)
ΔS° = 1 mol × (38.1 J/mol.K) + 1 mol × (213.8 J/mol.K) - 1 mol × (91.7 J/mol.K)
ΔS° = 160.2 J/K = 0.1602 kJ/K
Finally, we calculate the standard Gibbs free energy of the reaction at T = 25°C = 298 K.
ΔG° = ΔH° - T × ΔS°
ΔG° = 179.2 kJ - 298 K × 0.1602 kJ/K
ΔG° = 131.5 kJ
