The ball's potential energy will decrease, and its kinetic energy will increase.
Explanation:
The ball's potential energy will decrease down the slope and its kinetic energy will increase.
Kinetic energy is the energy due to the motion of a body.
K. E = 
m v²
m is the mass of the ball
v is the velocity of the ball.
Potential energy is the energy due to the motion of a body down slope.
It is expressed as:
P.E = mgh
m is the mass of ball
g is the acceleration due to gravity of the body
h is the height.
We can see as the body drops down the slope, the height diminishes and the velocity increases.
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When we have this balanced equation for a reaction:
Fe(OH)2(s) ↔ Fe+2 + 2OH-
when Fe(OH)2 give 1 mole of Fe+2 & 2 mol of OH-
so we can assume [Fe+2] = X and [OH-] = 2 X
when Ksp = [Fe+2][OH-]^2
and have Ksp = 4.87x10^-17
[Fe+2]= X
[OH-] = 2X
so by substitution
4.87x10^-17 = X*(2X)^2
∴X^3 = 4.8x10^-17 / 4
∴the molar solubility X = 2.3x10^-6 M
Answer: C) Either benzene or oxygen may limit the amount of product that can be formed
Explanation: Benzene and oxygen are the reactants of the equation. What type and the amount of reactants there are in a chemical reaction affects the outcome. Therefore, seeing as benzene and oxygen are the reactants, the answer is C).
Answer:
a. 1.12 L
Explanation:
Step 1: Write the balanced equation for the photosynthesis
6 CO₂(g) + 6 H₂O(l) ⇒ C₆H₁₂O₆(s) + 6 O₂(g)
Step 2: Calculate the moles corresponding to 2.20 g of CO₂
The molar mass of CO₂ is 44.01 g/mol.
2.20 g × 1 mol/44.01 g = 0.0500 mol
Step 3: Calculate the moles of O₂ produced
The molar ratio of CO₂ to O₂ is 6:6. The moles of O₂ produced are 6/6 × 0.0500 mol = 0.0500 mol
Step 4: Calculate the volume occupied by 0.0500 moles of O₂ at STP
At STP, 1 mole of O₂ occupies 22.4 L.
0.0500 mol × 22.4 L/1 mol = 1.12 L
Answer:
ΔS° = -268.13 J/K
Explanation:
Let's consider the following balanced equation.
3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)
We can calculate the standard entropy change of a reaction (ΔS°) using the following expression:
ΔS° = ∑np.Sp° - ∑nr.Sr°
where,
ni are the moles of reactants and products
Si are the standard molar entropies of reactants and products
ΔS° = [2 mol × S°(HNO₃(l)) + 1 mol × S°(NO(g))] - [3 mol × S°(NO₂(g)) + 1 mol × S°(H₂O(l))]
ΔS° = [2 mol × 155.6 J/K.mol + 1 mol × 210.76 J/K.mol] - [3 mol × 240.06 J/K.mol + 1 mol × 69.91 J/k.mol]
ΔS° = -268.13 J/K
