Answer:
Newton's Third Law
Explanation:
Newton's Third Law stipulates that for every action there is an equal and opposite reaction.
So when the two players are tackling they exert a force on each other.
If player 1 tackles (exerts a force) player 2, player 2 will exert an equal and opposite reaction on player 1 as stated in Newton's Third Law.
Therefore when they tackle each other so hard they both experience reaction forces so powerful that they fly in opposite directions.
Thus this is an example of the Newton's Third Law.
Answer:
The correct answers are:
a) 180 g
b) 93.7 cm³
Explanation:
The density of a substance is the mass of the substance per unit of volume. So, it is calculated as follows:
density= mass/volume
From the data provided in the problem:
density = 0.8 g/cm³
a) Given: volume= 225 cm³
mass= density x volume = 0.8 g/cm³ x 225 cm³ = 180 g
b) Given: mass= 75.0 g
volume = mass/density = 75.0 g/(0.8 g/cm³)= 93.75 cm³≅ 93.7 cm³
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
Hello!
To solve this problem we are going to use the
Henderson-Hasselbach equation and clear for the molar ratio. Keep in mind that we need the value for Acetic Acid's pKa, which can be found in tables and is
4,76:
![pH=pKa + log ( \frac{[CH_3COONa]}{[CH_3COOH]} )](https://tex.z-dn.net/?f=pH%3DpKa%20%2B%20log%20%28%20%5Cfrac%7B%5BCH_3COONa%5D%7D%7B%5BCH_3COOH%5D%7D%20%29%20)
![\frac{[CH_3COOH]}{[CH_3COONa}= 10^{(pH-pKa)^{-1}}=10^{(4-4,76)^{-1}}=5,75](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BCH_3COOH%5D%7D%7B%5BCH_3COONa%7D%3D%2010%5E%7B%28pH-pKa%29%5E%7B-1%7D%7D%3D10%5E%7B%284-4%2C76%29%5E%7B-1%7D%7D%3D5%2C75%20)
So, the mole ratio of CH₃COOH to CH₃COONa is
5,75Have a nice day!
At the first reaction when 2HBr(g) ⇄ H2(g) + Br2(g)
So Kc = [H2] [Br2] / [HBr]^2
7.04X10^-2 = [H2][Br] / [HBr]^2
at the second reaction when 1/2 H2(g) + 1/2 Br2 (g) ⇄ HBr
Its Kc value will = [HBr] / [H2]^1/2*[Br2]^1/2
we will make the first formula of Kc upside down:
1/7.04X10^-2 = [HBr]^2/[H2][Br2]
and by taking the square root:
∴ √(1/7.04X10^-2)= [HBr] / [H2]^1/2*[Br]^1/2
∴ Kc for the second reaction = √(1/7.04X10^-2) = 3.769
