Answer:
4
Explanation:
Applying,
pH = 14-pOH.................... Equation 1
But,
pOH = -log(OH⁻)
Where OH⁻ = Hydroxyl ion concentration of NaOH.
From the question,
Given: OH⁻ = (1×10⁻¹⁰) M
Substitute these values into equation 1
pH = 14-log((1×10⁻¹⁰)
pH = 14-10
pH = 4
Hence the pH of NaOH with a molarity of 1.0 x 10-10M = 4
Density = mass / volume
= 69g / 23 ml
= 3 g / ml.
Thus, the density of the sample is 3 grams per ml or 3g/ ml
<u>Answer:</u> The activation energy of the reaction is 124.6 kJ/mol
<u>Explanation:</u>
To calculate activation energy of the reaction, we use Arrhenius equation, which is:
![\ln(\frac{K_{79^oC}}{K_{26^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BK_%7B79%5EoC%7D%7D%7BK_%7B26%5EoC%7D%7D%29%3D%5Cfrac%7BE_a%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= equilibrium constant at 79°C = 
= equilibrium constant at 26°C = 
= Activation energy of the reaction = ?
R = Gas constant = 8.314 J/mol K
= initial temperature = ![26^oC=[26+273]K=299K](https://tex.z-dn.net/?f=26%5EoC%3D%5B26%2B273%5DK%3D299K)
= final temperature = ![79^oC=[79+273]K=352K](https://tex.z-dn.net/?f=79%5EoC%3D%5B79%2B273%5DK%3D352K)
Putting values in above equation, we get:
![\ln(\frac{0.394}{2.08\times 10^{-4}})=\frac{E_a}{8.314J/mol.K}[\frac{1}{299}-\frac{1}{352}]\\\\E_a=124595J/mol=124.6kJ/mol](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7B0.394%7D%7B2.08%5Ctimes%2010%5E%7B-4%7D%7D%29%3D%5Cfrac%7BE_a%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B299%7D-%5Cfrac%7B1%7D%7B352%7D%5D%5C%5C%5C%5CE_a%3D124595J%2Fmol%3D124.6kJ%2Fmol)
Hence, the activation energy of the reaction is 124.6 kJ/mol
Answer:
a. 2nd order reaction.
b. The first step is the slow step.
Explanation:
r = k[NO][Cl₂]
a. The reaction is first-order in [NO] and first-order in [Cl₂], so it is second-order overall.
b. The first step is the slow step, because it predicts the correct rate law.
c. is wrong. Doubling [NO] would double the rate, because the reaction is first-order in [NO].
d. is wrong. Cutting [Cl₂] in half would halve the rate, because the reaction is first-order in [Cl₂].
e. is wrong. The molecularity is two, because two particles are colliding.
f. is wrong. Both steps are bimolecular.
Iron, but there are some amounts of other dense elements like gold, platinum, and uranium.
