Answer:
Flora and Fauna
Explanation:
therethere are four main regions which are each home to specific landforms, flora and fauna the. These are the upland, the Everglades, the Florida Keys and the Gulf Coast
Answer:
7.03 g
Explanation:
Step 1: Write the balanced synthesis reaction
N₂(g) + 3 H₂(g) ⇒ 2 NH₃(g)
Step 2: Calculate the moles corresponding to 32.5 g of N₂
The molar mass of N₂ is 28.01 g/mol.
32.5 g × 1 mol/28.01 g = 1.16 mol
Step 3: Calculate the number of moles of H₂ needed to react with 1.16 moles of N₂
The molar ratio of N₂ to H₂ is 1:3. The moles of H₂ needed are 3/1 × 1.16 mol = 3.48 mol.
Step 4: Calculate the mass corresponding to 3.48 moles of H₂
The molar mass of H₂ is 2.02 g/mol.
3.48 mol × 2.02 g/mol = 7.03 g
Answer:
Only Reaction 1
Explanation:
In reaction 1, there is a change in state from solid to liquid. Hence, there is an increase in number of ways particles and their energies could be arranged. As a result, entropy increases.
In reaction 2, there is a decrease in amount of gas particles (4 mol to 2 mol). Hence there is a decrease in the number of ways particles and their energies could be arranged. As a result entropy decreases
a) After adding 10 mL of HCl
first, we need to get moles of (CH3)3N = molarity * volume
= 0.29 m * 0.025 L
= 0.00725M moles
then, we need to get moles of HCl = molarity * volume
= 0.3625 m * 0.01L
= 0.003625 moles
so moles of (CH3)3N remaining = moles of (CH3)3N - moles of HCl
= 0.00725 - 0.003625
= 0.003625 moles
and when the total volume = 0.01 L + 0.025L = 0.035 L
∴ [(CH3)3N] = moles remaining / total volume
= 0.003625 moles / 0.035L
= 0.104 M
when we have Pkb so we can get Kb :
pKb = - ㏒Kb
4.19 = -㏒Kb
∴Kb = 6.5 x 10^-5
when Kb = [(CH3)3NH+] [OH-] / [(CH3)3N]
and by using ICE table we assume we have:
[(CH3)3NH+] = X & [OH] = X
and [(CH3)3N] = 0.104 -X
by substitution:
∴ 6.5 x 10^-5 = X^2 / (0.104-X) by solving for X
∴X = 0.00257 M
∴[OH-] = X = 0.00257 M
∴POH = -㏒[OH]
= -㏒0.00257
= 2.5
∴ PH = 14 - POH
= 14 - 2.5
= 11.5
b) after adding 20ML of HCL:
moles of HCl = molarity * volume
= 0.3625 m * 0.02 L
= 0.00725 moles
the complete neutralizes of (CH3)3N we make 0.003625 moles of (CH3)3NH+ So, now we need the Ka value of (CH3)3NH+:
and when the total volume = 0.02L + 0.025 = 0.045L
∴ [ (CH3)3NH+] = moles / total volume
= 0.003625 / 0.045L
= 0.08 M
when Ka = Kw / Kb
and we have Kb = 6.5 x 10^-5 & we know that Kw = 1 x 10^-14
so, by substitution:
Ka = (1 x 10^-14) / (6.5 x 10^-5)
= 1.5 x 10^-10
when Ka expression = [(CH3)3N][H+] / [(CH3)3NH+]
by substitution:
∴ 1.5 x 10^-10 = X^2 / (0.08 - X) by solving for X
∴X = 3.5 x 10^-6 M
∴ [H+]= X = 3.5 x 10^-6 M
∴PH = -㏒[H+]
= -㏒(3.5 x 10^-6)
= 5.5
C) after adding 30ML of HCl:
moles of HCl = molarity * volume
= 0.3625m * 0.03L
= 0.011 moles
and when moles of (CH3)3N neutralized = 0.003625 moles
∴ moles of HCl remaining = moles HCl - moles (CH3)3N neutralized
= 0.011moles - 0.003625moles
= 0.007375 moles
when total volume = 0.025L + 0.03L
= 0.055L
∴[H+] = moles / total volume
= 0.007375 mol / 0.055L
= 0.134 M
∴PH = -㏒[H+]
= -㏒ 0.134
= 0.87
Answer:
ΔGº = --29 kJ
Explanation:
The equation to use is ΔGº = -nFEºcell
where n is the number of mole of electrons exchanged in the redox equation
F is Faraday´s constant 96485 C/mole
Eº cell potential
Ni²⁺(aq) + 2e⁻ ⇒ Ni (s) Eºred = -0.25 V
Cd(s) ⇒ Cd²⁺ + 2e⁻ Eºox = +0.40 V
Eºcell = 0.40 V -0.25 V = 0.15 V
ΔGº = - 2 mole x 96485 C/mol x 0.15 V = -2.9 x 10⁴J = -29 kJ
(remember 1 Coulomb x 1 Volt = 1 Joule)
