The question is incomplete, the table of the question is given below
Answer:
I) xA= 0.34, yA= 0.55
ii) 76.2 mole % vapor
iii) Percentage of vapor volume = 98%
Explanation:
i) xA= 0.34, yA= 0.55
xA= 0.34, yA= 0.55
ii) 0.50 = 0.55 nv + 0.34 nL
Therefore, nV = 0.762 mol vapor and nL = 0.238 mol liquid
This shows 76.2 mole % vapor
iii) ρA= 0.791 g/cm3 and, ρE = 0.789 g/cm3
Therefore, ρ = 0.790 g/cm3
Now, we have:
MA = 58.08 g/mol and ME= 46.07 g/mol
So Ml = (0.34 x 58.08)+[(1 -0.34) x 46.07] = 50.15 g/mol
1 mol liquid = (0.762 mol vapor/0.238 mol liquid) = 3.2 mol vapor
Liquid volume = Vl= [1 mol x (50.15 g/mol)] / (0.790 g/cm3) = 63.48 cm3
Vapour volume = Vv = 3.2 mol x(22400 cm3/mol) x [(65+273)/273] = 88747 cm3
Therefore, percentage of vapour volume = 88747 / (88747+63.48) = 99.9 %
Answer:
1x10^-8 M
Explanation:
Since the solution turns blue, it mean the solution is a base.
Now, to know which option is correct, we need to determine the pH of each solution. This is illustrated below:
1. Concentration of Hydrogen ion, [H+] = 1x10^-2 M
pH =..?
pH = - log [H+]
pH = - log 1x10^-2
pH = 2
2. Concentration of Hydrogen ion, [H+] = 5x10-2 M
pH =..?
pH = - log [H+]
pH = - log 5x10^-2
pH = 1.3
3. Concentration of Hydrogen ion, [H+] = 5x10 M
pH =..?
pH = - log [H+]
pH = - log 5x10
pH = - 1.7
4. Concentration of Hydrogen ion, [H+] = 1x10-8 M
pH =..?
pH = - log [H+]
pH = - log 1x10^-8
pH = 8
A pH reading shows if the solution is acidic or basic. A pH reading between 0 and 6 indicates an acidic solution, a pH reading of 7 indicates a neutral solution while a pH reading between 8 and 14 indicates a basic solution.
From the above calculations, the pH reading indicates a basic solution when the hydrogen ion concentration was 1x10^-8 M.
1.062 mol/kg.
<em>Step 1</em>. Write the balanced equation for the neutralization.
MM = 204.22 40.00
KHC8H4O4 + NaOH → KNaC8H4O4 + H2O
<em>Step 2</em>. Calculate the moles of potassium hydrogen phthalate (KHP)
Moles of KHP = 824 mg KHP × (1 mmol KHP/204.22 mg KHP)
= 4.035 mmol KHP
<em>Step 3</em>. Calculate the moles of NaOH
Moles of NaOH = 4.035 mmol KHP × (1 mmol NaOH/(1 mmol KHP)
= 4.035 mmol NaOH
<em>Step 4</em>. Calculate the mass of the NaOH
Mass of NaOH = 4.035 mmol NaOH × (40.00 mg NaOH/1 mmol NaOH)
= 161 mg NaOH
<em>Step 5</em>. Calculate the mass of the water
Mass of water = mass of solution – mass of NaOH = 38.134 g - 0.161 g
= 37.973 g
<em>Step 6</em>. Calculate the molal concentration of the NaOH
<em>b</em> = moles of NaOH/kg of water = 0.040 35 mol/0.037 973 kg = 1.062 mol/kg
Explanation:
At one end, new crusts are being produced, at other end the crust is being destroyed and this strikes a unique balance.
At the mid-ocean ridges, the lithospheric plates are diverging. This is implies that the earth is pulling apart here. When the earth is pulling apart, new materials from the asthenosphere comes to the surface thereby creating new lithospheric plate.
As new plates are formed, they push back against the old one. New plates are found very close to the margin and it begins to age away from the margin.
On the other end, old plates are taken away from this center to ocean trenches. At oceanic trenches subduction is occurring.
In a subduction, the lithospheric plate plunges deep into the asthenosphere where they are being melted back.
This is a covergent margin.
This process continues in a dynamic manner to cycle matter on earth.
learn more:
Sea floor spreading brainly.com/question/9912731
#learnwithBrainly
Answer:
Q = 233.42 J
Explanation:
Given data:
Mass of lead = 175 g
Initial temperature = 125.0°C
Final temperature = 22.0°C
Specific heat capacity of lead = 0.01295 J/g.°C
Heat absorbed by water = ?
Solution:
Heat absorbed by water is actually the heat lost by the metal.
Thus, we will calculate the heat lost by metal.
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = T2 - T1
ΔT = 22.0°C - 125.0°C
ΔT = -103°C
Q = 175 g × 0.01295 J/g.°C×-103°C
Q = -233.42 J
Heat absorbed by the water is 233.42 J.
