The answer is everything.
False, it is an example of an Ionic solid
Answer:
Explanation:
H3PO4(aq) + 3NaOH(aq) → Na3PO4(aq) + 3H2O(l)
mole of NaOH = 23.6 * 10 ⁻³L * 0.2M
= 0.00472mole
let x be the no of mole of H3PO4 required of 0.00472mole of NaOH
3 mole of NaOH required ------- 1 mole of H3PO4
0.00472mole of NaOH ----------x
cross multiply
3x = 0.0472
x = 0.00157mole
[H3PO4] = mole of H3PO4 / Vol. of H3PO4
= 0.00157mole / (10*10⁻³l)
= 0.157M
<h3>The concentration of unknown phosphoric acid is 0.157M</h3>
<h3>
Answer:</h3>
B. 0.33 mol
<h3>
Explanation:</h3>
We are given;
Gauge pressure, P = 61 kPa (but 1 atm = 101.325 kPa)
= 0.602 atm
Volume, V = 5.2 liters
Temperature, T = 32°C, but K = °C + 273.15
thus, T = 305.15 K
We are required to determine the number of moles of air.
We are going to use the concept of ideal gas equation.
- According to the ideal gas equation, PV = nRT, where P is the pressure, V is the volume, R is the ideal gas constant, (0.082057 L.atm mol.K, n is the number of moles and T is the absolute temperature.
- Therefore, to find the number of moles we replace the variables in the equation.
- Note that the total ball pressure will be given by the sum of atmospheric pressure and the gauge
- Therefore;
- Total pressure = Atmospheric pressure + Gauge pressure
We know atmospheric pressure is 101.325 kPa or 1 atm
Total ball pressure = 1 atm + 0.602 atm
= 1.602 atm
That is;
PV = nRT
n = PV ÷ RT
therefore;
n = (1.602 atm× 5.2 L) ÷ (0.082057 × 305.15 K)
= 0.3326 moles
= 0.33 moles
Therefore, there are 0.33 moles of air in the ball.
Ksp of AgCl= 1.6×10⁻¹⁰
AgCl=Ag⁺ +Cl⁻
Ksp=[Ag⁺][Cl⁻]
Assume [Ag⁺]=[Cl⁻]=x
Ksp=x²
1.6×10⁻¹⁰=x²
x=0.000012
In FeCl₃:
FeCl₃------>Fe⁺³+ 3Cl⁻
as there is 0.010 M FeCl₃
So there will be ,
[Cl⁻]= 0.030
So
[Ag⁺]=Ksp/[Cl⁻]
=1.6×10⁻¹⁰/0.030
=5.3×10⁻⁹
so solubility of AgCl in FeCl₃ will be 5.3×10⁻⁹.
