Answer:
Explanation:
Fe⁺² (aq) + 2e⁻ = Fe (s) ; E⁰ = - .44 V
Fe⁺³ (aq) + e⁻ = ® Fe²⁺ (aq) ; E⁰ = + .77 V
Reduction potential of second reaction is more , so it will take place , ie Fe⁺³ will be reduced and Fe will be oxidised .
So reaction in the combined cell will be
2Fe⁺³ + Fe = 3Fe⁺²
cell potential = .77 - ( - .44 )
= 1.21 V .
The ore contains 55.4% calcium phosphate (related to the mineral apatite) so the amount of Ca3(PO4)2 is 55.4%x=1000g so x=1000/0.554= 1.805kg. Now for the % of P in this amount of calcium phosphate, use all the masses of the elements in Ca3PO4= Ca=40.078 x 3= 120.23 and (PO4)2= (30.974+64)2=189.95 (NB oxygen is 16 mass x 4 =64) so the total mass is 310.2 and we have 61.95 of P (Pmass x 2) so 61.95/3102.= 0.19 or 19% P. So of the 1.805 x 0.19= 0.34kg of phosphorus.
Answer:
Option A
Explanation:
Number of millimoles of Na3PO4 = 1 × 100 = 100
Number of millimoles of AgNO3 = 1 × 100 = 100
When 1 mole of Na3PO4 is dissociated we get 3 moles of sodium ions and 1 mole of phosphate ion
When 1 mole of AgNO3 is dissociated, we get 1 mole of Ag+ and 1 mole of NO3-
As Ag+ concentration is negligible, the dissociated Ag+ ion must have form the precipitate with phosphate ion and as number of moles of Ag+ and phosphate ion are same, therefore the concentration of phosphate ion must be negligible
Here as 100 millimoles of Na3PO4 is there, we get 300 millimoles of Na+ and 100 millimoles of PO43-
And as 100 millimoles of AgNO3 is there, we get 100 millimoles of Ag+ and 100 millimoles of NO3-
∴ Increasing order of concentration will be PO43- < NO3- < Na+
Answer:
The specific heat capacity of the metal is 0.843J/g°C
Explanation:
Hello,
To determine the specific heat capacity of the metal, we have to work on the principle of heat loss by the metal is equals to heat gained by the water.
Heat gained by the metal = heat loss by water + calorimeter
Data,
Mass of metal (M1) = 512g
Mass of water (M2) = 325g
Initial temperature of the metal (T1) = 15°C
Initial temperature of water (T2) = 98°C
Final temperature of the mixture (T3) = 78°C
Specific heat capacity of metal (C1) = ?
Specific heat capacity of water (C2) = 4.184J/g°C
Heat loss = heat gain
M2C2(T2 - T3) = M1C1(T3 - T1)
325 × 4.184 × (98 - 78) = 512 × C1 × (78 - 15)
1359.8 × 20 = 512C1 × 63
27196 = 32256C1
C1 = 27196 / 32256
C1 = 0.843J/g°C
The specific heat capacity of the metal is 0.843J/g°C
