Answer:
0.862 J/gºC
Explanation:
The following data were obtained from the question:
Mass of metal (Mₘ) = 50 g
Initial temperature of metal (Tₘ) = 100 °C
Mass of water (Mᵥᵥ) = 400 g
Initial temperature of water (Tᵥᵥ) = 20 °C
Equilibrium temperature (Tₑ) = 22 °C
Specific heat capacity of water (Cᵥᵥ) = 4.2 J/gºC
Specific heat capacity of metal (Cₘ) =?
The specific heat capacity of the metal can be obtained as follow:
Heat lost by metal = MₘCₘ(Tₘ – Tₑ)
= 50 × Cₘ × (100 – 22)
= 50 × Cₘ × 78
= 3900 × Cₘ
Heat gained by water = MᵥᵥCᵥᵥ(Tₑ – Tᵥᵥ)
= 400 × 4.2 × (22 – 20)
= 400 × 4.2 × 2
= 3360 J
Heat lost by metal = Heat gained by water
3900 × Cₘ = 3360
Divide both side by 3900
Cₘ = 3360 / 3900
Cₘ = 0.862 J/gºC
Therefore, the specific heat capacity of the metal is 0.862 J/gºC
Answer:

Explanation:
Complete ionic equation : In complete ionic equation, all the substance that are strong electrolyte and present in an aqueous are represented in the form of ions.
Net ionic equation : In the net ionic equations, we are not include the spectator ions in the equations.
Only the species which are present in aqueous state dissociate.
Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.
(a)
The balanced molecular equation will be,
The complete ionic equation in separated aqueous solution will be,
In this equation the species present are,
are the spectator ions.
Hence, the net ionic equation contains specie is

D
A- Harvard university professor
The ostrich runs 21 N then goes back by 25 S
The displacement is -4 S
21 - 25 = -4
The HCl added = 1.25 moles
and the moles of Na2HPO4 = 1 mole
Now when acid is added in the given solution of Na2HPO4
One mole of H+ will react with one mole of Na2HPO4 to given one mole of NaH2PO4
Na2HPO4 + H+ ---> NaH2PO4
Now this one mole formed NaH2PO4 will further react with 0.25 moles of H+ left to form 0.25 moles of H3PO4 and 0.75 moles of NaH2PO4 will remain in the solution
So this will result into formation of a buffer of phosphoric acid and NaH2PO4
NaH2PO4 + H+ ---> H3PO4
pKa of H3PO4 = 2.1
so pH = pKa + log [salt] / [acid] = 2.1 + log [0.75 / 0.25] = 2.58
so the pH will be in between 2.1 to 7.2