Answer:
<u>The temperature difference is</u> 
Explanation:
The formula that is to used is :
Δ
Δ
<em>where ,</em>
- <em>Δ
is the heat supplied in calories = 300cal</em> - <em>
is the mass of water taken = m (assumed)</em> - <em>Δ
is the change in temperature</em> - <em>
is the specific heat of water =
</em>
ΔT :

Answer:

Explanation:
Half reaction:
(1)
balance: 
(2)H and O balance in acidic medium:
(3) charge balance:
Hence balanced half-reaction:
First step is to balance the reaction equation. Hence we get
P4 + 5 O2 => 2 P2O5
Second, we calculate the amounts we start with
P4: 112 g = 112 g/ 124 g/mol – 0.903 mol
O2: 112 g = 112 g / 32 g/mol = 3.5 mol
Lastly, we calculate the amount of P2O5 produced.
2.5 mol of O2 will react with 0.7 mol of P2O5 to produce 1.4
mol of P2O5.
This is 1.4 * (31*2 + 16*5) = 198.8 g
The french broad? idk sorry
Answer:
(a) Potassium 3; phosphorus 1; oxygen 4
(b) Aluminium 3; oxygen 9; hydrogen 9
(c) Iron 10; sulfur 15; oxygen 60
Explanation:
(a) K₃PO₄
In one formula unit of K₃PO₄ , there are three atoms of potassium (K), one atom of phosphorus (P), and four atoms of oxygen (O).
In two formula units there are
Potassium — six atoms
Phosphorus — two atoms
Oxygen — eight atoms
(b) Al(OH)₃
In one formula unit of Al(OH)₃, there is one atom of aluminium (Al), three atoms of O, and three atoms of Hydrogen (H).
In three formula units there are
Aluminium — three atoms
Oxygen — nine atoms
Hydrogen — nine atoms
(c) Fe₂(SO₄)₃
In one formula unit of Fe₂(SO₄)₃, there are two atoms of iron (Fe), three atoms of sulfur (S), and 12 atoms of O.
In five formula units there are
Iron — 10 atoms
Sulfur — 15 atoms
Oxygen — 60 atoms