KE=.5mv^2
M=mass
v=velocity
.5(4)(100)=200
That should be the answer.
Answer:
0.144 nm
Explanation:
Silver's electronic configuration is (Kr)(4d)10(5s)1, and it has an atomic radius of 0.144 nm.
When mixture of NaCl and Al₂(SO₄)₃ is allowed to react with excess NaOH, only Al₂(SO₄)₃ reacts with it and NaCl does not react with NaOH due to presence of common ion (Na⁺). On reaction gelatinous precipitate of aluminium hydroxide [Al(OH)₃] is produced. The balanced chemical reaction is represented as-
Al₂(SO₄)₃ + 6NaOH → 2Al(OH)₃ + 3Na₂SO₄
On this reaction, 0.495 g = 0.495/78 moles =6.346 X 10⁻³ moles of Al(OH)₃.
As per balanced reaction, two moles of Al(OH)₃ is produced from one mole Al₂(SO₄)₃. So, 6.346 X 10⁻³ moles of Al(OH)₃ is produced from (6.346 X 10⁻³)/2 moles=3.173 X 10⁻³ moles of Al₂(SO₄)₃= 3.173 X 10⁻³ X 342 g of Al₂(SO₄)₃=1.085 g of Al₂(SO₄)₃.
So, mass percentage of Al₂(SO₄)₃ is= (amount of Al₂(SO₄)₃/total amount of mixture)X100 =
=74.8 %.
Answer:
Approximately
under standard conditions.
Explanation:
Equation for the overall reaction:
.
Write down the ionic equation for this reaction:
.
The net ionic equation for this reaction would be:
.
In this reaction:
- Zinc loses electrons and was oxidized (at the anode):
. - Copper gains electrons and was reduced (at the cathode):
.
Look up the standard potentials for each half-reaction on a table of standard reduction potentials.
Notice that
is oxidation and is likely not on the table of standard reduction potentials. However, the reverse reaction,
, is reduction and is likely on the table.
The reduction potential of
would be
, the opposite of the reverse reaction
.
The standard potential of the overall reaction would be the sum of the standard potentials of the two half-reactions:
.
Answer 12.2g is the answer