Answer:
7.54%
Explanation:
A hydrate is a salt that has molecules of water incorporated into the crystals. It is represented by the molecular formula of the salt followed by how many molecules of water it has: XY.nH₂O.
So, the mass of water in the sample will be the difference between the hydrate and the salt without water:
mass of water = 1.034 - 0.956 = 0.078 g
The mass percentage is the mass of water divided by the total mass, and then multiplied by 100%:
(0.078/1.034)x100%
7.54%
<em> $ 434.40</em>
<em>41 miles ............ 1 gallon</em>
<em>4700 miles .......x gallon</em>
<em>x = 4700×1/41 = 114.63 gallons</em>
<em />
<em>114.63 gallons×$3.79 ≈ $ 434.40</em>
Answer:
2 moles of Sn are produced when 4 moles of H2(g) are consumed completely
Explanation:
to determine the number of moles of sn (l) produced when 4.0 moles of H2 (g) is consumed completely.
First, find the number of moles of H2 consumed by taking this as limiting reagent.
Then find the moles of Sn (l) taking into account the stoichiometric relationship between H2(g) and Sn(l). 2:1
(s) + 2
(g) ⇒ Sn(l) + 2
(g)
∴2 moles of Sn are produced when 4 moles of H2(g) are consumed completely.
(a) (i) Zinc blende is mainly Zinc sulphide and in industry it is converted to ZnO by flames to burn off the sulphur as SO2 which is then used to make sulphuric acid.
(ii) Reduction of ZnO to Zn:-
2ZnO + C = CO2 + 2Zn
(b) This is called sacrificial anodic protection. The zinc ( which is more reactive than iron) when in contact with the oxygen in the air is oxidised and in the process loses its electrons. These travel through the electrolyte to the iron. The electrons then reduce any iron oxide to iron: fe2+ + 2e ---> Fe.
Answer:
18.94%.
Explanation:
- The decay of carbon-14 is a first order reaction.
- The rate constant of the reaction (k) in a first order reaction = ln (2)/half-life = 0.693/(5730 year) = 1.21 x 10⁻⁴ year⁻¹.
- The integration law of a first order reaction is:
<em>kt = ln [A₀]/[A]</em>
k is the rate constant = 1.21 x 10⁻⁴ year⁻¹.
t is the time = 13,750 years.
[A₀] is the initial percentage of carbon-14 = 100.0 %.
[A] is the remaining percentage of carbon-14 = ??? %.
∵ kt = ln [Ao]/[A]
∴ (1.21 x 10⁻⁴ year⁻¹)(13,750 years) = ln (100.0%)/[A]
1.664 = ln (100.0%)/[A]
Taking exponential for both sides:
5.279 = (100.0%)/[A]
<em>∴ [A]</em> = (100.0%)/5.279 = <em>18.94%.</em>
