Answer:
(a) 
(b)


(c) 
Explanation:
Hello,
(a) In this case, with the given formula we easily compute the mass of gold contained in the sovereign as shown below:

(b) Now, by knowing the density of gold and copper, 19.32 and 8.94 g/cm³ respectively, we compute each volume, by also knowing that the rest of the coin contains copper:


(c) Finally, the volume is computed by dividing the total mass over the total volume containing both gold and copper:

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Uses of nonmetals in our daily life: Oxygen which is 21% by volume helps in the respiration process.
Nonmetals used in fertilizers: Fertilizers contain nitrogen.
Nonmetals used in crackers: Sulphur and phosphorus are used in fireworks.
Alkanes are saturated hydrocarbon that contains only single bonds, whereas Alkenes and Alkynes are unsaturated hydrocarbons which contain one or more double bond and triple bonds.
<u>Explanation:</u>
- A saturated hydrocarbon with an only single bond is called alkanes. Ethane consisting of two carbon atoms that are bonded with a single bond and six hydrogen atoms sharing the other valence electron of carbon atoms. The molecular structure of alkane is CnH2n+2.
- An unsaturated hydrocarbon with a two bond is called alkenes. Ethene consisting of two carbon atoms double-bonded to each other. The molecular structure of alkene is CnH2n.
- An unsaturated hydrocarbon with a triple bond is called as alkynes. It involves sharing three pairs of electrons. The molecular structure of alkyne is CnH2n-2.
Answer:
Explanation:
Combustion reaction is given below,
C₂H₅OH(l) + 3O₂(g) ⇒ 2CO₂(g) + 3H₂O(g)
Provided that such a combustion has a normal enthalpy,
ΔH°rxn = -1270 kJ/mol
That would be 1 mol reacting to release of ethanol,
⇒ -1270 kJ of heat
Now,
0.383 Ethanol mol responds to release or unlock,
(c) Determine the final temperature of the air in the room after the combustion.
Given that :
specific heat c = 1.005 J/(g. °C)
m = 5.56 ×10⁴ g
Using the relation:
q = mcΔT
- 486.34 = 5.56 ×10⁴ × 1.005 × ΔT
ΔT= (486.34 × 1000 )/5.56×10⁴ × 1.005
ΔT= 836.88 °C
ΔT= T₂ - T₁
T₂ = ΔT + T₁
T₂ = 836.88 °C + 21.7°C
T₂ = 858.58 °C
Therefore, the final temperature of the air in the room after combustion is 858.58 °C