Answer: Option (b) is the correct answer.
Explanation:
Kinetic energy is defined as the energy obtained by the molecules of an object due to their motion.
Also, it is known that kinetic energy is directly proportional to temperature.
Mathematically, K.E =
where, T = temperature
Whereas potential energy is defined as the energy obtained by an object due to its position.
Mathematically, P.E = mgh
where, m = mass
g = acceleration due to gravity
h = height
Therefore, in the given curve when temperature remains constant then kinetic energy of molecules will also remain.
Hence, we can conclude that the segment QR represents an increase in the potential energy, but no change in the kinetic energy.
Answer:
(a) oxygen
(b) 154g (to 3sf)
(c) 79.9% (to 3sf)
Explanation:
mass (g) = moles × Mr/Ar
note: eqn means chemical equation
(a)
moles of P = 84.1 ÷ 30.973 = 2.7152 moles
moles of O2 = 85÷2(16) = 2.65625 moles
Assuming all the moles of P is used up,
moles of O2 / moles of phosphorus = 5/4 (according to balanced chemical eqn)
moles of O2 required = 5/4 × 2.7152moles = 3.394 moles (more than supplied which is 2.65625moles)
therefore there is insufficient moles of O2 and the limiting reactant is oxygen.
(b)
moles of P2O5 produced
= 2/5 (according to eqn) × 2.7152
= 1.08608moles
mass of P2O5 produced
= 1.08608 × [ 2(30.973) + 5(16) ]
= 154.164g
= approx. 154g to 3 sig. fig.
(c)
% yield = actual/theoretical yield × 100%
= 123/154 × 100%
= 79.870%
= approx. 79.9% (to 3sf)
<span>E.) In a chemical reaction, the final amount of the products is determined by the "None of the above"
[ Depends on all physical conditions & chemical situation ]
Hope this helps!</span>
Answer:
9.51 × 10⁴ kL
Explanation:
Step 1: Given data
Volume of the sample (V): 9.51 × 10⁹ cL
Step 2: Convert "V" to liters
We will use the conversion factor 1 L = 100 cL.
9.51 × 10⁹ cL × (1 L / 100 cL) = 9.51 × 10⁷ L
Step 3: Convert "V" to kL
We will use the conversion factor 1 kL = 1000 L.
9.51 × 10⁷ L × (1 kL / 1000 L) = 9.51 × 10⁴ kL
9.51 × 10⁹ cL is equal to 9.51 × 10⁴ kL.