Answer:
The 150 g Al will reach a higher temperature.
Explanation:
- The amount of heat added to a substance (Q) can be calculated from the relation:
Q = m.c.ΔT.
where, Q is the amount of heat added,
m is the mass of the substance,
c is the specific heat of the substance,
ΔT is the temperature difference (final T - initial T).
<em>Since, Q and c is constant, ΔT will depend only on the mass of the substance (m).</em>
∵ ΔT is inversely proportional to the mass of the substance.
∴ The piece with the lowest mass (150.0 g) will reach a higher temperature than that of a higher mass (250.0 g).
<em>So, the right choice is</em>: The 150 g Al will reach a higher temperature.
A.
elements C H N
percentage composition 74.03 8.70 17.27
Molecular mass 12 1 14
# of mole 6.17 8.70 1.23
÷smallest mole 5.0 7.0 1.0
mole ratio 5 : 7 ; 1
THE EMPERICAL FORMUKLA FOR A. IS C5H7O
NOTE: #of mole = percentage composition ÷ Mr
and the ÷ smallest mole is used to find the ratio...for the above question a. it is 1.23
and b. should be done using the same procedure
Answer:
( About ) 110 kilograms
Explanation:
Take a look at the attachment below,
Answer:
V₂ = 112.14 mL
Explanation:
Given data:
Initial volume = 500 mL
Initial pressure = 270 mmHg (270/760 =0.355 atm)
Initial temperature = 55 °C (55 +273 = 328 K)
Final temperature = 100°C (100+273 = 373 K)
Final volume = ?
Final pressure = 1.8 atm
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 0.355 atm × 500mL × 373 K / 328 K × 1.8 atm
V₂ = 66207.5 atm .mL. K / 590.4 K.atm
V₂ = 112.14 mL
