Answer is: 31,45%.
mrs₁(C₉H₁₆O₄-<span>azelaic acid) = 12g.
mr</span>₂(C₉H₁₆O₄) = 50g.
ω₂(C₉H₁₆O₄) = 15% = 0,15.
mrs₂(C₉H₁₆O₄) = mr₂·ω₂ = 50g·0,15 = 7,5g.
mrs₃(C₉H₁₆O₄) = mrs₁ + mr₂ = 12g + 7,5g = 19,5g.
mr₃ = mr₂ + mr₂ = 50g + 12g = 62g.
ω₃ = mrs₃÷mr₃ = 19,5g ÷ 62g = 31,45% = 0,3145.
Answer:
81°C.
Explanation:
To solve this problem, we can use the relation:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat released from water (Q = - 1200 J).
m is the mass of the water (m = 20.0 g).
c is the specific heat capacity of water (c of water = 4.186 J/g.°C).
ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = final T - 95.0°C).
∵ Q = m.c.ΔT
∴ (- 1200 J) = (20.0 g)(4.186 J/g.°C)(final T - 95.0°C ).
(- 1200 J) = 83.72 final T - 7953.
∴ final T = (- 1200 J + 7953)/83.72 = 80.67°C ≅ 81.0°C.
<em>So, the right choice is: 81°C.</em>
Answer:
The mass of water = 219.1 grams
Explanation:
Step 1: Data given
Mass of aluminium = 32.5 grams
specific heat capacity aluminium = 0.921 J/g°C
Temperature = 82.4 °C
Temperature of water = 22.3 °C
The final temperature = 24.2 °C
Step 2: Calculate the mass of water
Heat lost = heat gained
Qlost = -Qgained
Qaluminium = -Qwater
Q = m*c*ΔT
m(aluminium)*c(aluminium)*ΔT(aluminium) = -m(water)*c(water)*ΔT(water)
⇒with m(aluminium) = the mass of aluminium = 32.5 grams
⇒with c(aluminium) = the specific heat of aluminium = 0.921 J/g°C
⇒with ΔT(aluminium) = the change of temperature of aluminium = 24.2 °C - 82.4 °C = -58.2 °C
⇒with m(water) = the mass of water = TO BE DETERMINED
⇒with c(water) = 4.184 J/g°C
⇒with ΔT(water) = the change of temperature of water = 24.2 °C - 22.3 °C = 1.9 °C
32.5 * 0.921 * -58.2 = -m * 4.184 * 1.9
-1742.1 = -7.95m
m = 219.1 grams
The mass of water = 219.1 grams
Answer = c
Conservation of mass (mass is never lost or gained in chemical reactions), during chemical reaction no particles are created or destroyed, the atoms are rearranged from the reactants to the products.