Answer: D
Explanation: Keeping the pressure constant and increasing the temperature.
Answer:
(a) weight percent of Cu = 44.59%
(b) weight percent of Zn = 53.49%
(c) weight percent of Pb = 1.91%
Explanation:
Given: Alloy contains- Cu=93.1 g, Zn=111.7 g, Pb=4.0 g
Therefore, the total mass of the alloy (m) = mass of Cu (m₁) + mass of Zn (m₂)+ mass of Pb (m₃)
m= 93.1 g + 111.7 g + 4.0 g = 208.8 g
(a) weight percent of Cu = (m₁ ÷ m)× 100% = (93.1 g ÷ 208.8 g)× 100% =44.59%
(b) weight percent of Zn = (m₂ ÷ m)× 100% = (111.7 g ÷ 208.8 g)× 100% =53.49%
(c) weight percent of Pb = (m₃ ÷ m)× 100% = (4.0 g ÷ 208.8 g)× 100% =1.91%
Answer:
T₂ = 19.95°C
Explanation:
From the law of conservation of energy:
![Heat\ Lost\ by\ Copper = Heat\ Gained\ by\ Water\\m_cC_c\Delta T_c = m_wC_w\Delta T_w](https://tex.z-dn.net/?f=Heat%5C%20Lost%5C%20by%5C%20Copper%20%3D%20Heat%5C%20Gained%5C%20by%5C%20Water%5C%5Cm_cC_c%5CDelta%20T_c%20%3D%20m_wC_w%5CDelta%20T_w)
where,
mc = mass of copper = 37.2 g
Cc = specific heat of copper = 0.385 J/g.°C
mw = mass of water = 188 g
Cw = specific heat of water = 4.184 J/g.°C
ΔTc = Change in temperature of copper = 99.8°C - T₂
ΔTw = Change in temperature of water = T₂ - 18.5°C
T₂ = Final Temperature at Equilibrium = ?
Therefore,
![(37.2\ g)(0.385\ J/g.^oC)(99.8\ ^oC-T_2)=(188\ g)(4.184\ J/g.^oC)(T_2-18.5\ ^oC)\\99.8\ ^oC-T_2 = \frac{(188\ g)(4.184\ J/g.^oC)}{(37.2\ g)(0.385\ J/g.^oC)}(T_2-18.5\ ^oC)\\\\99.8\ ^oC-T_2 = (54.92) (T_2-18.5\ ^oC)\\54.92T_2+T_2 = 99.8\ ^oC + 1016.02\ ^oC\\\\T_2 = \frac{1115.82\ ^oC}{55.92}](https://tex.z-dn.net/?f=%2837.2%5C%20g%29%280.385%5C%20J%2Fg.%5EoC%29%2899.8%5C%20%5EoC-T_2%29%3D%28188%5C%20g%29%284.184%5C%20J%2Fg.%5EoC%29%28T_2-18.5%5C%20%5EoC%29%5C%5C99.8%5C%20%5EoC-T_2%20%3D%20%5Cfrac%7B%28188%5C%20g%29%284.184%5C%20J%2Fg.%5EoC%29%7D%7B%2837.2%5C%20g%29%280.385%5C%20J%2Fg.%5EoC%29%7D%28T_2-18.5%5C%20%5EoC%29%5C%5C%5C%5C99.8%5C%20%5EoC-T_2%20%3D%20%2854.92%29%20%28T_2-18.5%5C%20%5EoC%29%5C%5C54.92T_2%2BT_2%20%3D%2099.8%5C%20%5EoC%20%2B%201016.02%5C%20%5EoC%5C%5C%5C%5CT_2%20%3D%20%5Cfrac%7B1115.82%5C%20%5EoC%7D%7B55.92%7D)
<u>T₂ = 19.95°C</u>
Extrapolation allows you to estimate values greater than those in your actual data.
True