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Kitty [74]
3 years ago
11

A 100g sample of a metal was heated to 100oC and then quickly transferred to an insulated container holding 100g of water at 22o

C. The temperature of the water rose to reach a final temperature of 35oC. Which of the following can be concluded?
A) The metal temperature changed more than the water temperature did; therefore the metal
lost more thermal energy than the water gained.
B) The metal temperature changed more than the water temperature did, but the metal lost
the same amount of thermal energy as the water gained.
C) The metal temperature changed more than the water temperature did; therefore the heat
capacity of the metal must be greater than the heat capacity of the water.
D) The final temperature is less than the average starting temperature of the metal and the
water; therefore the total energy of the metal and water decreased.
Chemistry
1 answer:
Lady_Fox [76]3 years ago
4 0

Answer:

B) The metal temperature changed more than the water temperature did, but the metal lost

the same amount of thermal energy as the water gained.

Explanation:

Heat capacity or thermal capacity is defined as the amount of heat required by a given mass of a material to raise its temperature by one unit which means that the heat capacity of the water, that is the quantity of heat required to cause a rise from 22°C to 35°C that is a rise of 13°C is the quantity of heat that caused the drop in temperature of the metal from 100°C to 35°C a change of 65°C

The water has more capacity to absorb heat or a higher heat capacity than the metal

However, the first law of thermodynamics states that energy is neither created nor destroyed, but it changes from one form to another. In this case, the thermal energy lost by the metal is the same as the thermal or heat energy gained by the water

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3 years ago
1. NaOH mass of a solution of 200g in which its percentage is 25%. What mass of sulfuric acid solution is needed to completely n
Ede4ka [16]

Answer:

m_{H2SO4 = 61.25 g

m_{Na2SO4} = 88.75 g

Explanation:

m_{NaOH} = \frac{200 . 25 }{100} = 50 g

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2NaOH + H2SO4 ⇒ Na2SO4 + 2H2O

   2        :     1           :      1         :    2

 1.25                                                       (moles)

⇒  n_{H2SO4} = 1.25 × 1 ÷ 2 = 0.625 (moles) ⇒ m_{H2SO4} = 0.625 × 98 = 61.25 g

    n_{Na2SO4} = 1.25 × 1 ÷ 2 = 0.625 (moles) ⇒m_{Na2SO4} = 0.625 × 142 = 88.75 g

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hope it helps

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3 years ago
What is [h3o ] in a solution of 0.075 m hno2 and 0.030 m nano2?
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Answer is: concentration of hydrogenium ions is 9,54·10⁻⁵ M.
c(HNO₂) = 0,075 M.
c(NaNO₂) = 0,035 M.
Ka(HNO₂) = 4,5·10⁻⁵.
This is buffer solution, so use <span>Henderson–Hasselbalch equation:
pH = pKa + log(c(</span>NaNO₂) ÷ c(HNO₂)).
pH = -log(4,5·10⁻⁵) + log(0,035 M ÷ 0,075 M).
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pH = 4,02.
<span>[H</span>₃O⁺] = 10∧(-4,02).
<span>[H</span>₃O⁺] = 0,0000954 M = 9,54·10⁻⁵ M.
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