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1) The metal which reduces the other compound is the one higher in the reactivity. So in this case it is
.
2) The substance which brings about reduction while itself getting oxidised (that is losing electrons) is called a reducing agent. Here, $\mathrm{Zn}$ is the reducing agent and reduces Cobalt Oxide to Cobalt while itself getting oxidised to Zinc oxide.
Answer:
Hi.
The temperature is approximately zero degrees (0°C)
Explanation:
It is important to keep in mind that in the production of ice cream the decrease in the freezing point of the water present in the mixture is called the antifreeze power of the mixture. In ice cream, the freezing point decrease will be caused by each substance that is dissolved in the mixture: lactose, salts, sugars and any other substance. Each of these substances will contribute to the decrease in the freezing point of the mixture. The phase diagram attached in the file shows the sugar solutions in water. When a solution cools (point A), there comes a time when the freezing curve is reached (point B). At that moment ice begins to appear. As shown in the diagram this temperature is approximately zero degrees (0 ° C).
Answer:
The specific heat of the metal is 2.09899 J/g℃.
Explanation:
Given,
For Metal sample,
mass = 13 grams
T = 73°C
For Water sample,
mass = 60 grams
T = 22°C.
When the metal sample and water sample are mixed,
The addition of metal increases the temperature of the water, as the metal is at higher temperature, and the addition of water decreases the temperature of metal. Therefore, heat lost by metal is equal to the heat gained by water.
Since, heat lost by metal is equal to the heat gained by water,
Qlost = Qgain
However,
Q = (mass) (ΔT) (Cp)
(mass) (ΔT) (Cp) = (mass) (ΔT) (Cp)
After mixing both samples, their temperature changes to 27°C.
It implies that
, water sample temperature changed from 22°C to 27°C and metal sample temperature changed from 73°C to 27°C.
Since, Specific heat of water = 4.184 J/g°C
Let Cp be the specific heat of the metal.
Substituting values,
(13)(73°C - 27°C)(Cp) = (60)(27°C - 22℃)(4.184)
By solving, we get Cp =
Therefore, specific heat of the metal sample is 2.09899 J/g℃.