Dropping a 50 gram piece of metal with a temperature of 125° Celsius into 1000 grams of water at 20° Celsius would lead to a heat loss by the metal and an heat gain by the water. These loss and gain should be equal since energy cannot be created nor destroyed.
Answer:
0.886 J/g.°C
Explanation:
Step 1: Calculate the heat absorbed by the water
We will use the following expression
Q = c × m × ΔT
where,
- c: specific heat capacity
- ΔT: change in the temperature
Q(water) = c(water) × m(water) × ΔT(water)
Q(water) = 4.184 J/g.°C × 50.0 g × (34.4 °C - 25.36 °C) = 1.89 × 10³ J
According to the law of conservation of energy, the sum of the energy lost by the solid and the energy absorbed by the water is zero.
Q(water) + Q(solid) = 0
Q(solid) = -Q(water) = -1.89 × 10³ J
Step 2: Calculate the specific heat capacity of the solid
We will use the following expression.
Q(solid) = c(solid) × m(solid) × ΔT(solid)
c(solid) = Q(solid) / m(solid) × ΔT(solid)
c(solid) = (-1.89 × 10³ J) / 32.53 g × (34.4 °C - 100. °C) = 0.886 J/g.°C
17) 8.4 / 20 x 100
18) 20 . 0.5150
19) 6,50% because (as you said) the law of definite proportions states that regardless of the amount, a compound is always composed of the same elements in the same proportion by mass
Correct Answer: First Option
An oxidizing agent causes the oxidation of the other atom/element by itself being reduced. In simple words we can state that oxidizing agent gains electrons from the other atom/element i.e. the other atom loses electrons.
From the given options we have to find in which of them electrons are being removed. When the electrons are removed, the number of protons in the atom will be more than the number of electrons. As a result the net charge on the atom will be positive.
First option lists such a change. Initially charge on Al is neutral, 3 electrons are removed and it get +3 charge. This shows that Al is being oxidized, so it needs an oxidizing agent.
