Answer:
The correct option is;
d 4400
Explanation:
The given parameters are;
The mass of the ice = 55 g
The Heat of Fusion = 80 cal/g
The Heat of Vaporization = 540 cal/g
The specific heat capacity of water = 1 cal/g
The heat required to melt a given mass of ice = The Heat of Fusion × The mass of the ice
The heat required to melt the 55 g mass of ice = 540 cal/g × 55 g = 29700 cal
The heat required to raise the temperature of a given mass ice (water) = The mass of the ice (water) × The specific heat capacity of the ice (water) × The temperature change
The heat required to raise the temperature of the ice from 0°C to 100°C = 55 × 1 × (100 - 0) = 5,500 cal
The heat required to vaporize a given mass of ice = The Heat of Vaporization × The mass of the ice
The heat required to vaporize the 55 g mass of ice at 100°C = 80 cal/g × 55 g = 4,400 cal
The total heat required to boil 55 g of ice = 29700 cal + 5,500 cal + 4,400 cal = 39,600 cal
However, we note that the heat required to vaporize the 55 g mass of ice at 100°C = 80 cal/g × 55 g = 4,400 cal.
The heat required to vaporize the 55 g mass of ice at 100°C = 4,400 cal
Answer:
The mass stays the same only volume changes, the volume decreases
Explanation:
The ice shrinks (decreases volume) and becomes more dense. The weight will not (and cannot) change.
The reactant is what you begin with.
The product is what you end up with (so the answer is B)
Answer:
b)4.46 L/hr
Explanation:
To solve this question we need to convert the mL to liters (Using the conversion of 1000mL = 1L) and convert the time from seconds to hours (3600s = 1hr)
<em>mL to L:</em>
1.24mL/s * (1L / 1000mL) = 0.00124L/s
<em>seconds to hours:</em>
0.00124L/s * (3600s / 1hr) = 4.46L/hr
Right answer is:
<h3>b)4.46 L/hr
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