The effect of an insoluble impurity, such as sand, on the observed melting point of a compound would be none. It will not depress or elevate the melting point of the compound. Instead, it would affect the reading if you are trying to determine the melting point of the compound. This is because you might be missing the actual melting point of the compound since you will be waiting for the whole sample to liquify. You would not be able to determine exactly that temperature because of the insoluble impurity would have a different melting point than that of the compound.
Answer:
18 grams
Explanation:
Because 17,190 yrs divided by the amount of years it takes for a half-life to occur is 3. So 3 half life’s happen, so you divide 144/2 once, equaling 72. You divide 72/2 again (another half-life) and you get 36. You then calculate the third half life by dividing 36/2 which equals 18 grams.
Answer:
16974J of energy are required
Explanation:
The energy required is:
* The energy to heat solid water from -15°C to 0°C using:
q = m*S*ΔT
* The energy to convert the solid water to liquid water:
q = dH*m
* The energy required to increase the temperature of liquid water from 0°C to 75°C
q = m*S*ΔT
The first energy is:
q = m*S*ΔT
<em>m = Mass water = 25g</em>
<em>S is specific heat of ice = 2.03J/g°C</em>
<em>ΔT is change in temperature = 0°C - (-15°C) = 15°C</em>
q = 25g*2.03J/g°C*15°C
q = 761.3J
The second energy is:
q = dH*m
<em>m = Mass water = 25g</em>
<em>dH is heat of fusion of water = 80cal/g</em>
q = 80cal/g*25g
q = 2000cal * (4.184J/1cal) = 8368J
The third energy is:
q = m*S*ΔT
<em>m = Mass water = 25g</em>
<em>S is specific heat of water= 4.184J/g°C</em>
<em>ΔT is change in temperature = 75°C-0°C = 75°C</em>
q = 25g*4.184J/g°C*75°C
q = 7845J
The energy is: 7845J + 8368J + 761J =
16974J of energy are required
