The amount of heat (in calories) needed to raise the temperature is 324.885 calories
<h3>How to determine the temperature change </h3>
We'll begin by obtaining the temperature change. This can be obtained as followed:
- Initial temperature (T₁) = 14 °C
- Final temperature (T₂) = 25 °C
- Change in temperature (ΔT) = ?
ΔT = T₂ - T₁
ΔT = 25 - 14
ΔT = 11 °C
<h3>How to determine the heat (in Calories)</h3>
The amount of heat needed to raise the temperature can bee obtaimedals follow:
- Mass (M) = 33 g
- Change in temperature (ΔT) = 11 °C
- Specific heat capacity (C) of diethyl ether = 0.895 cal/g°C
- Heat (Q) =?
Q = MCΔT
Q = 33 × 0.895 × 11
Q = 324.885 calories
Thus, the amount of heat needed is 324.885 calories
Learn more about heat transfer:
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Answer:
The material moved by erosion is sediment. The agents of erosion lay down sediment in a process called deposition. Deposition builds new landforms. Weathering, erosion, and deposition act together in a cycle that wears down and builds up Earth's surface.
Explanation:
Answer:
no matter is destroyed or created, it merely changes form. In terms of atoms and bonds, there will be the same amount of atoms at the beginning of an experiment as the amount of atoms at the end of experiment. All that will have happened, is that during the reaction, bonds will have been broken and formed making new compounds. However, the amount of atoms remains exactly the same because matter can not be created or destroyed
Hope this helps!
The question is incomplete and complete question is :
It was important that the flask be completely dry before the unknown liquid was added so that water present would not vaporize when the flask was heated. A typical single drop of liquid water has a volume of approximately 0.050 mL. Assuming the density of liquid water is 1.00 g/mL, how many moles of water are in one drop of liquid, and what volume would this amount of water occupy when vaporized at 100°C and 1atm ?
Answer:
0.0028 moles of water are in one drop of liquid.
Volume of 0.0028 moles of water occupy when vaporized at 100°C and 1 atm is 86 mL.
Explanation:
Volume of of drop = v = 0.050 mL
Mass of drop = m
Density of water = d = 1.00 g/mL
Moles of water in drops:
0.0028 moles of water are in one drop of liquid.
Pressure at which 0.0028 moles are vaporized = P = 1 atm
Temperature at which 0.0028 moles are vaporized = T = 100°C = 100+273 K = 373 K atm
Volume of moles of water = V
Moles of water = n = 0.0028 mol
( ideal gas equation )
V = 0.086 L = 86 mL ( 1L = 1000 mL)
Volume of 0.0028 moles of water occupy when vaporized at 100°C and 1 atm is 86 mL.
