<u>Answer:</u> The amount of heat required to warm given amount of water is 470.9 kJ
<u>Explanation:</u>
To calculate the mass of water, we use the equation:
Density of water = 1 g/mL
Volume of water = 1.50 L = 1500 mL (Conversion factor: 1 L = 1000 mL)
Putting values in above equation, we get:
To calculate the heat absorbed by the water, we use the equation:
where,
q = heat absorbed
m = mass of water = 1500 g
c = heat capacity of water = 4.186 J/g°C
= change in temperature = 
Putting values in above equation, we get:
Hence, the amount of heat required to warm given amount of water is 470.9 kJ
Answer:
2: Moved faster and spread farther apart.
Explanation:
Restate the question: The movement of the liquid in a thermometer shows changes in temperature. An increase in temperature indicates the molecules in the liquid.
1. moved slower and closer together.
2.moved faster and spread farther apart.
3. contracted in size when heated.
4. expanded in size when heated.
Water that is cold does not have the energy to bounce of the walls, instead it is like a group of animals they group together for the warmth of the others when it gets really cold.
So it cant be 1.
We all know that power lines sag lower on a hot day (or a tire for a car, it has increases pressure). but those are different types of molecules.
So that rules out 3 and 4.
Which leaves you with 2.
The increase in temperature causes the water molecule to gain energy and move quickly, which resulted in water molecule that are farther apart and an increase in water volume.
Hope it helps!
Answer:
It's simpler than most questions :)
Explanation:
1) B
2) B
3) D
4) A
5) A
Hope this will help :)
The boiling point of water at 1 atm is 100 degrees celsius. However, when water is added with another substance the boiling point of it rises than when it is still a pure solvent. This called boiling point elevation, a colligative property. The equation for the boiling point elevation is expressed as the product of the ebullioscopic constant (0.52 degrees celsius / m) for water), the vant hoff factor and the concentration of solute (in terms of molality).
ΔT(CaCl2) = i x K x m = 3 x 0.52 x 0.25 = 0.39 °C
<span> ΔT(Sucrose) = 1 x 0.52 x 0.75 = 0.39 </span>°C<span>
</span><span> ΔT(Ethylene glycol) = 1 x 0.52 x 1 = 0.52 </span>°C<span>
</span><span> ΔT(CaCl2) = 3 x 0.52 x 0.50 = 0.78 </span>°C<span>
</span><span> ΔT(NaCl) = 2 x 0.52 x 0.25 = 0.26 </span>°C<span>
</span>
Thus, from the calculated values, we see that 0.75 mol sucrose dissolved on 1 kg water has the same boiling point with 0.25 mol CaCl2 dissolved in 1 kg water.
Because copper sulphate is not an ionic substance, no bonds are changed when it is dissolved. Thus it has to be a physical change
