Answer: The heat needed to be removed to freeze 45.0 g of water at 0.0 °C is 15.01 KJ.
Explanation:
- Firstly, we need to define the term <em>"latent heat"</em> which is the amount of energy required "absorbed or removed" to change the phase "physical state; solid, liquid and vapor" without changing the temperature.
- Types of latent heat: depends on the phases that the change occur between them;
- Liquid → vapor, <em>latent heat of vaporization</em> and energy is absorbed.
- Vapor → liquid, latent heat of liquification and the energy is removed.
- Liquid → solid, <em>latent heat of solidification</em> and the energy is removed.
- Solid → liquid, <em>latent heat of fusion</em> and the energy is absorbed.
- In our problem, we deals with latent heat of freezing "solidification" of water.
- The latent heat of freezing of water, ΔHf, = 333.55 J/g; which means that the energy required to be removed to convert 1.0 g of water from liquid to solid "freezing" is 333.55 g at 0.0 °C.
- Then the amount of energy needed to be removed to freeze 45.0 g of water at 0.0 °C is (ΔHf x no. of grams of water) = (333.55 J/g)(45.0 g) = 15009.75 J = 15.01 KJ.
Answer:
C
Explanation:
Lime water turns cloudy when it reacts Carbon dioxide
Answer:the initial composition of the reactants is
40cm^3 of CH4
40cm^3of H2
100cm^3 of H2O
Explanation:
Balanced reaction is
CH4 +H2+5/2O2______
CO2 +3H2O
Excess KOH at room temperature absorbs CO2 whose volume is given by 40cm^3 i.e the volume by which the solution decreases
So using Gay lussac combining ratio which states that gases combine in volumes that are in simple ratio to each other if gases.
Since CO2 in the equation is 1 mole
Means 1mole represent 40cm^3
So CH4:H2:O2 are in ratio of 1:1:5/2=(40:40:100)cm^3 respectively.
Answer: The air molecules move closer together.
Explanation:
