Answer:
A1+3
Explanation:
A1+3 is the most common ion of aluminum.
Answer: 3.54 x 10^{-9}
Explanation:
For this Case we have keep in mind the effect of the Common Ion, which is Cl.
You can find the amount of silver ions at equilibrium from Ksp since You already know the chloride ion concentration. Keep in mind that only a few amount of silver chloride will be dissolved. This means the chloride concentration will stay in 0.05 M, Then I can use Ksp to solve for the silver ion concentration. So you can use the following equation.
AgCl Ksp = Ksp/[Cl^-]= (1.77 x 10^{-10})/(0.05) = 3.54 x 10^{-9}
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:
Formaldehyde(CH
O) or methanal is your answer.
