Answer:
When heat gets transferred through electromagnetic waves that move through space
.
Explanation:
Radiation is the propagation of energy in the form of electromagnetic waves or subatomic particles through a vacuum or a material medium.
Answer:
pH = 11.9
Explanation:
First, we <u>determine the number of OH⁻ moles dissolved</u>:
<em>80% of Ca(OH)₂ is dissolved</em>:
- 0.0005 mol * 80/100 = 4x10⁻⁴ mol Ca(OH)₂
<em>There are two OH⁻ moles per Ca(OH)₂ mol</em>:
- 4x10⁻⁴ * 2 = 8x10⁻⁴ mol OH⁻
Now we can <u>calculate the molar concentration of OH⁻</u> (moles/L):
- 100 mL ⇒ 100/1000 = 0.1 L
- [OH⁻] = 8x10⁻⁴ mol / 0.1 L = 8x10⁻³ M
Then we <u>calculate the pOH of the solution</u>:
- pOH = -log[OH⁻] = -log(8x10⁻³ M) = 2.10
Finally, we can <u>calculate the pH of the solution</u> using the equation
The solution would not be able to conduct electricity because there would not be free flowing ions in the solution for electricity to travel through.
Answer:
O B. Convert the 10 g of NaCl to moles of NaCl.
Explanation:
The formula for finding the molality is m=moles of solute/kg of solvent. The solute for this question is NaCl and the solvent is water.
(10g NaCl)(1 mol NaCl/58.44g NaCl)=0.1711 mol NaCl
58.44 is the molar mass of NaCl
m=0.1711 mol NaCl/2 kg H2O
m=0.085557837
- <u>The reaction that takes place is:</u>
Hg(NO₃)₂(ac) + Na₂S(ac) → HgS(s) + 2Na⁺ + 2NO₃⁻
Now we calculate the moles of each reagent -using the molecular weights-, in order to determine the limiting reactant:
- Moles of mercury (II) nitrate = 85.14 g * =0.2622 moles.
- Moles of sodium sulfide = 14.334 g *=0.1837 moles.
Because the stoichiometric ratio between the reactants is 1:1, we compare the number of moles of each one upfront.
moles Hg(NO₃)₂ > moles Na₂S
<u>Thus Na₂S is the limiting reagent.</u>
So in order to find the mass of solid precipitate, we must calculate it using the moles of Na₂S:
The mass of the solid precipitate is 42.760 g.
- In order to calculate the grams of the reactant in excess that will remain after the reaction, we convert the moles that reacted into mass and substract them from the original mass:
Mass of Hg(NO₃)₂ remaining =
The mass of the remaning reactant in excess is 25.49 g.
- Because we assume complete precipitation, there are no more Hg⁺² or S⁻² ions in solution. The moles of NO₃⁻ and Na⁺ in solution remain the same during the reaction, so the number is calculated from the number added in the reactant:
Hg⁺²: 0 mol
NO₃⁻:
Na⁺:
S²⁻: 0 mol