Answer:
C. 0.20 M Mg ion & 0.40 M Cl ion
Explanation:
MgCl₂ is a ionic salt which is dissociated as this
MgCl₂ → Mg²⁺ + 2Cl⁻
First of all, we have a solution of 200 mL, with [MgCl₂] = 0.6M
Molarity . volume = moles.
0.6 mol/l . 0.2l = 0.12 mol
MgCl₂ → Mg²⁺ + 2Cl⁻
0.12mol 0.12 0.24
This moles are also in 400mL of water, so the new concentration is
[Mg²⁺] = 0.12 m/0.6L = 0.2M
[Cl⁻] = 0.24 m/0.6L = 0.4M
Remember we initially have 200mL and then, we add 400 mL, so we supose aditive volume. (600mL)
Please be specific. Which umbrella?
Stack effect? I'm not totally sure about this...
Explanation:
Large amount of tiny particles of water droplets, dust and smoke are present on a misty day. These tiny particles in the air scatter blue colour of white light passing through it. When this scattered light reaches our eyes, the smoke appears blue.
<span>The correct answer is 'freezing point depression'. Colligative properties depend on the concentration of molecules of a solute. Examples of other colligative properties are boiling point elevation or vapour pressure lowering. The salt causes ice on the side walk to melt because it lowers the freezing point. </span>
This doesn't need an ICE chart. Both will fully dissociate in water.
Assume HClO4 and KOH reacts with one another. All you need to do is determine how much HClO4 will remain after the reaction. Calculate pH.
Step 1:
write out balanced equation for the reaction
HClO4+KOH ⇔ KClO4 + H2O
the ratio of HClO4 to KOH is going to be 1:1. Each mole of KOH we add will fully react with 1 mole of HClO4
Step 2:
Determining the number of moles present in HClO4 and KOH
Use the molar concentration and the volume for each:
25 mL of 0.723 M HClO4
Covert volume from mL into L:
25 mL * 1L/1000mL = 0.025 L
Remember:
M = moles/L so we have 0.025 L of 0.723 moles/L HClO4
Multiply the volume in L by the molar concentration to get:
0.025L x 0.723mol/L = 0.0181 moles HClO4.
Add 66.2 mL KOH with conc.=0.273M
66.2mL*1L/1000mL = .0662 L
.0662L x 0.273mol/L = 0.0181 moles KOH
Step 3:
Determine how much HClO4 remains after reacting with the KOH.
Since both reactants fully dissociate and are used in a 1:1 ratio, we just subtract the number of moles of KOH from the number of moles of HClO4:
moles HClO4 = 0.0181; moles KOH = 0.0181, so 0.0181-0.0181 = 0
This means all of the HClO4 is used up in the reaction.
If all of the acid is fully reacted with the base, the pH will be neutral = 7.
Determine the H3O+ concentration:
pH = -log[H3O+]; [H3O+] = 10-pH = 10-7
The correct answer is 1.0x10-7.
