Volume= Mass/ Density= 20/184
= 0.108 cubic cm
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.
Natural abundance of oxygen I think
The much of the sample that would remain unchanged after 140 seconds is 2.813 g
Explanation
Half life is time taken for the quantity to reduce to half its original value.
if the half life for Scandium is 35 sec, then the number of half life in 140 seconds
=140 sec/ 35 s = 4 half life
Therefore 45 g after first half life = 45 x1/2 =22.5 g
22.5 g after second half life = 22.5 x 1/2 =11.25 g
11.25 g after third half life = 11.25 x 1/2 = 5.625 g
5.625 after fourth half life = 5.625 x 1/2 = 2.813
therefore 2.813 g of Scandium 47 remains unchanged.
Boyle Law says “the pressure of fixed amount of ideal gas which is at constant temperature is
inversely proportional to its volume".<span>
P = 1/V
<span>Where, P is pressure of the ideal gas and V is volume of the ideal gas.</span>
<span>For two situations, this law can be added as;
P</span>₁V₁ = P₂V₂<span>
</span><span>14 lb/in² x V₁ = 70 lb/in² x 500 mL</span><span>
</span><span>V₁ =
2500 mL</span><span>
Hence, the needed volume of atmospheric air = 2500
mL
<span>Here, we made two </span>assumptions. They are,
1. The
atmospheric air acts as ideal gas.
2.
Temperature is a constant.
<span>We didn't convert the units to SI units since
converting volume and pressure are products of two numbers, they will cut off. </span></span></span>
