Its quite alot of work, but here it is....
1. Silver Nitrate: AgNO3 (aq)
<span>Potassium Chloride: KCl </span>
<span>To do a double replacement reaction, switch the two metals around, in this case Silver and Potassium so you are left with: </span>
<span>Potassium Nitrate: KNO3 </span>
<span>Silver Chloride: AgCl </span>
<span>The chemical equation should be something like this: </span>
<span>AgNO3 (aq) + KCl => KNO3 + AgCl
</span>
2The chemical reaction is as follows:Al2(SO4)3 + 3Ca(OH)2 → 3CaSO4 + 2Al(OH)3 Explanation:1mol of Al2(SO4)3 will react with 3mol Ca(OH)2 to produce 3mol CaSO4 and 2 mol Al(OH)3.First we have to find the number of moles of Al2(SO4)3 :Number of moles = Mass/ Molar massMass of Al2(SO4)3 = 500gMolar mass of Al2(SO4)3 = 342.15 g/molNumber of moles = 500/342.15Number of moles = 1.461 mol Al2(SO4)3Multiplying the coeffecient of Ca(OH)2 with 1.461:= 3*1.461 = 4.383 mol Ca(OH)2 Now we have to find the number of moles of Ca(OH)2:Mass of Ca(OH)2 = 450gMolar mass of Ca(OH)2 = 74.09 g/mol Number of moles = 450/74.09Number of moles = 6.074 mol Ca(OH)2We need 4.383mol to react completely with the Al2(SO4)3, so the Ca(OH)2 is in excess, and the Al2(SO4)3 is the limiting reactant. Excess unreacted: 6.074-4.383 = 1.69mol Ca(OH)2 unreacted
a. displacement = 6 km
b. velocity = 4 km/h
<h3>Further explanation</h3>
Given
a morning paper route
Required
displacement
velocity
Solution
Displacement is a vector quantity that shows changes in the position of objects in a certain interval of time. Displacement has magnitude and direction
Can be simplified displacement = distanced traveled from starting point to ending point
a. Since we only consider the starting point and the end point, we just need to add up the distances to the north
3 km + 3 km = 6 km
b. Velocity = displacement changes with time
Answer:
Henderson Hasselbalch equation: pH = pKa + log [salt]/[acid]
You need to know the pKa for acetic acid. Looking it up one finds it to be 4.76
(a). pH = 4.76 + log [0.13]/[0.10]
= 4.76 + 0.11
= 4.87
(b) KOH + CH3COOH =>H2O + CH3COOK so (acid)goes down and (salt)goes up. Assuming no change in volume, you have 0.10 mol acid - 0.02 mol = 0.08 mol acid and 0.13 mol salt + 0.02 mol = 0.15 mol salt
pH = 4.76 + log [0.15]/[0.08]
= 4.76 + 0.27
= 5.03
Answer:
cinnamic acid - 150 mg
cis-stilbene - 100 μL
trans- stilbene - 100 mg
pyridinium tribromide - 200-385 mg
For this data:
moles of cinnamic acid = 0.150 g/148.16 g/mol = 0.001 mols
Theoretical mass of dibromoproduct formed = 0.001 mol x 307.97 g/mol = 0.312 g
cis-stilbene (100 ul = 0.1 ml)
moles of cis-stilbene = 0.1 ml x 1.01 g/mol/180.25 g/mol = 0.00056 mols
Theoretical mass of dibromoproduct formed = 0.00056 mol x 340.05 g/mol = 0.19 g
trans-stilbene
moles of tran-stilbene = 0.1 g/180.25 g/mol = 0.00055 mols
Theoretical mass of dibromoproduct formed = 0.00055 mol x 340.05 g/mol = 0.19 g
Explanation: