Yellow paint has 0.511 % PbCrO4 by mass
Mass of PbCrO4 in 1 kg of paint = (0.511 / 100) * 1 kg = 0.00511 kg = 5.11 g
Moles of PbCrO4 = 5.11 g/ 323.19 g/mol
= 0.0158 moles
Moles of K2CrO4 also = 0.0158 moles
Moles of FeCr2O4 = 0.0158 moles K2CrO4 * (4 mole FeCr2O4 / 8 moles K2CrO4)
= 0.0079 moles
Mass of FeCr2O4 (chromite) = 0.0079 moles * 223.83 g/mol
= 1.77 g
Paint is a colored liquid, liquefiable, or solid mastic composition that transforms into a thin solid film after being applied to a substrate. Most commonly used for protection, coloring, or adding texture. Paints come in a variety of colors and can be made in a variety of ways.
Paint consists of pigments, solvents, resins, and various additives. Pigments give the paint its color. Solvents make application easier. The resin helps dry. Additives range from fillers to antifungal agents. There are hundreds of different natural and synthetic pigments.
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Answer:
Exothermic reaction for the HCl, endothermic reaction for the water
Explanation:
Heat was lost by HCl as its temperature lowered, so it was an exothermic reaction for the HCL.
Heat was gained by water as its temperature increased, so it was an endothermic reaction for the water.
m = Mass of water = 100 g
c = Specific heat of water = 
= Change in temperature of water = 
Heat is given by
Heat gained by water is .
Which two solutions, when mixed together, will undergo a double replacement reaction and form a white, solid substance?
1. NaCl(aq) and LiNO3(aq)
2. KCl(aq) and AgNO3(aq) answer
3. KCl(aq) and LiCL(aq)
4. NaNO3(aq) and AgNO3(aq)
2 is the answer because AgCl is formed and that is a white ppt.
Answer is: <span>he boiling point of a 1.5 m aqueous solution of fructose is </span>100.7725°C.
The boiling point
elevation is directly proportional to the molality of the solution
according to the equation: ΔTb = Kb · b.<span>
ΔTb - the boiling point
elevation.
Kb - the ebullioscopic
constant. of water.
b - molality of the solution.
Kb = 0.515</span>°C/m.
b = 1.5 m.
ΔTb = 0.515°C/m · 1.5 m.
ΔTb = 0.7725°C.
Tb(solution) = Tb(water) + ΔTb.
Tb(solution) = 100°C + 0.7725°C = 100.7725°C.
Answer: -
0.1 ml of bleach should be added to each liter of test solution.
Explanation:-
Let the volume of bleach to be added is B ml.
Density of stock solution = 1.0 g/ml
Mass of stock solution = Volume of stock x density of stock
= B ml x 1.0 g/ml
= B g
Amount of NaOCl in this stock solution = 5% of B g
= 
x B g
= 0.05 B g
Now each test solution must be added 5 mg/l NaOCl.
Thus each liter of test solution must have 5 mg.
Thus 0.05 B g = 5 mg
= 0.005 g
B = 
= 0.1
Thus 0.1 ml of bleach should be added to each liter of test solution.
