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:
8L
Explanation:
Using Boyle's law which states that the volume of a given mass of gas is inversely proportional to the pressure, provided temperature remains constant
P1V1= P2V2
P1 = 2atm, V1 = 12L ,
P2 = 3atm , V2 =
12 × 2 = V2 × 3
Divide both sides by 3
V2 = 24 ÷ 3
V2 = 8L
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Answer:
1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)
Explanation:
Main reaction: 2Ag⁺(aq) + Mn(s) ⇄ 2Ag(s) + Mn²⁺(aq)
In the oxidation half reaction, the oxidation number increases:
Mn changes from 0, in the ground state to Mn²⁺.
The reduction half reaction occurs where the element decrease the oxidation number, because it is gaining electrons.
Silver changes from Ag⁺ to Ag.
1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)
To balance the hole reaction, we need to multiply by 2, the second half reaction:
Mn(s) ⇄ Mn²⁺(aq) + 2e⁻
(Ag⁺(aq) + 1e⁻ ⇄ Ag(s)) . 2
2Ag⁺(aq) + 2e⁻ ⇄ 2Ag(s)
Now we sum, and we can cancel the electrons:
2Ag⁺(aq) + Mn(s) + 2e⁻ ⇄ 2Ag(s) + Mn²⁺(aq) + 2e⁻
Answer:
The answer to the question is
The pressure of carbon dioxide after equilibrium is reached the second time is 0.27 atm rounded to 2 significant digits
Explanation:
To solve the question, we note that the mole ratio of the constituent is proportional to their partial pressure
At the first trial the mixture contains
3.6 atm CO
1.2 atm H₂O (g)
Total pressure = 3.6+1.2= 4.8 atm
which gives
3.36 atm CO
0.96 atm H₂O (g)
0.24 atm H₂ (g)
That is
CO+H₂O→CO(g)+H₂ (g)
therefore the mixture contained
0.24 atm CO₂ and the total pressure =
3.36+0.96+0.24+0.24 = 4.8 atm
when an extra 1.8 atm of CO is added we get Increase in the mole fraction of CO we have one mole of CO produces one mole of H₂
At equilibrium we have 0.24*0.24/(3.36*0.96) = 0.017857
adding 1.8 atm CO gives 4.46 atm hence we have
(0.24+x)(0.24+x)/(4.46-x)(0.96-x) = 0.017857
which gives x = 0.031 atm or x = -0.6183 atm
Dealing with only the positive values we have the pressure of carbon dioxide = 0.24+0.03 = 0.27 atm
<h3>Answer:</h3>
Correct Option-A (Ability to burn skin)
<h3>Explanation:</h3>
When skin tissues are exposed to Acids or Bases a chemical burn occurs as both of these substances are corrosive in nature. These burns occur without providing any heat, results from a very fast reaction, are extremely painful and causes damage to structures present under skin.
Option-B is incorrect because Acids taste sour, while, Bases taste bitter.
Option-C is incorrect because pH of Acids is less than 7 while, pH of Bases is greater than 7.
