Answer is: 0,250 atm.
T₁(chlorine) = 27°C = 300K.
T₂(chlorine) = 227°C = 500K.
T₂/T₁ = 1,66.
p1(chlorine) = 0,150 atm.
p2(chlorine) = ?
V(chlorine) = constant.
n(chlorine) = 0,0200 mol.
R - gas constant, R = 0.0821 atm·l/mol·K.
Ideal gas law: p·V = n·R·T.
p = n·R·T÷V.
If temperature was increased 1,66 times, pressure also increase 1,66 times: 1,66·0,150 atm = 0,250 atm.
Answer : The incorrect statements are:
(A) When two nonmetals react, the compound formed is ionic.
(B) Two nonmetals can undergo an oxidation-reduction reaction.
Explanation :
Covalent compound : It is defined as the compound which is formed by the sharing of electrons between the atoms forming a compound.
The covalent compound are usually formed when two non-metals react.
The two nonmetals can not undergo an oxidation-reduction reaction.
Ionic compound : It is defined as the compound which is formed when electron gets transferred from one atom to another atom.
Ionic compound are usually formed when a metal reacts with a non-metal.
The metal-nonmetal reaction can always be assumed to be an oxidation-reduction reaction.
For example : 
In this reaction, sodium shows oxidation because the oxidation state changes from (0) to (+1) and and chlorine shows reduction because the oxidation state changes from (0) to (-1).
Hence, the incorrect statements are, (A) and (B)
Answer:
Reactants: Ethanol (C₅H₅OH), Oxygen (O₂)
Products: Carbon dioxide (CO₂) and Water (H₂O)
Explanation:
Let’s rewrite each part of the described situation in chemical equation terms.
Ethanol - C₂H₅OH
is burned in the prescence of - +
oxygen - O₂
producing - →
carbon dioxide - CO₂
and - +
water - H₂O
Chemical equation:
C₂H₅OH + O₂ → CO₂ + H₂O
The terms on the left side of a chemical equation are reactants, and the terms on the right side of a chemical equation are products.
Therefore, since we know this, ethanol (C₂H₅OH) & oxygen (O₂) are the reactants, and carbon dioxide (CO₂) & water (H₂O) are the products.
Answer:
Solution A that will form a precipitate with Ksp = 2.3 x 10−4
Explanation:
Li₃PO₄ ⇄ 3 Li⁺(aq) + PO₄³⁻(aq)
3S S
Where S = Solubility(mole/lit) and Ksp = Solubility product
⇒ Ksp = (3S)³ x (S)
⇒ 27S⁴ = 2.3x10−4
⇒ S = 0.05 mol/lit
Concentration of Li₃PO₄ precipitate = 0.05
<u>Solution A </u>
0.500 lit of a 0.3 molar LiNO₃ contains 0.5 x 0.3 = 0.15 mole
0.4 lit of a 0.2 molar Na₃PO₄ contains = 3 x 0.4 x 0.2 = 0.24 mole
3 LiNO₃ + Na₃PO₄ → 3 NaNO₃ + Li₃PO₄
(Mole/Stoichiometry)

= 0.05 = 0.24
Since from (Mole/Stoichiometry) ratio we can conclude that LiNO₃ is limiting reagent.
So concentration of Li₃PO₄ is equal to 0.05.
The maximum oxygen uptake is known as the VO2 max.