Complete question:
ΔU for a van der Waals gas increases by 475 J in an expansion process, and the magnitude of w is 93.0 J. calculate the magnitude of q for the process.
Answer:
The magnitude of q for the process 568 J.
Explanation:
Given;
change in internal energy of the gas, ΔU = 475 J
work done by the gas, w = 93 J
heat added to the system, = q
During gas expansion process, heat is added to the gas.
Apply the first law of thermodynamic to determine the magnitude of heat added to the gas.
ΔU = q - w
q = ΔU + w
q = 475 J + 93 J
q = 568 J
Therefore, the magnitude of q for the process 568 J.
The answer is C. Life
Hope this helps! :)
Answer:
b) O₂
d) CH₄
Explanation:
In a chemical equation, the <u>reactants</u> are the substances that are written on the <u>left side</u>. The reactants react to produce the products. On the right side, the products are written.
The arrow => is used to separate the reactants from the products.
To the left of the arrow, you can see this written:
CH₄ + 2O₂
The plus sign + is used to separate each reactant.
In 2O₂ you can ignore the big 2 because it is a coefficient, which is written to balance the equation.
The reactants are CH₄ and O₂
Explanation:
A period 3 element is one of the chemical elements in the third row (or period) of the periodic table of the chemical elements. The periodic table is laid out in rows to illustrate recurring (periodic) trends in the chemical behaviour of the elements as their atomic number increases: a new row is begun when the periodic table skips a row and a chemical behaviour begins to repeat, meaning that elements with similar behaviour fall into the same vertical columns. The third period contains eight elements: sodium, magnesium, aluminium, silicon, phosphorus, sulfur, chlorine, and argon. The first two, sodium and magnesium, are members of the s-block of the periodic table, while the others are members of the p-block. All of the period 3 elements occur in nature and have at least one stable isotope.[1]
Answer is: pH of solution is 5,17.
Kb(NH₃) = 1,8·10⁻⁵.
c(NH₄Cl) = 0,084 M = 0,084 mol/L.
Chemical reaction: NH₄⁺ + H₂O → NH₃ + H₃O⁺.
Ka · Kb = 10⁻¹⁴.
Ka(NH₄⁺) = 10⁻¹⁴ ÷ 1,8·10⁻⁵.
Ka(NH₄⁺) = 5,55·10⁻¹⁰.
[H₃O⁺] = [NH₃] = x.
Ka(NH₄⁺) = [H₃O⁺] · [NH₃] ÷ [NH₄⁺].
5,55·10⁻¹⁰ = x² ÷ (0,084 M - x).
Solve quadratic equation: x = [H₃O⁺] = 6,8·10⁻⁶ M.
pH = -log[H₃O⁺].
pH = -log(6,8·10⁻⁶ M) = 5,17.
