Missing question: What is the rate constant for the reaction?
<span>[RS2](mol L-1) Rate (mol/(L·s))
0.150 0.0394
0.250 0.109
0.350 0.214
0.500 0.438</span>
Chemical reaction: 3RS₂ → 3R + 6S.
Compare second and fourth experiment, when concentration is doubled, rate of concentration is increaced by four. So rate is:
rate = k·[RS₂]².
k = 0,438 ÷ (0,500)².
k = 1,75 L/mol·s.
Answer:
4380 mmHg
Explanation:
Boyle's Law can be used to explain the relationship between pressure and volume of an ideal gas. The pressure is inversely related to volume, so if volume decrease the pressure will increase. It can be expressed in the equation as:
P1V1=P2V2
In this question, the first condition is 2L volume and 876 mmHg pressure. Then the system changed into the second condition where the volume is 400ml and the pressure is unknown. The pressure will be:
P1V1= P2V2
876 mmHg * 2L = P2 * 400ml /(1000ml/L)
P2= 876 mmHg * 2L / 0.4L
P2= 4380 mmHg
<span>A solution with a pH of 4 has ten times the concentration of H</span>⁺<span> present compared to a solution with a pH of 5.
</span>pH <span>is a numeric scale for the acidity or basicity of an aqueous solution. It is the negative of the base 10 logarithm of the molar concentration of hydrogen ions.
</span>[H⁺] = 10∧-pH.
pH = 4 → [H⁺]₁ = 10⁻⁴ M = 0,0001 M.
pH = 5 → [H⁺]₂ = 10⁻⁵ M = 0,00001 M.
[H⁺]₁ / [H⁺]₂ = 0,0001 M / 0,00001 M.
[H⁺]₁ / [H⁺]₂ = 10.