Answer:
9 × 10⁻³ mol·L⁻¹s⁻¹
Explanation:
Data:
k = 1 × 10⁻³ L·mol⁻¹s⁻¹
[A] = 3 mol·L⁻¹
Calculation:
rate = k[A]² = 1 × 10⁻³ L·mol⁻¹s⁻¹ × (3 mol·L⁻¹)² = 9 × 10⁻³ mol·L⁻¹s⁻¹
Search it up bro it’s on the internet lol
Answer:
Option (A) the solid X is ground to a fine powder.
Explanation:
X(s) + 2B(aq) → X+(aq) + B2(g)
In the reaction above, the rate of the reaction will be highest, when X being a solid is ground to fine powder.
Grounding X to fine powder simply means increasing the surface area of X.
An increase in surface area of reactants will definitely increase the rate of reaction because the particles of the solid will collide with the right orientation and hence speed up the reaction rate.
Use PV=nRT to solve the equation. You need to solve for n (number of moles). Don’t forget to convert the temperature to kelvins by adding 25+273. Use 0.082057 for R.
Yes, mass never changes. No exceptions.
