How does that mean that the number is a little too late and you don’t have a phone number so you please look at it
Answer:
The answer is Option a, that is "−9kJmole,5kJmole".
Explanation:
Please find the complete question in the attached file.
In the question, it uses the catalyst inside a process, which does not modify the process eigenvalues, however, it decreases the active energy with an enthalpy of -9kJmole, and also the power for activating decreases around 13 to 5 kJ mole, that's why the choice a is correct.
A volumetric flask is used to contain a predetermined volume of substance and only measures that volume, for example 250 ml.
Conical flasks can be used to measure the volume of substances but the accuracy they provide is usually up to 10ml. Conical flasks are used in titrations, reactions where the liquid may boil, and reactions which involve stirring.
Pippettes are of two types, volumetric and graduated. Pippettes are used where high accuracy is required and volumetric pippettes come in as little as 1 ml. Pippettes are usually used in titrations.
Graduated cylinders come in a wide variety of sizes and their accuracy can be down to as much as 1 ml. They are used to contain liquids.
Given parameters:
Mass of sucrose = 5g
Density of sucrose = 1.12g/mL
Percentage of sucrose per liter of cane juice = 12%
Unknown:
Volume of cane juice needed = ?
We need to establish the volume - density relationship. Density is the mass of a substance per unit volume.
Mathematically;
Density =
Now solve for the volume of sucrose;
1.12g/mL =
Volume =
= 4.46mL = 4.46 x 10⁻³L since 1000mL = 1L
Since 12% of 1 liter of cane juice is sucrose;
12% of x liter of cane juice = 4.46 x 10⁻³L
Volume of cane juice = 4.46 x 10⁻³ x
= 0.037L
Volume of cane juice is 0.037L
Answer:
The mass of Mg consumed is 21.42g
Explanation:
The reaction is

As per balanced equation, three moles of Mg will react with one mole of nitrogen to give one mole of magnesium nitride.
as given that mass of nitrogen reacted = 8.33g
So moles of nitrogen reacted = 
moles of Mg required = 3 X moles of nitrogen taken = 3X0.2975 = 0.8925mol
Mass of Mg required = moles X molar mass = 0.8925 X 24 = 21.42 g