<span>Activation energy for a particular chemical reaction is
</span><span>a. the energy that must be added to get a reaction started, which is recovered as the reaction proceeds
b. changed by the enzyme that catalyzes the reaction
A and B are correct, so to answer you have to choose
</span><span>e. more than one of the available answers is correct</span>
Answer:
The rate is 
Explanation:
Stoichiometry
Kinetics
![-r_{A}=k \times [CH_{3}Cl] \times [NaOH]](https://tex.z-dn.net/?f=-r_%7BA%7D%3Dk%20%5Ctimes%20%5BCH_%7B3%7DCl%5D%20%5Ctimes%20%5BNaOH%5D%20)
The rate constant K can be calculated by replacing with the initial data
![1 \times 10^{-4}\frac{mole}{Ls}=k \times [0,2M] \times [1,0M] =5 \times 10^{-4}\frac{L}{mole s}](https://tex.z-dn.net/?f=%201%20%5Ctimes%2010%5E%7B-4%7D%5Cfrac%7Bmole%7D%7BLs%7D%3Dk%20%5Ctimes%20%5B0%2C2M%5D%20%5Ctimes%20%5B1%2C0M%5D%20%20%3D5%20%5Ctimes%2010%5E%7B-4%7D%5Cfrac%7BL%7D%7Bmole%20s%7D)
Taking as a base of calculus 1L, when half of the 
is consumed the mixture is composed by
(half is consumed)
(by stoicheometry)
Then, the rate is
The reaction rate decreases because there’s a smaller concentration of reactives.
Answer:
315mL
Explanation:
Data obtained from the question include the following:
Molarity of stock solution (M1) = 0.135 M
Volume of stock solution needed (V1) =?
Molarity of diluted solution (M2) = 0.0851 M
Volume of diluted solution (V2) = 500mL
The volume of the stock solution needed can be obtain as follow:
M1V1 = M2V2
0.135 x V1 = 0.0851 x 500
Divide both side by 0.135
V1 = (0.0851 x 500) / 0.135
V1 = 315mL
Therefore, the volume of the stock solution needed is 315mL
Answer:
Glucose(Sugars) or C6H12O6
Explanation:
Answer:
1.32 moles.
Explanation:
From the question given above, the following data were obtained:
Density of Al = 2.70 g/cm³
Volume of Al = 13.2 cm³
Number of mole of Al =.?
Next, we shall determine the mass of Al.
This can be obtained as follow:
Density of Al = 2.70 g/cm³
Volume of Al = 13.2 cm³
Mass of Al =?
Density = mass / volume
2.7 = mass of Al / 13.2
Cross multiply
Mass of Al = 2.7 × 13.2
Mass of Al = 35.64 g
Finally, we shall determine the number of mole of Al. This can be obtained as follow:
Mass of Al = 35.64 g
Molar mass of Al = 27 g/mol
Number of mole of Al =?
Mole = mass / molar mass
Number of mole of Al = 35.64 / 27
Number of mole of Al = 1.32 moles
Thus, 1.32 moles of aluminum are present in the block of the metal.
