Answer:
so the reaction rate increases by a factor 6.
Explanation:
For the given equation the reaction is first order with respect to both ester and sodium hydroxide
So we can say that the rate law is
![Rate(initial)=K[NaOH][CH_{3}COOC_{2}H_{5}]](https://tex.z-dn.net/?f=Rate%28initial%29%3DK%5BNaOH%5D%5BCH_%7B3%7DCOOC_%7B2%7DH_%7B5%7D%5D)
now as per given conditions the concentration of ester is increased by half it means that the new concentration is 1.5 times of old concentration
The concentration of NaOH is quadrupled means the new concentration is 4 times of old concentration.
The new rate law is
![Rate(final)=K[1.5XNaOH][4XCH_{3}COOC_{2}H_{5}]](https://tex.z-dn.net/?f=Rate%28final%29%3DK%5B1.5XNaOH%5D%5B4XCH_%7B3%7DCOOC_%7B2%7DH_%7B5%7D%5D)
the final rate = 6 X initial rate
so the reaction rate increases by a factor 6.
Answer:
Barium precipitation : in this method we will add Barium chloride to the wastewater to precipitate Barium sulphate precipitate from the waste water
Explanation:
The measures that can be taken for the recovery of sulfates and other contaminants from wastewater includes : ion exchange, precipitation and Reverse osmosis. of all these measures i will discuss precipitation
Precipitation measures can be performed using different methodologies the methodology that i will discuss is :
Barium precipitation : in this method we will add Barium chloride to the wastewater to precipitate Barium sulphate precipitate from the waste water
The sulphate contained in the wastewater can be reduced to as low as 200 mg/l using this method
Temperature : the increase in temperature of the waste water will cause the reduction in formation of sulphate in the waste water
PH changes : for the precipitation of sulphate to be effective the required PH level is 4.5
Neither plate subducts because the crust have approximately the same density. Instead, the rocks are smashed together causing them to recrystallize due to the intense heat and pressure from the colliding plates.
Answer:
Mass = 1.33 g
Explanation:
Given data:
Mass of argon required = ?
Volume of bulb = 0.745 L
Temperature and pressure = standard
Solution:
We will calculate the number of moles of argon first.
Formula:
PV = nRT
R = general gas constant = 0.0821 atm.L/mol.K
By putting values,
1 atm ×0.745 L = n × 0.0821 atm.L/mol.K× 273.15 K
0.745 atm. L = n × 22.43 atm.L/mol
n = 0.745 atm. L / 22.43 atm.L/mol
n = 0.0332 mol
Mass of argon:
Mass = number of moles × molar mass
Mass = 0.0332 mol × 39.95 g/mol
Mass = 1.33 g
2 atoms of Al
3 of S
4*3=12 of O
so total= 17
