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8_murik_8 [283]
2 years ago
11

If you flow a solution of Mg CO3-- in water over a Cation resin, what will be in the outlet stream leaving the bed

Chemistry
1 answer:
vodka [1.7K]2 years ago
8 0

In a a cation-exchange resin, the outlet stream leaving the bed will contain H^{+} and CO_3 ^{2-}.

<h3>What is cation-exchange resin?</h3>
  • A resin or polymer that serves as a medium for ion exchange is known as an ion-exchange resin or cation-exchange resin.
  • It is an insoluble matrix (or support structure) made from an organic polymer substrate, typically appearing as tiny (0.25-1.43 mm radius) microbeads that are white or yellowish in color.
  • The process is known as cation-exchange resin because the beads are often porous, providing a wide surface area on and inside them where the trapping of ions takes place along with the concomitant release of other ions.
  • cation-exchange resin comes in many different varieties. Polystyrene sulfonate is the main ingredient in most commercial resins. Many diverse separation, purification, and decontamination techniques use cation-exchange resin.
  • The most typical examples are water filtration and water softening.

To learn more about cation-exchange resin with the given link

brainly.com/question/21052225

#SPJ4

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What is the pressure in a 27.0-L cylinder filled with 44.9 g of oxygen gas at a temperature of 315 KK
steposvetlana [31]

Answer:

1.343 atm

Explanation:

We are given the following;

Pressure, p = ?

Volume v = 27 L

Mass of oxygen = 44.9 g

Temperature, T = 315 K

The formular relating all these variables is the equation;

PV = nRT

where R = gas constant = 0.08206 L atm / mole K

To obtain n, we use;

number of moles, n = Mass /  molar mass =  44.9 / 32 = 1.403 moles

From the ideal gas equation;

P = nRT / V

P = 1.403 * 0.08206 * 315 / 27

P = 36.27 / 27 = 1.343 atm

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For the reaction A +B+ C D E, the initial reaction rate was measured for various initial concentrations of reactants. The follow
lora16 [44]

Answer:

Rate constant of the reaction is 3.3\times 10^{-3} M^{-2} s^{-1}.

Explanation:

A + B + C → D + E

Let the balanced reaction be ;

aA + bB + cC → dD + eE

Expression of rate law of the reaction will be written as:

R=k[A]^a[B]^b[C]^c

Rate(R) of the reaction in trail 1 ,when :

[A]=0.30 M,[B]=0.30 M,[C]=0.30 M

R=9.0\times 10^{-5} M/s

9.0\times 10^{-5} M/s=k[0.30 M]^a[0.30 M]^b[0.30 M]^c...[1]

Rate(R) of the reaction in trail 2 ,when :

[A]=0.30 M,[B]=0.30 M,[C]=0.90 M

R=2.7\times 10^{-4} M/s

2.7\times 10^{-4} M/s=k[0.30 M]^a[0.30 M]^b[0.90 M]^c...[2]

Rate(R) of the reaction in trail 3 ,when :

[A]=0.60 M,[B]=0.30 M,[C]=0.30 M

R=3.6\times 10^{-4} M/s

3.6\times 10^{-4} M/s=k[0.60 M]^a[0.30 M]^b[0.30 M]^c...[3]

Rate(R) of the reaction in trail 4 ,when :

[A]=0.60 M,[B]=0.60 M,[C]=0.30 M

R=3.6\times 10^{-4} M/s

3.6\times 10^{-4} M/s=k[0.60 M]^a[0.60 M]^b[0.30 M]^c...[4]

By [1] ÷ [2], we get value of c ;

c = 1

By [3] ÷ [4], we get value of b ;

b = 0

By [2] ÷ [3], we get value of a ;

a = 2

Rate law of reaction is :

R=k[A]^2[B]^0[C]^1

Rate constant of the reaction = k

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k=\frac{9.0\times 10^{-5} M/s}{[0.30 M]^2[0.30 M]^0[0.30 M]^1}

k=3.3\times 10^{-3} M^{-2} s^{-1}

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