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PtichkaEL [24]
2 years ago
13

A student is given two 10g samples, each a mixture of only nacl(s) and kcl(s) but in different proportions. Which of the followi

ng pieces of information could be used to determine which mixture has the higher proportion of KCl(s) ?
A. The volume of each mixture
B. The mass of Cl in each mixture
C. The number of isotopes of Na and K
D. The reaction of each mixture with water
Chemistry
1 answer:
earnstyle [38]2 years ago
4 0

Answer:

a

Explanation:

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2 years ago
How many liters of O2 at 298 K and 1.00 bar are produced in 1.50 hr in an electrolytic cell operating at a current of 0.0200 A?
stellarik [79]

Answer: 0.0069L

Explanation:

2H2O(l) ---->O2(g) + 4H+(aq) + 4e-

no of moles= it/eF

NO of moles of O2 produced = (Current in Ampere x Time in second)/ (Faraday constant x Number of electrons required)

Moles of O2 produced = (0.02x (60 x 60X1.5 s)/(96485 x 4)

= 0.0002798 moles= 2.798x 10 ^-4moles

Using  ideal gas equation,

P V = n R T

Where, P is the pressure,

V is the volume,

n is the number of moles,

R is the gas constant, and T is the temperature

We have, 1 bar = 0.986923 atm

Substituting the values,

V = nRT/P = (2.798 x 10-4moles x 0.08205 L atm mol K x 298 K)/ 0.986923 atm = 0.0069L

Volume of O2 produced = 0.0069L

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3 years ago
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Answer:

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Explanation:

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2 years ago
What is the final concentration of cl- ion when 250 ml of 0.20 m cacl2 solution is mixed with 250 ml of 0.40 m kcl solution? (as
serious [3.7K]

CaCl2 and KCl are both salts which dissociate in water when dissolved. Assuming that the dissolution of the two salts are 100 percent, the half reactions are:

<span>CaCl2 ---> Ca2+  +  2 Cl-</span>

KCl ---> K+ + Cl-

Therefore the total Cl- ion concentration would be coming from both salts. First, we calculate the Cl- from each salt by using stoichiometric ratio:

Cl- from CaCl2 = (0.2 moles CaCl2/ L) (0.25 L) (2 moles Cl / 1 mole CaCl2)

Cl- from CaCl2 = 0.1 moles

 

Cl- from KCl = (0.4 moles KCl/ L) (0.25 L) (1 mole Cl / 1 mole KCl)

Cl- from KCl = 0.1 moles

 

Therefore the final concentration of Cl- in the solution mixture is:

Cl- = (0.1 moles + 0.1 moles) / (0.25 L + 0.25 L)

Cl- = 0.2 moles / 0.5 moles

<span>Cl- = 0.4 moles             (ANSWER)</span>

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