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Colt1911 [192]
3 years ago
12

Which of the following solutions will have the highest concentration of chloride ions?

Chemistry
1 answer:
Mnenie [13.5K]3 years ago
4 0

Answer:

B = 0.10 M of AlCl₃

Explanation:

A = 0.10 M NaCl

NaCl → Na⁺ + Cl⁻

sodium chloride dissociate into sodium and chlorine ions. The one mole of NaCl produced on mole of Na⁺ and one mole of Cl⁻, so concentration of chloride ions will be 0.10 M.

B = 0.10 M of AlCl₃

AlCl₃  →    Al³⁺ + 3Cl⁻

AlCl₃ dissociate into aluminium cation and chlorine anion. one mole of AlCl₃ dissociate into one mole of Al³⁺ and three mole of Cl⁻.

so concentration of chloride ion in 0.10 M solution is

0.10 × 3 = 0.3 M

C = 0.10 M MgCl₂

MgCl₂   → Mg²⁺ +  2Cl⁻

MgCl₂ dissociate into one mole of Mg²⁺ and two mole of Cl⁻.

so concentration of chloride ions in 0.10 M solution of MgCl₂ is,

2× 0.10 M = 0.20 M

d = 0.05 M CaCl₂  

CaCl₂  →   Ca²⁺ +  2Cl⁻

CaCl₂ dissociate into one mole of  Ca²⁺ and two mole of Cl⁻, So concentration of chloride ions in 0.05 M CaCl₂ is

2× 0.05 = 0.1 M

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An element's atomic number is 60. How many protons would an atom of this element have? __protons
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3 years ago
A. Based on the activation energies and frequency factors, rank the following reactions from fastest to slowest reaction rate, a
deff fn [24]

Answer:

A) E_{a} = 350KJ/mol, E_{a} = 50KJ/mol, E_{a} = 50KJ/mol

     A = 1.5×10^{-7}s^{-1}, A = 1.9×10^{-7} s^{-1}, A=1.5×10^{-7} s^{-1}

B) 4.469

Explanation:

From Arrhenius equation

      K=Ae^{\frac{E_{a} }{RT} }

where; K = Rate of constant

            A = Pre exponetial factor

            E_{a} = Activation Energy

             R = Universal constant

             T = Temperature in Kelvin

Given parameters:

E_{a} =165KJ/mol

T_{1}=505K

T_{2}=525K

R=8.314JK^{-1}mol^{-1}

taking logarithm on both sides of the equation we have;

InK=InA-\frac{E_{a} }{RT}

since we have the rate of two different temperature the equation can be derived as:

In(\frac{K_{2} }{K_{1} } )=\frac{E_{a} }{R}(\frac{1}{T_{1} } -\frac{1}{T_{2} } )

In(\frac{K_{2} }{K_{1} } )=\frac{165000J/mol}{8.314JK^{-1}mol^{-1}  }.(\frac{1}{505} -\frac{1}{525} )

In(\frac{K_{2} }{K_{1} } )= 19846.04×7.544×10^{-5} = 1.497

\frac{K_{2} }{K_{1} } =e^{1.497} = 4.469

 

6 0
3 years ago
Calculate the molarity of an HCl solution if 23.88 mL of it reacts with 6.5287 grams of sodium carbonate (106 g/mol) according t
Ivanshal [37]

Answer:

5.158 mol/L

Explanation:

To find the molarity, you need to use the formula:

Molarity (M) = moles / volume (L)

You have been grams sodium carbonate. You need to (1) convert grams Na₂CO₃ to moles (via molar mass), then (2) convert moles Na₂CO₃ to moles HCl (via mole-to-mole ratio from equation), then (3) convert mL to L (by dividing by 1,000), and then (4) use the molarity equation.

<u>Steps 1 - 2:</u>

2 HCl + 1 Na₂CO₃ ----> 2 NaCl + H₂O + CO₂

6.5287 g Na₂CO₃         1 mole            2 moles HCl
--------------------------  x  -------------  x  -------------------------  =  0.12318 mole HCl
                                      106 g           1 mole Na₂CO₃

<u>Step 3:</u>

23.88 mL / 1,000 = 0.02388 L

<u>Step 4:</u>

Molarity = moles / volume

Molarity = 0.12318 mole / 0.02388 L

Molarity = 5.158 mole/L

**mole/L is equal to M**

7 0
2 years ago
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