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Alex787 [66]
2 years ago
9

When potassium hydroxide and barium chloride react, potassium chloride and barium hydroxide are formed. The balanced equation fo

r this reaction is __________.A. KH + BaCl --> KCl + BaH
B. KOH + BaCl --> KCl + BaOH
C. 2KOH + BaCl₂ --> 2KCl + Ba(OH)₂
D. KOH + BaCl₂ --> KCl₂ + BaOH
Chemistry
2 answers:
romanna [79]2 years ago
8 0

Answer: C.

2KOH + BaCl2 => 2KCl + Ba(OH)2

Explanation:

There are 2 atoms of O and H in Ba(OH)2 in the product so to balance we put 2 on KOH. Then balance K by adding 2 in KCl to have equal number of atoms for K and Cl in both the reactant and product side. Barium have 1 atom on the reactant and in the product.

Mashcka [7]2 years ago
7 0

Answer:

The answer to your question is letter C.

Explanation:

Reaction

Potassium hydroxide = KOH

Barium chloride = BaCl₂

Potassium chloride = KCl

Barium hydroxide = Ba(OH)₂

           KOH   +   BaCl₂    ⇒    KCl   +   Ba(OH)₂

         Reactant        Elements       Products

               1                     K                      1

               1                     Ba                    1

               2                    Cl                     1  

                1                     H                    2

                1                     O                    2

The reaction is unbalanced

        2KOH   +   BaCl₂    ⇒   2KCl   +   Ba(OH)₂

         Reactant        Elements       Products

               2                     K                     2

               1                     Ba                    1

               2                    Cl                     2  

               2                      H                    2

               2                     O                    2

Now, the reaction is balanced

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0.0277 M.

Explanation:

The integral rate law of a first order reaction:

<em>Kt = ln ([A₀]/[A]),</em>

where, k is the rate constant of the reaction <em>(k = 3.36 × 10⁻⁵ s⁻¹)</em>,

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[A₀] is the initial concentration of cyclopropane <em>([A₀] = 0.0445 M)</em>

<em>∵ Kt = ln ([A₀]/[A]),</em>

∴ (3.36 × 10⁻⁵ s⁻¹)(14100 s) = ln (0.0445 M)/[A]

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3 years ago
What were the four elements that ancient Greeks believed to exist
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For each of the esters provided, identify the alcohol and the carboxylic acid that reacted.
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Answer:

52. The alcohol USED => methanol, CH3OH

The carboxylic acid USED => propanoic acid, CH3CH2COOH.

53. The alcohol USED => Ethanol, CH2CH3OH

The carboxylic acid USED => Formic acid, HCOOH.

Explanation:

52. To obtain Methyl propanoate, CH3CH2COOCH3, we simply react propanoic, CH3CH2COOH and methanol, CH3OH together as shown below:

CH3CH2COOH + CH3OH —> CH3CH2COOCH3 + H2O

The alcohol used: methanol, CH3OH

The carboxylic acid used: propanoic acid, CH3CH2COOH.

53. To obtain Ethyl methanoate, HCOOCH2CH3, we simply react

Formic acid, HCOOH and ethanol, CH3CH2OH together as show below:

HCOOH + CH3CH2OH —> HCOOCH2CH3 + H2O

The alcohol USED => Ethanol, CH2CH3OH

The carboxylic acid USED => Formic acid, HCOOH.

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3 years ago
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Answer:

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Explanation:

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To calculate the molality, we need to know the number of moles of urea and the volume of solution in liters. We assume 100 grams of solution.

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This solution contains 0.619 moles of urea in 96.9 mL of solution. To express it in molarity, we need to calculate the moles present in 1000 mL (1 L) of the solution.

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