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3 years ago

Identify the products formed in this Brønsted-Lowry reaction. HCO−3+HBrO↽−−⇀acid+base

Chemistry

1 answer:

sweet-ann [11.9K]3 years ago

3 0

Answer:

$HCO^-_3+HBrO\rightleftharpoons H_2CO_3+BrO^-$

Explanation:

Hello there!

In this case, according to the Brønsted-Lowry theory of acids, it is possible to realize that HBrO acts as the Brønsted-Lowry acid as it has a free hydrogen to give away to HCO₃⁻ which acts as the Brønsted-Lowry base able to accept the incoming hydrogen; therefore, the Brønsted-Lowry reaction is:

$HCO^-_3+HBrO\rightleftharpoons H_2CO_3+BrO^-$

Best regards!

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3 years ago

10/28 11. If the subscript indicating the quantity of a particular element in an ionic formula would be one, the numerical value

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3 years ago

What is the Percent yield is 4.65 got copper is produced by 1.87 percent yield is got copper is produced when 1.87 grams of alum

konstantin123 [22]

Answer:

The percent yield is 70.486%

Explanation:

$3CuSO_{4}+2Al\longrightarrow Al_{2}(SO_{4})_{3}+3Cu\\\\Weight\:of\:Al=1.87g\\54g\:of\:Al\:will\:give\:190.5g\:of\:Cu\\$

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4 years ago

What mass of CaCl2 (in g ) should the chemist use?

solmaris [256]

The mass of the solute required is 250.25 g.

<h3>What is the mass of the solute?</h3>

We know that the number of moles of the solute can be used to obtain the mass of the solute that is required. We can now try to find the mass of the solute that is required.

Concentration of the solution = 0.350M

Volume of the solution = 6.5 L

Number of moles of the solute = 0.350M * 6.5 L

= 2.275 moles

We now have the mass of the solute as;

2.275 moles * 110 g/mol

= 250.25 g

Th measured mass of the solute that we would have to use is 250.25 g.

Learn more about solute:brainly.com/question/7932885

#SPJ1

Missing parts;

A chemist wants to make 6.5 L of a .350M CaCl2 solution. What mass of CaCl2(in g) should the chemist use?

3 0

1 year ago

Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Cr2 ] = 0.892 M and [Fe2 ] = 0.0150

Effectus [21]

<u>Given:</u>

Concentration of Cr2+ = 0.892 M

Concentration of Fe2+ = 0.0150 M

<u>To determine:</u>

The cell potential, Ecell

<u>Explanation:</u>

The half cell reactions for the given cell are:

Anode: Oxidation

Cr(s) ↔ Cr2+(aq) + 2e⁻ E⁰ = -0.91 V

Cathode: Reduction

Fe2+ (aq) + 2e⁻ ↔ Fe (s) E⁰ = -0.44 V

------------------------------------------

Net reaction: Cr(s) + Fe2+(aq) ↔ Cr2+(aq) + Fe(s)

E°cell = E°cathode - E°anode = -0.44 - (-0.91) = 0.47 V

The cell potential can be deduced from the Nernst equation as follows:

Ecell = E°cell - (0.0591/n)log[Cr2+]/[Fe2+]

Here, n = number of electrons = 2

Ecell = 0.47 - 0.0591/2 * log[0.892]/[0.0150] = 0.418 V

Ans: The cell potential is 0.418 V

8 0

3 years ago

Read 2 more answers