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Mariana [72]
2 years ago
10

If you add a solution of NaOH to a solution of H₂CO₃, two reactions occur, one after the other. Write the chemical equations for

these two reactions. (Hint: NaOH dissociate into Na+ and OH-, and the hydroxide ion is the actual base).
Chemistry
1 answer:
Crazy boy [7]2 years ago
6 0

We have a solution of NaOH and H₂CO₃

First, NaOH will dissociate into Na⁺ and OH⁻ ions

The Na⁺ ion will substitute one of the Hydrogen atoms on H₂CO₃ to form NaHCO₃

The H⁺ released from the substitution will bond with the OH⁻ ion to form a water molecule

If there were to be another NaOH molecule, a similar substitution will take place, substituting the second hydrogen from H₂CO₃ as well to form Na₂CO₃

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A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a
Morgarella [4.7K]

Answer : The percent abundance of the heaviest isotope is, 78 %

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = \frac{x}{100}

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = \frac{10}{100}

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] %  = (90 - x) %

Fractional abundance of middle-weight isotope = \frac{(90-x)}{100}

Now put all the given values in above formula, we get:

48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]

x=78\%

Therefore, the percent abundance of the heaviest isotope is, 78 %

5 0
3 years ago
Read 2 more answers
Determine the number of ions produced in the dissociation of the compound listed. AlF3
scoray [572]
Answer is: the number of ions produced in the dissociation of aluminium fluoride is 4.
<span>
Chemical dissociation of aluminium fluoride in water:
AlF</span>₃(aq) → Al³⁺(aq) + 3F⁻(aq).<span>
There are four ions, one aluminium cation and three fluoride anions.
Aluminium has oxidation +3, because it lost three electrons, to have electron configuration as noble gas neon and fluorine has oxidation -1, because it gain one electron to </span>have electron configuration as noble gas neon.
6 0
3 years ago
A sample of vinegar was found to have an acetic acid concentration of 0.8846 m. What is the acetic acid % by mass? Assume the de
jenyasd209 [6]

Answer:

5.3%

Explanation:

Let the volume be 1 L

volume , V = 1 L

use:

number of mol,

n = Molarity * Volume

= 0.8846*1

= 0.8846 mol

Molar mass of CH3COOH,

MM = 2*MM(C) + 4*MM(H) + 2*MM(O)

= 2*12.01 + 4*1.008 + 2*16.0

= 60.052 g/mol

use:

mass of CH3COOH,

m = number of mol * molar mass

= 0.8846 mol * 60.05 g/mol

= 53.12 g

volume of solution = 1 L = 1000 mL

density of solution = 1.00 g/mL

Use:

mass of solution = density * volume

= 1.00 g/mL * 1000 mL

= 1000 g

Now use:

mass % of acetic acid = mass of acetic acid * 100 / mass of solution

= 53.12 * 100 / 1000

= 5.312 %

≅ 5.3%

3 0
3 years ago
For a particular redox reaction, Cr is oxidized to CrO 2 − 4 and Ag + is reduced to Ag . Complete and balance the equation for t
Sunny_sXe [5.5K]

Answer:

6Ag⁺ + Cr + 8OH⁻ → 6Ag + CrO₄²⁻ + 4H₂O

Explanation:

We can balance the redox reaction of Cr and Ag⁺, in terms of two half-reactions, one for Ag⁺ and other for Cr:

Ag⁺   →   Ag      

In the above equation we need to balance the number of electrons, we know that the Ag⁺ is being reduced to Ag, so the reaction is:

Ag⁺ + e⁻ →  Ag   (1)

Now, we need to balance the half-reaction of Cr:

Cr   →  CrO₄²⁻  

From above, we know that the Cr is being oxidated to CrO₄²⁻, so we need to balance the number of electrons and the number of oxygen atoms. The Cr⁰ is being oxidated to Cr⁶⁺, so for the electron balance, we need to add 6e⁻ to the right side of the equation. Since the reaction is in a basic medium, the oxygen atoms will be balanced with OH⁻ ions as follows:          

Cr + OH⁻ →  CrO₄²⁻ + 6e⁻  

The hydrogen atoms will be balanced using H₂O molecules:  

Cr + OH⁻ →  CrO₄²⁻ + 6e⁻ + H₂O    

The balanced equation is:

Cr + 8OH⁻ →  CrO₄²⁻ + 6e⁻ + 4H₂O   (2)

Since the reaction (1) involves 1 electron and the reaction (2) involves 6 electrons, by increasing the reaction (1) six times and by the addition of the two reactions (1 and 2) we can have the net redox reaction:

6*(Ag⁺ + e⁻ →  Ag)  

<u>Cr + 8OH⁻ →  CrO₄²⁻ + 6e⁻ + 4H₂O</u>

6Ag⁺ + Cr + 8OH⁻ → 6Ag + CrO₄²⁻ + 4H₂O                  

Therefore, the net equation is: 6Ag⁺ + Cr + 8OH⁻ → 6Ag + CrO₄²⁻ + 4H₂O.

I hope it helps you!

7 0
3 years ago
When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 16.8g of carbon were burned in the presence of
Mrrafil [7]
61.6 grams of carbon dioxide was produced
4 0
3 years ago
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