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Olegator [25]
3 years ago
11

Give the formula of each of the following (don't worry about subscripts or superscripts, example: HCO3- enter as HCO3-): the con

jugate acid of HSO3- H2SO3 the conjugate base of H2O the conjugate acid of NH3 the conjugate base of CH3NH2
Chemistry
2 answers:
azamat3 years ago
8 0

Answer:

See explanation below

Explanation:

In this case, for you to understand, let's do the acid base reaction, so you can see the conjugate base or acid.

For the case of the HSO3-, if you do the reaction:

HSO3- + H2O ------> H2SO3 + OH-

In this case, the negative charge, acts as a base, and took an hydrogen from the water. This forms the OH- and the conjugate acid of the HSO3-. The H2SO3 is called sulfurous acid.

For the case of H2SO3:

H2SO3 + H2O ---------> HSO3- + H3O+

The H2SO3 acts like acid, so the conjugate base would be the HSO3-

For the H2O:

2H2O --------> H3O+ + OH-

The water acts as base and acid at a time. In this case, when it acts like a base, took one hydrogen atom from the other molecule of water, and forms the H3O+. This would be the conjugate acid. The conjugate base would be the OH-

For the NH3:

NH3 + H2O --------> NH4+ + OH-

The atom of nitrogen in the NH3, took an atom of hydrogen to form the NH4+, which would be the conjugate acid.

For the CH3NH2:

CH3NH2 + H2O -------> CH3NH- + H3O+

The water took an atom of hydrogen, so, the water is the base and CH3NH2 is the acid, so the conjugate base is the CH3NH-

Regards

vovangra [49]3 years ago
3 0

Answer:

The relative conjugate acids and bases are listed below:

CH3NH2 → CH3NH3+

H2SO3→ HSO3-

NH3→ NH4+

Explanation:

In a Brønsted-Lowry acid-base reaction, a conjugate acid is the species resulting from a base accepting a proton; likewise, a conjugate base is the species formed after an acid has donated a hydrogen atom (proton).

To this end:

  • HSO3- is the conjugate acid of H2SO3 i.e sulfuric acid has lost a proton (H+)
  • NH4+ is the conjugate acid of NH3 i.e the base ammonia has gained a proton (H+)
  • OH- is the conjugate base of H20
  • CH3NH3+ is the conjugate base of the base CH3NH2 methylamine
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13) Find the mass, in grams, of 5.75 liters of N2.
faltersainse [42]
Molecular weight of N2
gram Atomic weight of N is 14 g.
gram Molecular weight = 14 g×2
= 28 g
Liter concepts
1 liter of gas always occupies 1 gram molecular weight .
Application of the concept
5.75 liter gives 0.257g


1 mole will occupy 28 g

0.257 will occupy 28 g × 0.257
= 7.196 g

The mass of the gas is 7.196 g.

Hope it helps you
8 0
2 years ago
Hess’s law
Delvig [45]

From the statement of Hess' law, the enthalpy of the reaction A---> C is +90 kJ

<h3>What is Hess' law?</h3>

Hess' law of constant heat summation states that for a multistep reaction, the standard enthalpy of reaction is always constant and is independent of the pathway or intermediate routes taken.

From Hess' law, the enthalpy change for the reaction A ----> C is calculated as follows:

A---> C = A ---> B + B ---> C

ΔH of A---> C = 30 kJ + 60 kJ

ΔH = 90 kJ

Therefore, the enthalpy of the reaction A---> C is +90 kJ

The above reaction A---> C can be shown in the enthalpy diagram below:

A -------------------> C (ΔH = +90 kJ)

\ /

\ / (ΔH = +60 kJ)

(ΔH = +30 J) \ /

> B

Learn more about enthalpy and Hess law at: brainly.com/question/9328637

8 0
2 years ago
When nahco3 completely decomposes, it can follow this balanced chemical equation: 2nahco3 → na2co3 h2co3 determine the theoretic
BigorU [14]

Theoretical yield = 2.397

The product could be sodium carbonate

percent yield = 98.456%

When nahco3 completely decomposes, it can follow this balanced chemical equation:

2nahco3 → na2co3 h2co3

If the mass of the NaHCO3 sample is 3.80 g, we must use stoichiometry to calculate the theoretical yields of each of the products.

mass of NaHCO₃ = 3.80 g

molar mass of NaHCO₃ = 84 g/mol

so the no of moles of NaHCO₃ = 3.80/84 =  0.0452 mol

You see, one mole of sodium carbonate and one mole of hydrogen carbonate are produced from two moles of sodium bicarbonate.

so, the no of moles of sodium carbonate = 0.0452/2 = 0.0226 mol

∴ mass of sodium carbonate ( Na₂CO₃) = no of moles of Na₂CO₃ × molar mass of Na₂CO₃

=  0.0226 × 106 ≈ 2.397 g

no of moles of hydrogen carbonate = 0.0452/2 = 0.0226 mol

mass of the hydrogen carbonate ( H₂CO₃) = no of moles of H₂CO₃ × molar mass of H₂CO₃

= 0.0226 × 62 g = 1.401 g

mass of one of the products was measured to be 2.36 g , from above data, we can say it must be sodium carbonate because value is the nearest of 2.397 g.

percentage yield = experimental yield/theoretical yield × 100

here experimental yield of Na₂CO₃ = 2.36 g

and theoretical yield of Na₂CO₃ = 2.397 g

∴ % yield = 2.36/2.397 × 100 ≈ 98.456%

Therefore the percentage yield of the product is 98.456%

To learn more about percentage yield visit:

brainly.com/question/22257659

#SPJ4

6 0
1 year ago
The element silver exists in nature as two isotopes: 107Ag has a mass of 106.9051 u, and 109Ag has a mass of 108.9048 u. The ave
Lilit [14]

Answer:

107Ag has abundance of 51.7%

109Ag has abundance of 48.3%

Explanation: Please see attachment for explanation

3 0
3 years ago
Define Condensation .​
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Answer:

Condensation is the change of the physical state of matter from the gas phase into the liquid phase.

6 0
2 years ago
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