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2 years ago

What is the conjugate acid of each of the following? What is the conjugate base of each?

Chemistry

1 answer:

Lilit [14]2 years ago

3 0

Answer:

a. H₂O (conjugate acid) ; b. OH⁻ (conjugate base), H₃O⁺ (conjugate acid) ; c. H₂CO₃ (conjugate acid), CO₃⁻² (conjugate base) ; d. NH₄⁺ (conjugate strong acid) e. H₂SO₄ (conjugate acid), SO₄⁻² (conjugate base) ; f. No conjugate acid either base; g. H₂S (conjugate acid), S⁻² (conjugate base);

h. H₄N₂ (conjugate base)

Explanation:

a. OH⁻ + H⁺ ⇄ H₂O

The hydroxide acts like a Bronsted Lory base, so it can catch a proton. Water will be the conjugate acid.

b. H₂O, is an amphoterus compound. It can act as an acid or a base. If it is a base, the conjugate acid is the H₃O⁺. If it is an acid, the conjugate base is the OH⁻.

c. HCO₃⁻ + H⁺ ⇄ H₂CO₃

HCO₃⁻ + H₂O ⇄ CO₃⁻² + H₃O⁺

The bicarbonate is also amphoteric. When it catches the proton, the carbonic acid is the conjugate acid, cause it works as a base.

When the HCO₃⁻ (acid) release the proton, the conjugate base is the carbonate.

d. Ammonia is a weak base, so the conjugate strong acid is the ammonium.

NH₃ + H₂O ⇄ NH₄⁺ + OH⁻

e. Another amphoteric compound. The acid sulfate acts an acid and a base.

(like bicarbonate). When it is a base, the conjugate acid is the sulfuric acid, when it is an acid, the conjugate base is the sulfate.

HSO₄⁻ + H₂O ⇄ SO₄⁻² + H₃O⁺

HSO₄⁻ + H⁺ ⇄ H₂SO₄

f. H₂O₂ does not recieve H⁺ or OH⁻, and it does not release H⁺. It is a neutral compound and it doesn't act as a base or acid.

g. HS⁻ is amphoterous.

HS⁻ + H⁺ ⇄ H₂S

HS⁻ + H₂O ⇄ S⁻² + H₃O⁺

Same case as bicarbonate or acid sulfate.

h. H₅N₂⁺ + H₂O ⇄ H₄N₂ + H₃O⁺

Hidrazinium acts an acid, so, the conjugate base will be the hidrazine.

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What mass of NH3 can be made from 1221g of H2 and excess N2

expeople1 [14]

The mass of NH3 which can be made from 1221g of H2 and excess N2 is 6867g NH3

Explanation:

NH3 mass is 6867g
H2 mass is 1221g
N2 in excess

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2 years ago

The name "penicillin" is used for several closely-related antibiotics. A 1.2177 g sample of one of these compounds is burned, pr

masya89 [10]

Answer: The empirical formula for the given organic compound is $C_7H_{11}O_2SN$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$S_vN_wC_xH_yO_z+O_2\rightarrow CO_2+H_2O+SO_2+NO$

where, 'v', 'w' 'x', 'y' and 'z' are the subscripts of sulfur, nitrogen, carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=0.2829g$

Mass of $H_2O=0.1159g$

Mass of $SO_2=0.4503g$

Mass of NO = 0.2109 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

Molar mass of sulfur dioxide = 64 g/mol

Molar mass of nitrogen monoxide = 30 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 2.1654 g of carbon dioxide, $\frac{12}{44}\times 2.1654=0.590g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 0.6965 g of water, $\frac{2}{18}\times 0.6965=0.077g$ of hydrogen will be contained.

For calculating the mass of sulfur:

In 64 g of sulfur dioxide, 32 g of sulfur is contained.

So, in 0.4503 g of sulfur dioxide, $\frac{32}{64}\times 0.4503=0.225g$ of sulfur will be contained.

For calculating the mass of nitrogen:

In 30 g of nitrogen monoxide, 14 g of nitrogen is contained.

So, in 0.2109 g of nitrogen monoxide, $\frac{14}{30}\times 0.2109=0.098g$ of nitrogen will be contained.

Mass of oxygen in the compound = (1.2177) - (0.590 + 0.077 + 0.225 + 0.098) = 0.2277 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.590g}{12g/mole}=0.049moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.077g}{1g/mole}=0.077moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.2277g}{16g/mole}=0.0142moles$

Moles of Sulfur = $\frac{\text{Given mass of Sulfur}}{\text{Molar mass of sulfur}}=\frac{0.225g}{32g/mole}=0.007moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{0.098g}{14g/mole}=0.007moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.007 moles.

For Carbon = $\frac{0.049}{0.007}=7$

For Hydrogen = $\frac{0.077}{0.007}=11$

For Oxygen = $\frac{0.0142}{0.007}=2.03\approx 2$

For Sulfur = $\frac{0.007}{0.007}=1$

For Nitrogen = $\frac{0.007}{0.007}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O : S : N = 7 : 11 : 2 : 1 : 1

Hence, the empirical formula for the given compound is $C_7H_{11}O_2S_1N_1=C_7H_{11}O_2SN$

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2 years ago

Write a balanced half-reaction describing the oxidation of gaseous dihydrogen to aqueous hydrogen cations.

xz_007 [3.2K]

The oxidation state of hydrogen gas is 0 and oxidation state of hydrogen cation is +1.
There’s an increase in oxidation number therefore it’s an oxidation reaction.
Oxidation reactions give out electrons. The masses and charges on both sides should be balanced
Half reaction is
H2 —> 2H+ +2e

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2 years ago

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Empirical Formula of P3O4H2?

puteri [66]

Answer:

H2 P4 O1. Explanation: In order to calculate the Empirical formula , we will assume that we have started with 10 g of the compound.

Explanation:

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2 years ago

How many milliliters of 0.200 M NH4OH are needed to react with 12.0 mL of 0.550 M FeCl3?

trapecia [35]

Answer:

9.9 ml of 0.200M NH₄OH(aq)

Explanation:

3NH₄OH(Iaq) + FeCl₃(aq) => NH₄Cl(aq) + Fe(OH)₃(s)

?ml of 0.200M NH₄OH(aq) reacts completely with 12ml of 0.550M FeCl₃(aq)

1 x Molarity NH₄OH x Volume Am-OH Solution(L) = 2 x Molarity FeCl₃ x Volume FeCl₃ Solution

1(0.200M)(Vol Am-OH Soln) = 3(0.550M)(0.012L)

=> Vol Am-OH Soln = 3(0.550M)(0.012L)/1(0.200M) = 0.0099 Liter = 9.9 milliliters

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2 years ago