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4 years ago

In tne equation below identify the bronsted lowry acid bronsted lowry base conjugate acid conjugate base

Chemistry

1 answer:

Ad libitum [116K]4 years ago

6 0

Answer:

Brønsted-Lowry acid: HNO₂

Brønsted-Lowry base: NH₃

Conjugate acid: NH₄⁺

Conjugate base: NO₂⁻

Explanation:

The equation is:

$HNO_2(aq)+NH_3(aq)\rightarrow NH4^+(aq) + NO2^-(aq)$

Brønsted-Lowry acids are H⁺ donors.

Brønsted-Lowry bases are H⁺ acceptors.

Thus, on the left side, HNO₂ is the acid and NH₃ is the base.

The conjugate acids and conjugate bases are on the right side of the equation.

The conjugate acid is the spieces that is formed after a base accepts the proton; thus it is NH₄⁺. A conjugate acid contains one more H atom and one more + charge than the base that formed it.

The conjugate base is the species that is formed after the acid donates its proton; thus, NO₂⁻ is the conjugate base. A conjugate base contains one less H atom and one more - charge than the acid that formed it.

Summarizing:

Brønsted-Lowry acid: HNO₂

Brønsted-Lowry base: NH₃

Conjugate acid: NH₄⁺

Conjugate base: NO₂⁻

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3 years ago

A green rat snake that lives in the grass and a brown rat snake that lives in the desert is a form of what ?

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6 0

3 years ago

What is the mass of the solid NH4Cl formed when 75.5 g of NH3 is mixed with an equal mass of HCl? What is the volume of the gas

Gekata [30.6K]

Answer : The volume of the gas remaining is 56.5 liters.

The gas is hydrochloric acid and the formula of the gas is HCl.

The mass of $NH_4Cl$ produced is, 110.7 grams.

Explanation :

The balanced chemical reaction will be:

$NH_3+HCl\rightarrow NH_4Cl$

First we have to calculate the moles of $NH_3$ and HCl

$\text{Moles of }NH_3=\frac{\text{Mass of }NH_3}{\text{Molar mass of }NH_3}$

Molar mass of NH_3 = 17 g/mole

$\text{Moles of }NH_3=\frac{75.5g}{17g/mole}=4.44mole$

and,

$\text{Moles of }HCl=\frac{\text{Mass of }HCl}{\text{Molar mass of }HCl}$

Molar mass of HCl = 36.5 g/mole

$\text{Moles of }HCl=\frac{75.5g}{36.5g/mole}=2.07mole$

Now we have to calculate the limiting and excess reagent.

From the balanced reaction we conclude that

As, 1 mole of HCl react with 1 mole of $NH_3$

So, 2.07 mole of HCl react with 2.07 mole of $NH_3$

From this we conclude that, $NH_3$ is an excess reagent because the given moles are greater than the required moles and HCl is a limiting reagent and it limits the formation of product.

The remaining moles of HCl gas = 4.44 - 2.07 = 2.37 moles

Now we have to calculate the volume of the gas remaining.

Using ideal gas equation :

PV = nRT

where,

P = Pressure of gas = 752 mmHg = 0.989 atm (1 atm = 760 mmHg)

V = Volume of gas = ?

n = number of moles of gas = 2.37 moles

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of gas = $14.0^oC=273+14.0=287K$

Putting values in above equation, we get:

$0.989atm\times V=2.37mole\times (0.0821L.atm/mol.K)\times 287K$

V = 56.5 L

Now we have to calculate the moles of $NH_4Cl$

As, 1 mole of HCl react with 1 mole of $NH_4Cl$

So, 2.07 mole of HCl react with 2.07 mole of $NH_4Cl$

Now we have to calculate the mass of $NH_4Cl$

$\text{ Mass of }NH_4Cl=\text{ Moles of }NH_4Cl\times \text{ Molar mass of }NH_4Cl$

Molar mass of $NH_4Cl$ = 53.5 g/mole

$\text{ Mass of }NH_4Cl=(2.07moles)\times (53.5g/mole)=110.7g$

Thus, the volume of the gas remaining is 56.5 liters.

The gas is hydrochloric acid and the formula of the gas is HCl.

The mass of $NH_4Cl$ produced is, 110.7 grams.

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