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3 years ago

Which two compounds are classified as bases by the Brønsted-Lowry definition, but not by the Arrhenius definition, and why?

Chemistry

1 answer:

Rainbow [258]3 years ago

5 0

Answer:

b) Ammonia (NH3) and methane (CH4) because they contain hydrogen but do not dissociate to form hydrogen ions.

Explanation:

A substance will be classified as a base according to Arrhenius theory if it produces OH (-) hydroxide ions in water. Ammonia (NH3) is a basic substance, but it is not an Arrhenius base because it does not release hydroxide ions into solution, the same happens with methane (CH4), where all four hydrogens are bound very tightly to the carbon and are not going anywhere.

You might be interested in

Write the chemical symbol for three different atoms or atomic cations with 22 electrons

damaskus [11]

Answer:

Explanation:

To write the chemical symbol for an atom or cation with 22 electrons, we must understand some few things about an atom.

An atom is made up of positively charged paticles called protons which are located in the nucleus. The neutrons do not have any charges and are they reside together in the nucleus with the protons.

Electrons are negatively charged particles which orbits the nucleus.

Titanium in its neutral state is the 22nd element on the periodic table and it has 22 electrons.

Vanadium is the the 23rd element on the periodic table. If it loses an electron, it becomes positively charged and it is a cation.

Chromium is the 24th element on the periodic table. A loss of two electrons makes the net electrons remaining on it 22.

The symbols are:

₂₂Ti

₂₃V¹⁺

₂₄Cr²⁺

6 0

4 years ago

Calculate the volume of the gas when the pressure of the gas is 1.60 atm at a temperature of 298 K. There are 160. mol of gas in

LekaFEV [45]

Answer:

2445 L

Explanation:

Given:

Pressure = 1.60 atm

Temperature = 298 K

Volume = ?

n = 160 mol

Using ideal gas equation as:

$PV=nRT$

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 08206 L.atm/K.mol

Applying the equation as:

1.60 atm × V = 160 mol × 0.08206 L.atm/K.mol × 298 K

Answer to four significant digits, Volume = 2445 L

6 0

3 years ago

What does a chemist do

Mama L [17]

A chemist (from Greek chēm (ía) alchemy; replacing chemist from Medieval Latin alchimista[1]) is a scientist trained in the study of chemistry. Chemists study the composition of matter and its properties. Chemists carefully describe the properties they study in terms of quantities, with detail on the level of molecules and their component atoms. Chemists carefully measure substance proportions, reaction rates, and other chemical properties. The word 'chemist' is also used to address Pharmacists in Commonwealth English.

3 0

4 years ago

The equilibrium constant for the reaction

Hitman42 [59]

The question is incomplete, here is the complete question:

The equilibrium constant for the reaction

N₂O₄(g)⇌2NO₂ at 2°C is Kc = 2.0

If each yellow sphere represents 1 mol of N₂O₄(g) and each gray sphere 1 mol of NO₂ which of the following 1.0 L containers represents the equilibrium mixture at 2°C?

The image is attached below.

<u>Answer:</u> The system which represents the equilibrium having value of $K_c=2.0$ is system (b)

<u>Explanation:</u>

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_c$

For a general chemical reaction:

$aA+bB\rightarrow cC+dD$

The expression for $K_{c}$ is written as:

$K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}$

For the given chemical equation:

$N_2O_4(g)\rightleftharpoons 2NO_2$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$ .......(1)

We are given:

Volume of the container = 1.0 L

Value of $K_c$ = 2.0

Molarity of the substance is calculated by using the equation:

$\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume}}$

For the given images:

<u>For a:</u>

Number of Gray spheres = 8 moles

Number of yellow spheres = 4 moles

Putting values in expression 1, we get:

$K_c=\frac{(8/1)^2}{(4/1)}\\\\K_c=16$

<u>For b:</u>

Number of Gray spheres = 4 moles

Number of yellow spheres = 8 moles

Putting values in expression 1, we get:

$K_c=\frac{(4/1)^2}{(8/1)}\\\\K_c=2$

<u>For c:</u>

Number of Gray spheres = 6 moles

Number of yellow spheres = 6 moles

Putting values in expression 1, we get:

$K_c=\frac{(6/1)^2}{(6/1)}\\\\K_c=6$

<u>For d:</u>

Number of Gray spheres = 2 moles

Number of yellow spheres = 8 moles

Putting values in expression 1, we get:

$K_c=\frac{(2/1)^2}{(8/1)}\\\\K_c=\frac{1}{2}$

Hence, the system which represents the equilibrium having value of $K_c=2.0$ is system (b)

3 0

3 years ago

Chem quiz please help

GalinKa [24]

The theoretical and percentage yield for the reaction are:

The theoretical yield is 21 g
The percentage yield is 119%

<h3>Balanced equation </h3>

CH₄ + 2O₂ —> CO₂ + 2H₂O

Molar mass of CH₄ = 16 g/mol

Mass of CH₄ from the balanced equation = 1 × 16 = 16 g

Molar mass of O₂ = 32 g/mol

Mass of O₂ from the balanced equation = 2 × 32 = 64 g

Molar mass of CO₂ = 44 g/mol

Mass of CO₂ from the balanced equation = 1 × 44 = 44 g

SUMMARY

From the balanced equation above,

16 g of CH₄ reacted with 64 g of O₂ to produce 44 g of CO₂

<h3>How to determine the limiting reactant</h3>

From the balanced equation above,

16 g of CH₄ reacted with 64 g of O₂

Therefore,

20 g of CH₄ will react with = (20 × 64 ) / 16 = 80 g of O₂

From the above calculation, a higher mass (i.e 80 g) of O₂ than what was given (i.e 30 g) is needed to react completely with 20 g of CH₄.

Therefore, O₂ is the limiting reactant

<h3>How to determine the theoretical yield of CO₂</h3>

From the balanced equation above,

64 g of O₂ reacted to produce 44 g of CO₂

Therefore,

30g of O₂ will react to produce = (30 × 44) / 64 = 21 g of CO₂

<h3>How to determine the percentage yield </h3>

Actual yield of CO₂ = 25 g
Theoretical yield of CO₂ = 21 g
Percentage yield =?

Percentage yield = (Actual / Theoretical) × 100

Percentage yield = (25 / 21) ×100

Percentage yield = 119%

Learn more about stoichiometry:

brainly.com/question/14735801

#SPJ1

4 0

2 years ago