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kykrilka [37]
3 years ago
7

Explain how to label a compound as a Bronstead-Lowry acid or base .

Chemistry
1 answer:
-BARSIC- [3]3 years ago
7 0

Explanation:

According to the Bronsted-Lowry conjugate acid-base theory:

An acid is defined as a substance which looses donates protons and thus forming conjugate base.

A base is defined as a substance which accepts protons and thus forming conjugate acid.

For example:

HCl(aq)+NH_3(aq)\rightarrow Cl^-(aq)+NH_4^{+}(aq)

Here, HCl is loosing a proton, thus it is considered as an acid and after losing a proton, it forms Cl^- which is a conjugate base.

And, NH_3 is gaining a proton, thus it is considered as a base and after gaining a proton, it forms NH_4^{+} which is a conjugate acid.

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How does the temperature of absolute zero relate to the kinetic energy of a substance's molecules?
skelet666 [1.2K]

Answer:

I will say D. The molecules become arranged into regular cubic arrangement. but i'm not 100% sure

Explanation:

8 0
3 years ago
What is a secondary pollutant in the troposphere
Komok [63]
Secondary air pollutants are the ones that are formed as a result of reactions
between primary pollutants and other elements in the atmosphere, such as  ozone
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3 years ago
A sample of metal has a mass of 24.54 g, and a volume of 5.02 mL. What is the density of this metal? g/cm
zmey [24]

Answer:

4.88 g / cm³

Explanation:

Density of a substance is given by the mass of the substance divided by the volume of the substance .

Hence , d = m / V

V = volume

m = mass ,

d = density ,

From the question ,

The mass of the metal = 24.54 g

The volume of the metal = 5.02 mL

Hence , by using the above formula ,and putting the corresponding values , the density is calculated as -

d = m / V

d = 24.54 g / 5.02 mL

d = 4.88 g /mL

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Hence ,

d = 4.88 g / cm³

4 0
4 years ago
A sample consisting of 1.0 mol of perfect gas molecules with CV = 20.8 J K−1 is initially at 4.25 atm and 300 K. It undergoes re
Marat540 [252]

Answer : The value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

Explanation :

First we have to calculate the value of \gamma.

\gamma=\frac{C_p}{C_v}

As, C_p=R+C_v

So, \gamma=\frac{R+C_v}{C_v}

Given :

C_v=20.8J/K\\\\R=8.314J/K

\gamma=\frac{8.314+20.8}{20.8}=1.4

Now we have to calculate the initial volume of gas.

Formula used :

P_1V_1=nRT_1

where,

P_1 = initial pressure of gas = 4.25 atm

V_1 = initial volume of gas = ?

T_1 = initial temperature of gas = 300 K

n = number of moles of gas = 1.0 mol

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

(4.25atm)\times V_1=(1.0mol)\times (0.0821L.atm/mol.K)\times (300K)

V_1=5.80L

Now we have to calculate the final volume of gas by using reversible adiabatic expansion.

P_1V_1^{\gamma}=P_2V_2^{\gamma}

where,

P_1 = initial pressure of gas = 4.25 atm

P_2 = final pressure of gas = 2.50 atm

V_1 = initial volume of gas = 5.80 L

V_2 = final volume of gas = ?

\gamma = 1.4

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

(4.25atm)\times (5.80L)^{1.4}=(2.50atm)\times V_2^{1.4}

V_2=8.47L

Now we have to calculate the final temperature of gas.

Formula used :

P_2V_2=nRT_2

where,

P_2 = final pressure of gas = 2.50 atm

V_2 = final volume of gas = 8.47 L

T_2 = final temperature of gas = ?

n = number of moles of gas = 1.0 mol

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

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

(2.50atm)\times (8.47L)=(1.0mol)\times (0.0821L.atm/mol.K)\times T_2

T_2=257.9K\approx 258K

Now we have to calculate the work done.

w=nC_v(T_2-T_1)

where,

w = work done = ?

n = number of moles of gas =1.0 mol

T_1 = initial temperature of gas = 300 K

T_2 = final temperature of gas = 258 K

C_v=20.8J/K

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

w=(1.0mol)\times (20.8J/K)\times (258-300)K

w=-873.6J

Therefore, the value of final volume, temperature and the work done is, 8.47 L, 258 K and -873.6 J

8 0
3 years ago
Anyone know how to do this???
IgorC [24]

Answer:

nope

Explanation:

cuz i cant do math

8 0
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