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ohaa [14]
4 years ago
10

Define and explain auto-ionization of water??? Explain please??

Chemistry
2 answers:
Oduvanchick [21]4 years ago
6 0

Answer:

in pure water, H2O molecules undergo a reaction in which two molecules react in such a way that one molecule give its H+ ion to other molecule and become a hydroxyl ion(OH-) and molecule which gain H+ ion become H3O+ ion.

Explanation:

H2O + H2O ⇄ ( H3O+) +( OH-)

icang [17]4 years ago
5 0

Answer:

Hey there!

Auto-ionization of water is an ionization reaction in pure water or in another aqueous solution, in which a water molecule, H2O, loses the nucleus of one of its hydrogen atoms to become a hydroxide ion, OH−.

Let me know if this helps :)

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How long does it take for a 12.62g sample of ammonia to heat from 209K to 367K if heated at a constant rate of 6.0kj/min? The me
Georgia [21]
First, consider the steps to heat the sample from 209 K to 367K.

1) Heating in liquid state from 209 K to 239.82 K

2) Vaporaizing at 239.82 K

3) Heating in gaseous state from 239.82 K to 367 K.


Second, calculate the amount of heat required for each step.

1) Liquid heating

Ammonia = NH3 => molar mass = 14.0 g/mol + 3*1g/mol = 17g/mol

=> number of moles = 12.62 g / 17 g/mol = 0.742 mol

Heat1 = #moles * heat capacity * ΔT

Heat1 = 0.742 mol * 80.8 J/mol*K * (239.82K - 209K) = 1,847.77 J

2) Vaporization

Heat2 = # moles * H vap

Heat2 = 0.742 mol * 23.33 kJ/mol = 17.31 kJ = 17310 J

3) Vapor heating

Heat3 = #moles * heat capacity * ΔT

Heat3 = 0.742 mol * 35.06 J / (mol*K) * (367K - 239.82K) = 3,308.53 J

Third, add up the heats for every steps:

Total heat = 1,847.77 J + 17,310 J + 3,308.53 J = 22,466.3 J

Fourth, divide the total heat by the heat rate:

Time = 22,466.3 J / (6000.0 J/min) = 3.7 min

Answer: 3.7 min


3 0
3 years ago
Please show some work For the reaction: NO(g) + 1/2 O2(g) → NO2(g) ΔH°rxn is -114.14 kJ/mol. Calculate ΔH°f of gaseous nitrogen
uranmaximum [27]

Answer:

148.04 kJ/mol

Explanation:

Let's consider the following thermochemical equation.

NO(g) + 1/2 O₂(g) → NO₂(g)      ΔH°rxn = -114.14 kJ/mol

We can find the standard enthalpy of formation (ΔH°f) of NO(g) using the following expression.

ΔH°rxn = 1 mol × ΔH°f(NO₂(g)) - 1 mol × ΔH°f(NO(g)) - 1/2 mol × ΔH°f(O₂(g))

ΔH°f(NO(g)) = 1 mol × ΔH°f(NO₂(g)) - ΔH°rxn - 1/2 mol × ΔH°f(O₂(g)) / 1 mol

ΔH°f(NO(g)) = 1 mol × 33.90 kJ/mol - (-114.14 kJ) - 1/2 mol × 0 kJ/mol / 1 mol

ΔH°f(NO(g)) = 148.04 kJ/mol

8 0
3 years ago
Given an unknown mixture consisting of two substances, explain how a scientist could use lab techniques to determine whether the
Katyanochek1 [597]
<span>Pass the mixture through filter paper. The large particles in the suspension will filter out. to tell the difference between a solution and a colloid, shine a beam of light through the mixture, if it reflects then it is a colloid, if it doesn't then it is a solution</span>
4 0
3 years ago
A helium-filled balloon has a volume of 2.48 L and a pressure of 150 kPa. The volume of the balloon increases to 2.98 L after yo
Tema [17]

Answer:

THE NEW PRESSURE OF THE HELIUM GAS AT 2.98 L VOLUME IS 124.8 kPa.

AT AN INCREASE ALTITUDE, THERE IS A LOWER PRESSURE ENVIRONMENT AND THE HELIUM GAS PRESSURE DECREASES AND HENCE AN INCREASE IN VOLUME.

Explanation:

The question above follows Boyle's law of the gas law as the temperature is kept constant.

Boyle's law states that the pressure of a fixed mass of gas is inversely proportional to the volume, provided the temperature remains constant.

Mathematically, P1 V1 = P2 V2

P1 = 150 kPa = 150 *10^3 Pa

V1 = 2.48 L

V2 = 2.98 L

P2 = ?

Rearranging the equation, we obtain;

P2 = P1 V1 / V2

P2 = 150 kPa * 2.48 / 2.98

P2 = 372 *10 ^3 / 2.98

P2 = 124.8 kPa.

The new pressure of the gas when at a height which increases the volume of the helium gas to 2.98 L is 124.8 kPa.

6 0
3 years ago
What is the relationship between melting/boiling point and intermolecular forces
sattari [20]
Boiling points are a measure of intermolecular forces. The intermolecular forces increase with increasing polarization of bonds. Boiling point increases with molecular weight, and with surface area.
8 0
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