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3 years ago

Write a chemical equation for Ba(OH )2 (aq) showing how it is an acid or a base according to the Arrhenius definition.

1 answer:

5 0

Balanced chemical reaction: Ba(OH)₂(aq) → Ba²⁺(aq) + 2OH⁻(aq).
According to the Arrhenius definition barium hydroxide is base.
According this theory acids are substances which produce hydrogen ions in water solution, bases are substances which produce hydroxide ions in solution. Acids and bases when react (neutralisation) produce salt and water.

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Why does a solution of acid and sodium thiosulfate go cloudy

klemol [59]

The soulustion. turn cloudy due to the precipitation of Sulphur.

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3 years ago

5.00 mol of ammonia are introduced into a 5.00 L reactor vessel in which it partially dissociates at high temperatures. 2NH 3(g)

allochka39001 [22]

Explanation:

system at equilibrium, will the reaction shift towards reactants ~

--?'

2. (2 Pts) Consider the reaction N2(g) + 3H2(g) =; 2NH3(g). The production of ammonia is an

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.1.+- w~t s~,H (o')l r'eo.c. e~ ei~i"liht-,·u.fn\ P~~,

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300 1.5x104

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e- ~ r fee, ct o. ~ 1<

900 3.4 X 10-3

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t d- IS o.,;r-. \4\a..i~1f't~ °the te.Y'il(lf1,:J'u.r-a a•~S. j lrvdu..c,,.) +~H~to{' '\

exothe-rnh't.-- ,.. ..,. (/.., ,~.

5. (4 Pts) Consider the reaction, N2(g) + 3H2(g) =; 2NH3(g). Kc= 4.2 at 600 K.

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4 years ago

Someone please help me!!

sesenic [268]

Answer:

the 3rd one (0.01 cm the one selected already)

Explanation:

copper wire isn't excessively big, and it wraps around the pencil because its malleable. I think that the most accurate would be 0.01 cm

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3 years ago

the heat of fusion of acetone is 5.7 kJ/mol. Calculate to two significant figures the entropy change when 6.3 mol of acetone mel

shtirl [24]

Answer: The entropy change of the process is $2.0\times 10^2J/K$

Explanation:

To calculate the entropy change for different phase at same temperature, we use the equation:

$\Delta S=n\times \frac{\Delta H_{f}}{T}$

where,

$\Delta S$ = Entropy change

n = moles of acetone = 6.3 moles

$\Delta H_{f}$ = enthalpy of fusion = 5.7 kJ/mol = 5700 J/mol (Conversion factor: 1 kJ = 1000 J)

T = temperature of the system = $-94.7^oC=[273-94.7]=178.3K$

Putting values in above equation, we get:

$\Delta S=\frac{6.3mol\times 5700J/mol}{178.3K}\\\\\Delta S=201.4J/K=2.0\times 10^2J/K$

Hence, the entropy change of the process is $2.0\times 10^2J/K$

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4 years ago

What is the charge of oxygen in Na2O?

Nuetrik [128]

In Na2O, what is the oxidation state of oxygen? In Na2O oxidation state of Na is 1+

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3 years ago

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