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4 years ago

What is the volume of an oxygen tank if it contains 12 moles of oxygen at 273 K under 75 kPa?

1 answer:

5 0

Given:
n = 12 moles of oxygen
T = 273 K, temperature
p = 75 kPa, pressure

Use the ideal gas law, given by
$pV=nRT \\ or \\ V= \frac{nRT}{p}$
where
V = volume
R = 8.3145 J/(mol-K), the gas constant

Therefore,
$V= \frac{(12\,mol)(8.3145\, \frac{J}{mol-K} )(273\,K)}{75 \times 10^{3} \, Pa}= 0.3632\,m^{3}$

Answer: 0.363 m³

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Katyanochek1 [597]

Ammonium hydroxide is considered a weak base because it produces few hydroxide ions in aqueous solution. (Option a)

<h3>Strength of electrolytes</h3>

Strong electrolytes: are completely ionized in aqueous solution.
Weak electrolytes: are partially ionized in aqueous solution.

Let's consider the ionization of ammonium hydroxide, which is a weak base.

NH₄OH ⇄ NH₄⁺ + OH⁻

Why is it considered a weak base?

a. it produces few hydroxide ions in aqueous solution. Yes, since it dissociates partially.

b. it produces few hydrogen ions in aqueous solution. No, since it does not produce hydrogen ions.

c. it does not conduct electricity in aqueous solution. No, since it conducts electricity, although poorly.

d. it will not take part in a neutralization reaction. No, since it will be neutralized by acids.

Ammonium hydroxide is considered a weak base because it produces few hydroxide ions in aqueous solution.

Learn more about weak bases here: brainly.com/question/322250

#SPJ1

5 0

2 years ago

Identify the atoms that are oxidized and reduced, the change in the oxidation state for each, and oxidising and reducing agents

KATRIN_1 [288]

Answer:

Explanation:

a) $Mg + NiCl_2 \rightarrow MgCl_2 + Ni$

Oxidation state of Mg changes to 0 to +2.

Oxidation state of Ni changes to +2 to 0.

Oxidation state of Cl does not change.

So, Mg is oxidized, Ni is reduced.

Mg acts as reducing agent and $NiCl_2$ acts as oxidizing agent.

b) $PCl_3 + Cl_2 \rightarrow PCl_5$

Oxidation state of P changes to +3 to +5.

Oxidation state of Cl changes to 0 to -1.

So,

P is oxidized and Cl is reduced.

$PCl_3$ acts as reducing agent and $Cl_2$ acts as oxidizing agent.

c) $C_2H_4 + 3O_2 \rightarrow 2CO_2 + 2H_2O$

Oxidation state of C changes to -2 to +4.

Oxidation state of O changes to 0 to -2.

Oxidation state of H does not change.

So,

C is oxidized and O is reduced.

$C_2H_4$ acts as reducing agent and $O_2$ acts as oxidizing agent.

d) $Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2$

Oxidation state of Zn changes to 0 to +2.

Oxidation state of H changes to +1 to 0.

Oxidation state of S and O do not change.

So, Zn is oxidized, H is reduced.

Zn acts as reducing agent and $H_2SO_4$ acts as oxidizing agent

e) $K_2S_2O_3 + I_2 \rightarrow K_2S_4_O6 + 2KI$

Oxidation state of S changes to +2 to +2.5.

Oxidation state of I changes to 0 to -1.

Oxidation state of K and O do not change.

So, S is oxidized, I is reduced.

$K_2S_2O_3$ acts as reducing agent and $I_2$ acts as oxidizing agent.

f) $3Cu + 8HNO_3 \rightarrow 3Cu(NO_3)_2 + 4H_2O+2NO$

Oxidation state of Cu changes to 0 to +2.

Oxidation state of N changes to N +5 to +2.

Oxidation state of H and O do not change.

So, Cu is oxidized, N is reduced.

Cu acts as reducing agent and $HNO_3$ acts as oxidizing agent

8 0

3 years ago

Why is there not a corresponding drop in ionization energies between Bi and Po in the sixth period as N and O?

Stels [109]

Answer:

See explanation

Explanation:

According to the Journal of Chemical Education, Volume 80, No.8 (2003); "The first ionization energy of bismuth appears to be anomalous......It has been claimed that spin– orbit coupling by the Russell–Saunders scheme would lower the ground state of Bi+ ..."

However, the involvement of d and f orbitals in Bi and Po implies that the outermost orbitals are poorly screened hence the drop between nitrogen and oxygen is not observed between Bi and Po.

The same argument could be extended to explain the reason why there not a corresponding drop between Ba and Tl is the sixth period even though they are in the same group as Be and B.

3 0

3 years ago

Calculate the enthalpy change for the reaction of hydrogen and chlorine using the bond energies below.

babunello [35]

Answer:

final-intial temperature= enthalpy change

8 0

3 years ago

Imagine a moving boat making waves in the middle of a pool of water. Also, imagine two observers, one behind the boat and one in

Simora [160]

The person who is behind the boat will see much powerful and bigger waves than the person who is in front of the boat. This phenomenon is caused by the Doppler effect which describes the changes in the observed frequency of a wave whenever there is relative motion between the wave source and the observer.

3 0

3 years ago