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Cloud [144]
3 years ago
5

Heat the oxygen to about 150 K. What state of matter is this?

Chemistry
1 answer:
adelina 88 [10]3 years ago
3 0

Answer:

  • <u>Gas</u>

Explanation:

It is easier if you convert the kelvin temperature into Celsius degrees:

  • ºC = T - 273.15 = 150 - 273.15 = -123.15ºC

Now, you know that that is a very cold temperature. Thus, may be the oxygen is not gas any more but it changed to liquid . . . or solid?

You must search for the boiling point and melting (freezing) point of oxygen in tables or the internet. At standard pressure (about 1 atm) they are:

  • Melting point: −218.79 °C,
  • Boiling point: −182.962 °C

That means that:

  • below -218.79ºC oxygen is solid (not our case).
  • between -218.79ºC and -182.962ºC oxygen is liquid (not our case)
  • over -182.962ºC oxygen is a gas. This is our case, because -123.15ºC is a higher temperature than -182.962ºC.

Hence, <em>the state of matter of oxygen at 150K</em>, and standard pressure, is gas.

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For the following reactions, write a balanced equation using half-reactions and calculate the voltage to be expected.
Iteru [2.4K]

a) 2Na(s) +2H_2O(I)→ 2Na^+(aq) + OH^-(aq)+ H_2(g)

b) 2Ag (s) +2H^(aq) → 2 Ag^+ (aq) +H_2(g)

<h3>What are half-reactions?</h3>

The half-reaction method is a way to balance redox reactions. It involves breaking the overall equation down into an oxidation part and a reduction part.

a)

2Na(s) +2H_2O(I)→ 2Na^+(aq) + OH^-(aq)+ H_2(g)

E^0 cell = E^0 (reduction)  - E^0 (oxidation)

= E^0(\frac{H_2O}{H_2}, OH^-) -E^0(Na^+/Na)

= -0.83 - (-2.71) =1.88V

b)

2Ag (s) +2H^(aq) → 2 Ag^+ (aq) +H_2(g)

E^0cell= E^0(H^+/H_2) -E^0(Ag^+/Ag)

E^0cell=-0. - (0.8) =-0.8V

Learn more about the half-reactions here:

https://brainly.in/question/18053421

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8 0
2 years ago
A gas is collected at 20.0 °C and 725.0 mm Hg. When the temperature is
krek1111 [17]

Answer:

676mmHg

Explanation:

Using the formula;

P1/T1 = P2/T2

Where;

P1 = initial pressure (mmHg)

P2 = final pressure (mmHg)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the information provided in this question;

P1 = 725.0mmHg

P2 = ?

T1 = 20°C = 20 + 273 = 293K

T2 = 0°C = 0 + 273 = 273K

Using P1/T1 = P2/T2

725/293 = P2/273

Cross multiply

725 × 273 = 293 × P2

197925 = 293P2

P2 = 197925 ÷ 293

P2 = 676mmHg.

The resulting pressure is 676mmHg

3 0
3 years ago
An element has a half-life of 30 years. if 1.0 mg of this element decays over a period of 90 years, how many mg of this element
labwork [276]

Answer:

144.6

Explanation:

6 0
3 years ago
Calculate the H3O+ concentration and then pH in a 0.100 M HC2H3O2 solution that is
Lana71 [14]

Answer:

Please check with the attachment posted.

Please do confirm the answer.

5 0
3 years ago
chegg write a net ionic equation describing the oxidation of no2 2 to no3 2 by o2 in a basic solution.
Marina86 [1]

When the same species undergoes both oxidation and reduction in a single redox reaction, this is referred to as a disproportionation. Therefore, divide it into two equal reactions.

NO2→NO^−3

NO2→NO

and do the usual changes

First, balance the two half reactions:

3. NO2 +H2O →NO^−3 + 2 H^+ + e−

4. NO2 +2 H^+ + 2e− → NO + H2O

Now multiply one or both half-reactions to ensure that each has the same number of electrons. Here, Eqn (3) x 2 results in each half-reaction having two electrons:

5. 2 NO2 + 2 H2O → 2 NO^−3 + 4H^+ + 2e−

Now add Eqn 4 and 5 (the electrons now cancel each other):

3NO2 + 2H^+ + 2H2O → NO + 2 NO−3 + H2O + 4H+

and cancel terms that’s common to both sides:

3NO2 + H2O → NO + 2NO^−3 + 2H+

This is the net ionic equation describing the oxidation of NO2 to NO3 in basic solution.

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7 0
1 year ago
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