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Fiesta28 [93]
3 years ago
13

How do you slow down the motion of a gas particle?

Chemistry
2 answers:
damaskus [11]3 years ago
7 0

Answer:

You need to turn down the heat that is creating the gas or turn down the temperature in the room the colder it is the slower the gas is.

Explanation:

jasenka [17]3 years ago
3 0
I think u turn down the heat not to sure
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Gt4rgbrbrtbr\t?<br>frehjkgbrtnvboror
3241004551 [841]

Answer:I can't see any question

8 0
3 years ago
write equations to show the chemical processes which occur when the first ionization and the second ionization energies of lithi
diamong [38]

Answer:

First ionization of lithium:

\text{Li}\;(g)\to \text{Li}^{+} \; (g) + \text{e}^{-}.

Second ionization of lithium:

\text{Li}^{+}\;(g) \to\text{Li}^{2+} \;(g) + \text{e}^{-}.

Explanation:

The ionization energy of an element is the energy required to remove the outermost electron from an atom or ion of the element in gaseous state. (Refer to your textbook for a more precise definition.) Some features of the equation:

  • Start with a gaseous atom (for the first ionization energy only) or a gaseous ion. Write the gaseous state symbol (g) next to any atom or ion in the equation.
  • The product shall contain one gaseous ion and one electron. The charge on the ion shall be the same as the order of the ionization energy. For the second ionization energy, the ion shall carry a charge of +2.
  • Charge shall balance on the two sides of the equation.

First Ionization Energy of Li:

  • The products shall contain a gaseous ion with charge +1 \text{Li}^{+}\;(g) as well as an electron \text{e}^{-}.
  • Charge shall balance on the two sides. There's no net charge on the product side. Neither shall there be a charge on the reactant side. The only reactant shall be a lithium atom which is both gaseous and neutral: \text{Li}\;(g).
  • Hence the equation: \text{Li}\;(g) \to \text{Li}^{+}\;(g) + \text{e}^{-}.

Second Ionization Energy of Li:

  • The product shall contain a gaseous ion with charge +2: \text{Li}^{2+}\;(g) as well as an electron \text{e}^{-}.
  • Charge shall balance on the two sides. What's the net charge on the product side? That shall also be the charge on the reactant side. What will be the reactant?
  • The equation for this process is \text{Li}^{+} \; (g) \to \text{Li}^{2+}\;(g) + \text{e}^{-}.
5 0
3 years ago
Enter the ions formed when (NH4)2S dissolves in water.
lisabon 2012 [21]

The ions formed are NH4(+) and S(2-)

The dissolution reaction of (NH4) 2S in water is as follows:


(NH4) 2S ==> 2 NH4 (+) + S (2-).



Ammonium sulfide is the ammonium salt of hydrogen sulfide. It has the formula (NH4) 2S and belongs to the sulfide family.


It is a relatively unstable compound (crystals decomposing at -18 ° C, but exists and is more stable in aqueous solution.) With a pKa exceeding 15, the hydrosulfide ion cannot be significantly deprotonated by ammonia. Thus, such solutions consist mainly of a mixture of ammonia and hydrosulphide of ammonium, it has a smell, close to that of hydrogen sulfide, and its aqueous solutions can be precisely by emitting H2S.

5 0
3 years ago
Read 2 more answers
The formation constant for the reaction ag (aq) 2nh3(aq) ag(nh3)2 (aq) is kf = 1.7 × 107 at 25°c. what is δg° at this temperatur
anzhelika [568]

The value of ΔG° at this temperature is -18034.18 J/mol

Calculation,

Given information

formation constant (Kf)= 1.7 × 10^{7}

Universal gas constant (R) = 8.314 J/K• mol

Temperature = 25° C = 25 °C + 273 = 300 K

Formula used:

ΔG° = -RT㏑Kf

By putting the valur of R,T, Kf we get the value of ΔG°

ΔG° = - 8.314 J/K• mol×300K㏑ 1.7 × 10^{7}

ΔG° = -2494.2㏑ 1.7 × 10^{7} = -18034.18 J/mol

So, change in standard Gibbs's free energy is -18034.18 J/mol

Learn about formation constant

brainly.com/question/14011682

#SPJ4

8 0
1 year ago
What would happen if hydrogen and nitrogen combine?
Tanzania [10]
<span>At room temperature and atmospheric pressure, nothing happens when the two gasses are mixed. However, at high temperature and pressure (450C, 200atm), in the presence of an iron oxide catalyst, the production of ammonia is thermodynamically advantageous.</span>
5 0
3 years ago
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