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Paul [167]
2 years ago
6

What is the general trend for atomic radius as you go down the noble gas family?

Chemistry
1 answer:
ANEK [815]2 years ago
7 0

Answer:

the atomic radius is larger as you go down the noble gas family

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Please write thoughtful, concise summaries of your findings in paragraph form. Include key calculation values and point out what
dedylja [7]
I am not sure how to do this
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3 years ago
A gaseous substance turns directly into a solid. Which term describes this change?
UkoKoshka [18]

Answer:

deposition

Explanation:

Sublmation- solid transforming into a gas, skipping the liquid stage.

eveporation- a liquid transformimg into a gas

melting- a solid transforming into a liquid

deposition- the opposite of sublimation (your anwser)

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3 years ago
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MgCl2 + AgNO3 → AgCl + Mg(NO3)2
Law Incorporation [45]

Answer:

MgCl2 + 2AgNO3 → 2AgCl + Mg(NO3)2

Explanation:

I'm assuming you want to balance it so...

The first thing I see is that there are two chlorines on the reactant side and one on the product side

Adding a coefficient of 2 would get 2AgCl2

Now there are two silvers on the reactant side, so add a 2 to AgNO3 on the products side. Now they are all balanced.

If that is not what you are looking for let me know!

6 0
3 years ago
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In which orbitals would the valence electrons for selenium (Se) be placed?
RSB [31]
Selenium falls in the same column with oxygen therefore it has the same number of valence electrons with oxygen which is 6. It fills the 3d orbital but only 4s and 4p are counted. The electronic configuration is [Ar]3d^10 4s^2.
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Ammonia will decompose into nitrogen and hydrogen at high temperature. An industrial chemist studying this reaction fills a 500.
patriot [66]

Explanation:

Chemical reaction equation for the give decomposition of NH_{3} is as follows:.

          2NH_{3}(g) \rightleftharpoons N_{2}(g) + 3H_{2}(g)

And, initially only NH_{3} is present.

The given data is as follows.

  P_{NH_{3}} = 2.3 atm at equilibrium

   P_{H_{2}} = 3 \times P_{N_{2}} = 0.69 atm

Therefore,

          P_{N_{2}} = \frac{0.69 atm}{3}

                        = 0.23 aatm

So, P_{NH_{3}} = 2.3 - 2(0.23)

                       = 1.84 atm

Now, expression for K_{p} will be as follows.

         K_{p} = \frac{(P_{N_{2}})(P^{3}_{H_{2}})}{(P^{2}_{NH_{3}})}

           K_{p} = \frac{(0.23) \times (0.69)^{3}}{(1.84)^{2}}

                      = \frac{0.23 \times 0.33}{3.39}

                     = 0.0224

or,           K_{p} = 2.2 \times 10^{-2}

Thus, we can conclude that  the pressure equilibrium constant for the decomposition of ammonia at the final temperature of the mixture is 2.2 \times 10^{-2}.

6 0
3 years ago
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