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3 years ago

What can be added to an atom to cause a nonvalence electron in the atom to temporarily become a valence electron?

Chemistry

2 answers:

Lilit [14]3 years ago

8 0

Answer:

Energy-

and it can be added to an atom to cause a non-valence electron in the atom to temporarily become a valence electron

vazorg [7]3 years ago

6 0

Answer:

What can be added to an atom to cause a nonvalence electron in the atom to temporarily become a valence electron is energy.

Explanation:

The normal state of the atoms, where all the electrons are occupying the lowest possible energy level, is called ground state.

The valence electrons are the electrons that occupy the outermost shell, this is the electrons in the highest main energy level (principal quantum number) of the atom.

So, a nonvalence electron occupies an orbital with less energy than what a valence electron does; in consequence, in order to a nonvalence electron jump from its lower energy level to the higher energy level of a valence electron, the former has to absorb (gain) energy.

This new state is called excited state and is temporary: the electron promoted to the higher energy level will emit the excess energy, in the form of light (photons), to come back to the lower energy level and so the atom return to the ground state.

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The atomic mass of magnesium is the weighted average of the atomic masses of

Debora [2.8K]

Answer:

2. All the naturally occurring isotopes of Mg.

Explanation:

You want to know the atomic mass of the magnesium you use in the lab. That’s “natural” magnesium. So, you must use the weighted average of all the naturally occurring isotopes in natural Mg.

1. and 3. are wrong. You won’t get the correct mass for natural Mg if you use only the artificial isotopes for your calculation.

4. is wrong. You must use all the naturally occurring isotopes. The two most abundant isotopes of Mg account for only 90 % of the atoms. If you ignore the other 10 %, your calculation will be wrong.

6 0

3 years ago

A chemical processing plant has a network of pipes that are used to transfer liquid chemicalproducts from one part of the plant

liq [111]

Answer:

Explanationis the long-distance transportation of a liquid or gas through a system of pipes—a pipeline—typically to a market area for consumption. The latest data from 2014 gives a total of slightly less than 2,175,000 miles (3,500,000 km) of pipeline in 120 countries of the world.[1] The United States had 65%, Russia had 8%, and Canada had 3%, thus 75% of all pipeline were in these three countries.[1]

Pipeline and Gas Journal's worldwide survey figures indicate that 118,623 miles (190,905 km) of pipelines are planned and under construction. Of these, 88,976 miles (143,193 km) represent projects in the planning and design phase; 29,647 miles (47,712 km) reflect pipelines in various stages of construction. Liquids and gases are transported in pipelines and any chemically stable substance can be sent through a pipeline.[2] Pipelines exist for the transport of crude and refined petroleum, fuels – such as oil, natural gas and biofuels – and other fluids including sewage, slurry, water, beer, hot water or steam for shorter distances. Pipelines are useful for transporting water for drinking or irrigation over long distances when it needs to move over hills, or where canals or channels are poor choices due to considerations of evaporation, pollution, or environmental impact.:

7 0

3 years ago

The acid-dissociation constant for benzoic acid (C6H5COOH) is 6.3×10−5. Calculate the equilibrium concentration of H3O+ in the s

raketka [301]

Answer : The equilibrium concentration of $H_3O^+$ in the solution is, $2.1\times 10^{-3}M$

Explanation :

The dissociation of acid reaction is:

$C_6H_5COOH+H_2O\rightarrow H_3O^++C_6H_5COO^-$

Initial conc. c 0 0

At eqm. c-x x x

Given:

c = $7.0\times 10^{-2}M$

$K_a=6.3\times 10^{-5}$

The expression of dissociation constant of acid is:

$K_a=\frac{[H_3O^+][C_6H_5COO^-]}{[C_6H_5COOH]}$

$K_a=\frac{(x)\times (x)}{(c-x)}$

Now put all the given values in this expression, we get:

$6.3\times 10^{-5}=\frac{(x)\times (x)}{[(7.0\times 10^{-2})-x]}$

$x=2.1\times 10^{-3}M$

Thus, the equilibrium concentration of $H_3O^+$ in the solution is, $2.1\times 10^{-3}M$

4 0

3 years ago

If the H+ in a solution is 1x10^-1 mol/L what is the OH-

damaskus [11]

PH scale is used to determine how acidic or basic a solution is.
we have been given the hydrogen ion concentration. Using this we can calculate pH,
pH = - log[H⁺]
pH = - log (1 x 10⁻¹ M)
pH = 1
using pH can calculate pOH
pH + pOH = 14
pOH = 14 - 1
pOH = 13
using pOH we can calculate the hydroxide ion concentration
pOH = - log [OH⁻]
[OH⁻] = antilog(-pOH)
[OH⁻] = 10⁻¹³ M
hydroxide ion concentration is 10⁻¹³ M

8 0

4 years ago

What is very fine sediment called

Molodets [167]

Loess, which can be carried by wind over long distances.

3 0

4 years ago