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sergejj [24]
3 years ago
10

What is the formula for the ion when aluminum achieves noble-gas electron configuration

Chemistry
2 answers:
amm18123 years ago
7 0

Answer:

K+

Explanation:

Potassium(an alkali metal) belongs to element in group 1 and period 4 in the periodic table. It has to release one electron to attain noble configuration.

beks73 [17]3 years ago
7 0

Answer: Al3+

Explanation:

Al belongs to group 3. It releases 3 electrons to attain the noble gas configuration

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Hydrogen gas and fluorine gas will react to form hydrogen fluoride gas. What is the standard free energy change for this reactio
Marta_Voda [28]

Answer:

\Delta G=-541.4kJ/mol

Explanation:

Hello there!

In this case, according to the given information, it turns out firstly necessary to write out the described chemical reaction as shown below:

H_2+F_2\rightarrow 2HF

Now, we set up the expression for the calculation of the standard free energy change, considering the free energy of formation of each species, specially those of H2 and F2 which are both 0 because they are pure elements:

\Delta G=2\Delta G_f^{HF}-(\Delta G_f^{H_2}+\Delta G_f^{F_2})\\\\\Delta G=2*-270.70kJ/mol-(0kJ/mol+0kJ/mol)\\\\\Delta G=-541.4kJ/mol

Regards!

4 0
3 years ago
Write an expression for "7 less than y."
victus00 [196]

Answer:

y-7

Explanation:

4 0
3 years ago
Read 2 more answers
How many valence electrons are in the element gallium? explain how you determined the number.
Natasha2012 [34]

Answer:

three valence electrons

Explanation:

Gallium has three electrons in the outer energy level and therefore has three valence electrons. The identification of valence electrons is vital because the chemical behavior of an element is determined primarily by the arrangement of the electrons in the valence shell.

4 0
3 years ago
What is the empirical formula for ribose (C5H10O5)?
Dafna1 [17]

What is the empirical formula for ribose (C5H10O5)?

C. CH20

3 0
3 years ago
An important reaction in the formation of photochemical smog is the photodissociation of NO2: NO2 + hv >>> NO(g) + O(g)
vaieri [72.5K]

Explanation:

NO_2 + hv\rightarrow NO(g) + O(g)

The maximum wavelength of light that can cause this reaction is 420 nm.

a) The wavelength given lies in the range of visible light range that is from 400 nano meters to 700 nano meters.

The light with wavelength of 420 nm is found in the range of visible light.

b)The maximum strength of a bond :

E=\frac{h\times c}{\lambda}

where,

E = energy of photon = Energy required to break single molecule of nitrogen dioxide

h = Planck's constant = 6.626\times 10^{-34}Js

c = speed of light = 3\times 10^8m/s

\lambda = wavelength = 420 nm=420\times 10^{-9}m

E=\frac{6.626\times 10^{-34}Js\times 3\times 10^8 m/s}{420\times 10^{-9}m}

E=4.733\times 10^{-19} J

Energy required to break 1 mole of nitrogen dioxide molecules:

= E\times N_A=4.733\times 10^{-19} J\times 6.022\times 10^{23} mol^{-1}

=285,012.66 J/mol=285.13 kJ/mol

(1 J = 0.001 kJ )

285.13 is the maximum strength of a bond, in kJ/mol, that can be broken by absorption of a photon of 420-nm light.

c) the photodissociation reaction showing Lewis-dot structures is given in an image attached.

6 0
4 years ago
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