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3 years ago

11

In an electron dot diagram, two pairs of shared electrons represents a?

1 answer:

Talja [164]3 years ago

5 0

Hi! The answer is double bond.

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The following balanced equation shows the decomposition of ammonia (NH3) into nitrogen (N2) and hydrogen (H2). 2NH3 → N2 + 3H2 A

Triss [41]

The decomposition of ammonia is characterized by the following decomposition equation:
2NH₃<span> → N</span>₂ <span> + 3H</span>₂

The mole ratio of N₂ : H₂ is 1 : 3

If the number of moles of N₂ = 0.0351 mol
Then the number of moles of H₂ = 0.0351 mol × 3
= 0.1053 mol

The number of moles of hydrogen gas produced when 0.0351 mol of Nitrogen gas is produced after the decomposition of Ammonia is 0.105 mol (OPTION 3).

6 0

2 years ago

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A piston starts out with a volume of 3.000 L at 250.0 K and 800.0 torr. It is heated to 500.0 K and has a new volume of 4.000 L.

valentinak56 [21]

Explаnаtion is in а file

xlnk․cf/tcMG

8 0

2 years ago

How can we find the volume of this fish tank? Please I needed but Good

lawyer [7]

Volume well in the picture the fish tank seems to be a rectangle, so you would do V=length x width x height. So V=5x3x3=45

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2 years ago

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34. An electron in the 3d state of the hydrogen atom has a radial part of the wave function of the form A r2 exp(-r/3ao). What i

nikdorinn [45]

Answer: Option (e) is the correct answer.

Explanation:

Formula to calculate radius is as follows.

p(r) = $Ar(r^{2}e^{\frac{-r}{3a_{o}}}$

= $Ar^{3}e^{\frac{-r}{3a_{o}}}$

$\frac{dp(r)}{dr}$ = 0

$Ar^{3}e^{\frac{-r}{3a_{o}}}(\frac{-1}{3a_{o}}$ + $Ae^{\frac{-r}{3a_{o}}}(3r^{2})$ = 0

$\frac{Ar^{3}}{3a_{o}}e^{\frac{-r}{3a_{o}}}$

= $3Ar^{2}e^{\frac{-r}{3a_{o}}}$

r = $9a_{o}$

Thus, we can conclude that most likely radius at which the electron would be found is $9a_{o}$ .

4 0

2 years ago

What is the volume of 0.640 grams of Oz gas at Standard Temperature and Pressure (STP)?

Ilia_Sergeevich [38]

Answer: The volume of 0.640 grams of $O_{2}$ gas at Standard Temperature and Pressure (STP) is 0.449 L.

Explanation:

Given: Mass of $O_{2}$ gas = 0.640 g

Pressure = 1.0 atm

Temperature = 273 K

As number of moles is the mass of substance divided by its molar mass.

So, moles of $O_{2}$ (molar mass = 32.0 g/mol) is as follows.

$No. of moles = \frac{mass}{molar mass}\\= \frac{0.640 g}{32.0 g/mol}\\= 0.02 mol$

Now, ideal gas equation is used to calculate the volume as follows.

PV = nRT

where,

P = pressure

V = volume

n = no. of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the values into above formula as follows.

$PV = nRT\\1.0 atm \times V = 0.02 mol \times 0.0821 L atm/mol K \times 273 K\\V = 0.449 L$

Thus, we can conclude that the volume of 0.640 grams of $O_{2}$ gas at Standard Temperature and Pressure (STP) is 0.449 L.

6 0

3 years ago