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3 years ago

Compare the three atom diagrams. which one shows the electron with the highest potential energy

Chemistry

1 answer:

lara31 [8.8K]3 years ago

5 0

Answer : The one diagram which shows the electron with the highest potential energy is attached below.

Explanation : One can easily find the highest potential energy of the atom just by looking at the diagram, the electron which is from farthest distance from the atomic nucleus will have the highest potential energy in the electron.

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manganese has an atomic number of 25 and an atomic mass of 55 amu. How many particles are found in its nucleus

saw5 [17]

The particles in the nucleus is 55

6 0

3 years ago

Which isotope has the greatest number of protons?

tino4ka555 [31]

Answer:

The isotope with the greatest number of protons is:

<u>option D: Pu-239, with 94 protons</u>

Explanation:

The number of <em>protons</em> is the atomic number and is a unique number for each type of element.

You can tell the number of protons searching the element in a periodic table and reading its atomic number.

Thus, this is how you tell the number of protons or each isotope

Sample Chemical symbol Element atomic number # of protons

A Pa-238 Pa protactinium 91 91

B U-240 U uranium 92 92

C Np-238 Np neptunium 93 93

D Pu-239 Pu plutonium 94 94

8 0

3 years ago

Ammonium nitrate, a common fertilizer, is used as an explosive in fireworks and by terrorists. It was the material used in the t

Sladkaya [172]

Explanation:

The given balanced reaction is as follows.

$2NH_{4}NO_{3} \rightarrow 2N_{2} + O_{2} + 4H_{2}O$

It is given that mass of ammonium nitrate is 86.0 kg.

As 1 kg = 1000 g. So, 86.0 kg = 86000 g.

Hence, moles of $NH_{4}NO_{3}$ present will be as follows.

Moles of $NH_{4}NO_{3}$ = $\frac{mass given}{\text{molar mass of NH_{4}NO_{3}}}$

= $\frac{86000 g}{80.043 g/mol}$

= 1074.42 mol

Therefore, moles of $N_{2}$ , $O_{2}$ and $H_{2}O$ produced by 1074.42 mole of $NH_{4}NO_{3}$ will be as follows.

Moles of $O_{2}$ = $\frac{1}{2} \times 1074.42 mol$

= 537.21 mol

Moles of $N_{2}$ = $\frac{2}{2} \times 1074.42 mol$

= 1074.42 mol

Moles of $H_{2}O$ = $\frac{4}{2} \times 1074.42 mol$

= 2148.84 mol

Therefore, total number of moles will be as follows.

537.21 mol + 1074.42 mol + 2148.84 mol

= 3760.47 mol

According to ideal gas equation, PV = nRT. Hence, calculate the volume as follows.

PV = nRT

1 atm \times V = 3760.47 mol \times 0.0821 L atm/mol K \times 580 K[/tex] (as $307^{o}C$ = 307 + 273 = 580 K)

V = 179066.06 L

Thus, we can conclude that total volume of the gas is 179066.06 L.

3 0

3 years ago

For the reaction 6 Li + N2 → 2 Li3N , what is the maximum amount of Li3N (34.8297 g/mol) which could be formed from 14.18 mol (6

Yuki888 [10]

Moles Li = 3.50 g / 6.941 g/mol= 0.504
the ratio between Li and N2 is 6 : 1
moles N2 required = 0.504 /6=0.0840
we have 3.50 g / 28.0134 g/mol=0.125 moles of N2 so N2 is in excess
the ratio between Li and Li3N is 6 : 2
moles Li3N = 0.504 x 2 /6=0.168
mass Li3N = 0.168 mol x 34.8297 g/mol=5.85 g

4 0

3 years ago

Tris has a molecular weight of 121 g/mol. How many grams of Tris would you need to make 100 mL of a 100 mM solution of Tris

Zepler [3.9K]

Answer:

1.21 g of Tris

Explanation:

Our solution if made of a solute named Tris

Molecular weight of Tris is 121 g/mol

[Tris] = 100 mM

This is the concentration of solution:

(100 mmoles of Tris in 1 mL of solution) . 1000

Notice that mM = M . 1000 We convert from mM to M

100 mM . 1 M / 1000 mM = 0.1 M

M = molarity (moles of solute in 1 L of solution, or mmoles of solute in 1 mL of solution). Let's determine the mmoles of Tris

0.1 M = mmoles of Tris / 100 mL

mmoles of Tris = 100 mL . 0.1 M → 10 mmoles

We convert mmoles to moles → 10 mmol . 1mol / 1000mmoles = 0.010 mol

And now we determine the mass of solute, by molecular weight

0.010 mol . 121 g /mol = 1.21 g

8 0

3 years ago