All categories

3 years ago

How do the number of protons in U-235 and U-238 compare?

1 answer:

3 0

The number of protons among the 2 isotopes remain the same, what varies is the number of neutrons between the 2 isotopes. The isotope with the larger mass number possess a greater number of neutrons compared to the other one.

You might be interested in

Why is the enthalpy of formation of oxygen zero?

madreJ [45]

All elements in their standard states (oxygen gas, solid carbon in the form of graphite, etc.) have a standard enthalpy of formation of zero, as there is no change involved in their formation.

4 0

2 years ago

Use the periodic table to identify the element with the electron configuration 1s²2s²2p⁴. Write its orbital diagram, and give th

natta225 [31]

Answer:

1. Orbital diagram

2p⁴ ║ ↑↓ ║ "↑" ║ ↑

2s² ║ ↑↓ ║

1s² ║ ↑↓ ║

2. Quantum numbers

n = 2,
l = 1,

$m_{l}$ = 0,

$m_{s}$ = +1/2

Explanation:

The fill in rule is:

Follow shell number: from the inner most shell to the outer most shell, our case from shell 1 to 2

Follow the The Aufbau principle, 1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p<6s<4f<5d<6p<7s<5f<6d<7p
Hunds' rule: Every orbital in a sublevel is singly occupied before any orbital is doubly occupied. All of the electrons in singly occupied orbitals have the same spin (to maximize total spin).

So, the orbital diagram of given element is as below and the sixth electron is marked between " "

2p⁴ ║ ↑↓ ║ "↑" ║ ↑

2s² ║ ↑↓ ║

1s² ║ ↑↓ ║

The quantum number of an electron consists of four number:

n (shell number, - 1, 2, 3...)
l (subshell number or orbital number, 0 - orbital s, 1 - orbital p, 2 - orbital d...)
$m_{l}$ (orbital energy, or "which box the electron is in"). For example, orbital p (l = 1) has 3 "boxes", it was number from -1, 0, 1. Orbital d (l = 2) has 5 "boxes", numbered -2, -1, 0, 1, 2
$m_{s}$ (spin of electron), either -1/2 or +1/2

In our case, the electron marked with " " has quantum number

n = 2, shell number 2,
l = 1, subshell or orbital p,

$m_{l}$ = 0, 2nd "box" in the range -1, 0, 1

$m_{s}$ = +1/2, single electron always has +1/2

3 0

3 years ago

Why is zinc not extracted from ZnO through reduction using CO?

Jet001 [13]

The standard Gibbs free energy of formation of ZnO from Zn is lower than that of CO2 from CO. Therefore, CO cannot reduce ZnO to Zn. Hence, Zn is not extracted from ZnO through reduction using CO

6 0

1 year ago

How many moles of hydrogen are in the sample? Round your answer to 4 significant digits.

finlep [7]

Answer:

1.56 mol H₂

Explanation:

Mg₃(Si₂O₅)₂(OH)₂

There are 4 Si moles per Mg₃(Si₂O₅)₂(OH)₂ mol. With that in mind we can calculate how many Mg₃(Si₂O₅)₂(OH)₂ moles are there in the sample, using the given number of silicon moles:

3.120 mol Si * $\frac{1molMg_3(Si_2O_5)_2(OH)_2}{4molSi}$ = 0.78 mol Mg₃(Si₂O₅)₂(OH)₂

Then we can convert Mg₃(Si₂O₅)₂(OH)₂ moles into hydrogen moles, keeping in mind that there are 2 hydrogen moles per Mg₃(Si₂O₅)₂(OH)₂ mol:

0.78 mol Mg₃(Si₂O₅)₂(OH)₂ * 2 = 1.56 mol H₂

8 0

3 years ago

we make 200 mL of a solution with a 0.025 M concentration of Ca(OH)2 The solution is then diluted to 1.00 L by adding additional

ExtremeBDS [4]

Answer:

12.00

Explanation:

4 0

3 years ago