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3 years ago

Which of the following substances can be decomposed by chemical change?

Chemistry

1 answer:

vovangra [49]3 years ago

8 0

NaCl can be decomposed by electrolysis

Na, Cl2 and K are in the simplest form and they cannot be decomposed

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When 92.0 g of ethanol (c2h5oh are vaporized at its boiling point of 78.3°c, it requires 78.6 kj of energy. what is the approxim

Juli2301 [7.4K]

The solution would be like this for this specific problem:

(78.6 kJ) / (92.0 g / (46.0684 g C2H5OH/mol)) = 39.4 kJ/mol

39.3

So the approximate molar heat of vaporization of ethanol in kJ/mol is 39.3.

I hope this answers your question.

7 0

3 years ago

Read 2 more answers

What is the wavelength of an X-ray that has a frequency of 7.8 x 1017 Hz

SVEN [57.7K]

Answer:

λ = 0.38 ×10⁻⁹ m

Explanation:

Given data:

Wavelength of xray = ?

Frequency of xray = 7.8 ×10¹⁷ Hz

Solution:

Formula:

Speed of light = wavelength × frequency

speed of light = 3×10⁸ m/s

Now we will put the values in formula.

3×10⁸ m/s = λ × 7.8 ×10¹⁷ Hz

λ = 3×10⁸ m/s / 7.8 ×10¹⁷ Hz

Hz = s⁻¹

λ = 3×10⁸ m/s / 7.8 ×10¹⁷s⁻¹

λ = 0.38 ×10⁻⁹ m

3 0

2 years ago

Please HELP will give Brainliest!!!!!

dedylja [7]

4 mol / 205g H2O = 4/.205 = 19.5 mol/kg boiling point = 100 + 19.5 • 0.51 = 109 ºC

7 0

3 years ago

What is the diagram of 2CO2

Fynjy0 [20]

Answer:

hope i helped!

pls mark brainliest if i am correct

7 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