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lyudmila [28]
2 years ago
9

An electron ______ is an area around the nucleus of an atom where an electron is likely to be found.

Chemistry
1 answer:
jek_recluse [69]2 years ago
4 0

Answer:

orbital

hope this helps! <3

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Is colour a chemical or physical
dexar [7]

Answer:

Physical property

Explanation:

hope this helped!

6 0
3 years ago
A 32.14 gram sample of a hydrate of MnSO4 was heated thoroughly in a porcelain crucible, until its weight remained constant. Aft
amid [387]

Answer:

MnSO₄.7H₂O

Explanation:

To solve this question, we need to convert the mass of the dehydrated MnSO₄. The difference between mass of the hydrate and dehydrated compound is the mass of water. With the mass we can find the moles of water and the formula of the hydrate:

<em>Moles MnSO₄ -Molar mass: 151g/mol-:</em>

17.51g * (1mol / 151g) = 0.116 moles

<em>Moles H₂O -Molar mass: 18g/mol-:</em>

32.14g-17.51g = 14.63g * (1mol / 18g) = 0.813 moles

The ratio of moles MnSO₄: Moles H₂O represent the amount of water molecules in the hydrate:

0.813mol / 0.116mol = 7 molecules of water.

The hydrate formula is:

<h3>MnSO₄.7H₂O</h3>
8 0
2 years ago
What are some good movies to watch on HBOMAX
Marina CMI [18]

Answer:

My Neighbor Totoro

Princess Mononoke

Pom Poko

Only Yesterday

Kiki's Delivery Service

Nausicaä of the Valley of the Wind

7 0
2 years ago
Read 2 more answers
In an acid-base titration,
Lemur [1.5K]

Answer:

b

Explanation:

An acid-base titration is an experimental procedure used to determined the unknown concentration of an acid or base by precisely neutralizing it with an acid or base of known concentration. ... It is filled with a solution of strong acid (or base) of known concentration.

7 0
3 years ago
What is the density of lead (in g/cm^3 3 ) if a rectangular bar measuring 0.500 cm in height, 1.55 cm in width, and 25.00 cm in
Viktor [21]

Answer:

Density = 11.4 g/cm³

Explanation:

Given data:

Density of lead = ?

Height of lead bar = 0.500 cm

Width of lead bar = 1.55 cm

Length of lead bar = 25.00 cm

Mass of lead bar = 220.9 g

Solution:

Density = mass/ volume

Volume of bar = length × width × height

Volume of bar = 25.00 cm × 1.55 cm × 0.500 cm

Volume of bar = 19.4 cm³

Density of bar:

Density = 220.9 g/ 19.4 cm³

Density = 11.4 g/cm³

3 0
2 years ago
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