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Nata [24]
3 years ago
8

A unit of matter which cannot be broken down into other substances by ordinary chemical methods is:

Chemistry
1 answer:
Shkiper50 [21]3 years ago
6 0
a. ELEMENT: A substance<span> that </span>cannot<span> be decomposed (</span>broken down<span>) </span>into simpler substances by ordinary chemical<span> means.</span>
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There are two naturally occurring isotopes of indium: indium-113 and indium-115. How many neutrons are in a single atom of indiu
KonstantinChe [14]
There are 66 neutrons in a single atom of indium-115. The atomic number of indium-115 is 49, meaning there are 49 protons. Then the atomic mass is 115, so 115-49 = 66. 
5 0
3 years ago
Which compound is expected to exhibit hydrogen bonding forces?
Likurg_2 [28]
A hydrogen bond<span> is the electrostatic attraction between two polar groups that occurs when a </span>hydrogen<span> (H) atom covalently bound to a highly electronegative atom such as nitrogen (N), oxygen (O), or fluorine (F) experiences the electrostatic field of another highly electronegative atom nearby.  examples h20</span>
7 0
3 years ago
What is the cell potential for the reaction mg(s+fe2+(aq?mg2+(aq+fe(s at 77 ?c when [fe2+]= 3.40 m and [mg2+]= 0.210 m . express
Galina-37 [17]
First, you need to calculate the standard cell potential using standard reduction potential from a textbook or online. Since Mg becomes Mg+2, magnesium is being oxidized because it is losing electrons, you need to flip its potential

Fe+2 + 2e- --> Fe                  potential= -0.44
Mg+2 + 2e- --> Mg                potential= -2.37


Cell potential= (-0.44) + (+2.37)= 1.93 V

Now, you need to use Nernst formula to get the answer. I have attached a PDF with the work.
Download pdf
8 0
3 years ago
A disturbance in matter that carries energy from one place to another is called
siniylev [52]
The answer you're looking for is: a wave.
6 0
3 years ago
Cesium has a radius of 272 pm and crystallizes in a body-centered cubic structure. What is the edge length of the unit cell
olchik [2.2K]

Answer: Edge length of the unit cell = 628pm

Explanation: For a body centred cubic structured system, the relationship between the edge length of the unit cell and radius of the atoms in the structure is

Edge length of Unit cell (a) = (4R)/(√3)

R = 272pm = (272 × (10^-12))m = (2.72 × (10^-10))m

a = (4 × (2.72 × (10^-10)))/(√3)

a = (6.28157 × (10^-10))m = 628pm

4 0
3 years ago
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