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Pani-rosa [81]
3 years ago
13

A species whose oxidation number increases in a reaction is

Chemistry
1 answer:
Marta_Voda [28]3 years ago
7 0
A species whose oxidation number increases in a reaction is oxidized (reducing reagents or reactant).
For example, balanced chemical reaction: 2KClO₃ → 2KCl + 3O₂<span>.
</span><span>Oxygen in the reactant state has oxidation number -2 (+1 + 5 + 3 · x = 0) and it is oxidized to 0 (in molecule of oxygen) in the product state.</span>
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Oxana [17]

Answer:

It should be 1, a.

Explanation:

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3 years ago
Only the _____ word in an organism's complete scientific name has its first letter capitalized
dimaraw [331]
Only the first word in an organism's complete scientific name has its first letter capitalized.

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What metal will have properties most similar to those of chromium (Cr)? Why?
garri49 [273]
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5 0
3 years ago
Which reaction results in only a physical change?
Advocard [28]
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4 0
3 years ago
Low concentrations of EDTA near the detection limit gave the following dimensionless instrument readings: 175, 104, 164, 193, 13
Mila [183]

Answer:

Following are the solution to these question:

Explanation:

Calculating the mean:

\bar{x}=\frac{175+104+164+193+131+189+155+133+151+176}{10}\\\\

  =\frac{1571}{10}\\\\=157.1

Calculating the standardn:

\sigma=\sqrt{\frac{\Sigma(x_i-\bar{x})^2}{n-1}}\\\\

Please find the correct equation in the attached file.

=28.195

For point a:

=3s+yblank \\\\=3 \times 28.195+50\\\\=84.585+50\\\\=134.585\\

For point b:

=3 \ \frac{s}{m}\\\\ = \frac{(3 \times 28.195)}{1.75 \times 10^9 \ M^{-1}}\\\\= 4.833 \times 10^{-8} \ M

For point c:

= 10 \frac{s}{m} \\\\= \frac{(10 \times 28.195)}{1.75 x 10^9 \ M^{-1}}\\\\ = 1.611 \times 10^{-7}\  M

It is calculated by using the slope value that is 1.75 \times 10^9 M^{-1}. The slope value 1.75 \times 10^9 M^{-1}is ambiguous.

7 0
2 years ago
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