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zloy xaker [14]
3 years ago
11

In the following reaction, which element is oxidized and which is reduced?

Chemistry
1 answer:
bazaltina [42]3 years ago
3 0

Answer:

The answer to your question is:    Iron oxidizes and Copper reduces

Explanation:

An element oxidizes when it loses electrons

An element reduces when it gains electrons

Then

             Fe    ⇒    Fe⁺²       Now, is more positive, it loses electrons

            Cu⁺²  ⇒    Cu         Now, is more negative, it gains electrons

         Iron oxidizes and Copper reduces

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Balance the combustion reaction between butane and oxygen. 2C4H10 + O2 → CO2 + H2O
dimaraw [331]
<h2>Answer:</h2><h3>Balanced equation:</h3>

2C4H10 + 8O2 → 8CO2 + 10H2O

<h3>Description:</h3>

There should be 8 molecules of oxygen which react with 2 molecules of butane to produce 8 molecules of carbon dioxide and 10 molecules of water.


6 0
3 years ago
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A topographic map would be most useful for which activity?
Ainat [17]


the answer is D. because a topographic map measures the lowest and highest points in a certain place

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3 years ago
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reativity of alkali metals increases down the group while reativity decreases down the group in helogens ​
BigorU [14]

Complete question is;

Chemical reactivity of alkali metals increases down the group while reactivity of halogens decreases down the group. Give reasons

Answer:

Explained below

Explanation:

Alkali metals exhibit reactivity due to their electropositivity. Now, for alkalis, their electro-positivity increases down their group. Since their reactivity increases with increase in electropositivity, it means their reactivity also increases down the group.

Whereas, the reactivity of halogens occurs as a result of their electronegativity. Now, electronegativity for halogens decreases down the group. Since their reactivity decreases with decrease in electronegativity, it means that their reactivity will also decrease down the group.

4 0
3 years ago
How many moles are in 1.51x10^26 atoms of xenon (Xe)? Please and thank you :)!!
RoseWind [281]
<h3>Answer:</h3>

251 mol Xe

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 1.51 × 10²⁶ atoms Xe

[Solve] moles Xe

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                     \displaystyle 1.51 \cdot 10^{26} \ atoms \ Xe(\frac{1 \ mol \ Xe}{6.022 \cdot 10^{23} \ atoms \ Xe})
  2. [DA] Multiply/Divide [Cancel out units]:                                                         \displaystyle 250.747 \ mol \ Xe

<u>Step 4: Check</u>

<em>Follow sig fig rule and round. We are given 3 sig figs.</em>

250.747 mol Xe ≈ 251 mol Xe

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