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RideAnS [48]
3 years ago
13

Please Help! 30 PTS. for correct answer and best explanation!

Chemistry
2 answers:
MaRussiya [10]3 years ago
8 0

Answer:

\huge \boxed{\mathrm{CH_4+ 2O_2 \Rightarrow CO_2 +2 H_2O}}

\rule[225]{225}{2}

Explanation:

\sf CH_4+ O_2 \Rightarrow CO_2 + H_2O

Balancing the Hydrogen atoms on the right side,

\sf CH_4+ O_2 \Rightarrow CO_2 +2 H_2O

Balancing the Oxygen atoms on the left side,

\sf CH_4+ 2O_2 \Rightarrow CO_2 +2 H_2O

In a balanced equation, the product side and the reactant side must have the same mass of matter, since matter cannot be destroyed or created according to the law.

\rule[225]{225}{2}

irakobra [83]3 years ago
6 0

Look kid, this question has been up for 17 hours think its time to let it go......... hoped i helped. ✔verified✔

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Place the following elements in order of increasing electronegativity.K Cs PA. K &lt; P &lt; CsB. P &lt; Cs &lt; KC. Cs &lt; K &
skad [1K]

Answer:

C. Cs < K < P

Explanation:

Electronegativity is the basically the tendency of an atom to attract electrons. The pauling scale is used to quantify electronegativity, the most electronegative element; fluorine has a value of 4.0 and francium, the least electronegative element has a value of 0.7

The electronegative values of the following elements are given as;

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Arranging in order of increasing electronegativity, we have;

Cs < K < P. The correct option is;

C. Cs < K < P

5 0
3 years ago
At standard temperature and pressure. 0.500 mole of xenon gas occupies
ANEK [815]

Answer:

0.500 mole of Xe (g) occupies 11.2 L at STP.

General Formulas and Concepts:

<u>Gas Laws</u>

  • STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

<u>Stoichiometry</u>

  • Mole ratio
  • Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

<em>Identify.</em>

0.500 mole Xe (g)

<u>Step 2: Convert</u>

  1. [DA] Set up:                                                                                                  \displaystyle 0.500 \ \text{mole Xe} \bigg( \frac{22.4 \ \text{L Xe}}{1 \ \text{mole Xe}} \bigg)
  2. [DA] Evaluate:                                                                                               \displaystyle 0.500 \ \text{mole Xe} \bigg( \frac{22.4 \ \text{L Xe}}{1 \ \text{mole Xe}} \bigg) = 11.2 \ \text{L Xe}

Topic: AP Chemistry

Unit: Stoichiometry

3 0
2 years ago
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