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3 years ago

How many moles of methane are in 7.31*10^25 molecules?

Chemistry

1 answer:

GrogVix [38]3 years ago

7 0

<h3>Answer:</h3>

121 mol CH₄

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

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Organic</u>

Writing chemical compounds
Writing organic structures
Prefixes
Alkanes, Alkenes, Alkynes

<u>Atomic Structure</u>

Using Dimensional Analysis
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<h3>Explanation:</h3>

<u>Step 1: Define</u>

7.31 × 10²⁵ molecules CH₄

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

Avogadro's Number

<u>Step 3: Convert</u>

<u /> $\displaystyle 7.31 \cdot 10^{25} \ molecules \ CH_4(\frac{1 \ mol \ CH_4}{6.022 \cdot 10^{23} \ molecules \ CH_4} ) = 121.388 \ mol \ CH_4$ <u />

<u />

<u>Step 4: Check</u>

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

121.388 mol CH₄ ≈ 121 mol CH₄

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4 years ago

A reaction mixture initially contains 0.140 MCO and 0.140 MH2O. What will be the equilibrium concentration of CO?

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Answer:

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Explanation:

From the question, we can make the following deductions; we are given mixture that contains two compounds, that is A and B, 0.140 M CO and 0.140 M H2O respectively. Then, we are asked to find the equilibrium concentration of Carbonmonoxide,CO.

So, the equation for the reaction is given below;

CO + H2O <-----------------> CO2 + H2.

Initially: we have 0.14M of CO, 0.14M of H2O and zero (0) concentration of CO2 and H2.

At time,t = CO =0.14 - x , H2O = 0.14 - x, CO2 and H2 = x.

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==> 10.1= x/0.14 - x.

====> 0.141 - 10.1 x = x.

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