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3 years ago

At 189k a sample of gas has a volume of 32.0cm. what does the gas occupy at 242k

Chemistry

1 answer:

sattari [20]3 years ago

5 0

<h2>Answer:</h2>

At 189 K volume = 32.0 cm

As volume of gas is directly proportional to temperature.

As the temperature increase the volume of gas will also increase.

Volume at 1 K = 32/189

Volume at 242 K = 32/189 * 242 = 40.97 cm.

Gas occupy 40.97 cm volume at 242 K.

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Give the empirical formula for C↓8H↓8

mr_godi [17]

Answer:

Empirical formula of C₈H₈ = CH

Explanation:

Data Given:

Molecular Formula = C₈H₈

Empirical Formula = ?

Solution

Empirical Formula:

Empirical formula is the simplest ration of atoms in the molecule but not all numbers of atoms in a compound.

So,

tha ration of the molecular formula should be divided by whole number to get the simplest ratio of molecule

C₈H₈ Consist of Carbon (C), and Hydrogen (H)

Now

Look at the ratio of these two atoms in the compound

C : H

8 : 8

Divide the ratio by two to get simplest ratio

C : H

8/8 : 8/8

1 : 1

So for the empirical formula is the simplest ratio of carbon to hydrogen 1 : 1

So the empirical formula will be

Empirical formula of C₈H₈ = CH

4 0

3 years ago

Help please for chemistry class

IRISSAK [1]

Answer:

option c is correct

Explanation:

the addition of catalyst does not effect the position or equilibrium constant and increase the forward and backward reaction in equal rates so no effect would be observed

3 0

3 years ago

CH4 + 202 → CO2 + 2H2O<br> How many grams of O2 needed to produce 36 grams of H2O?

Kipish [7]

<h3>Answer:</h3>

64 g O₂

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

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] CH₄ + 2O₂ → CO₂ + 2H₂O

[Given] 36 g H₂O

[Solve] x g O₂

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

[RxN] 2 mol O₂ → 2 mol H₂O

[PT] Molar Mass of O - 16.00 g/mol

[PT] Molar Mas of H - 1.01 g/mol

Molar Mass of O₂ - 2(16.00) = 32.00 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

<u>Step 3: Stoichiometry</u>

Set up conversion: $\displaystyle 36 \ g \ H_2O(\frac{1 \ mol \ H_2O}{18.02 \ g \ H_2O})(\frac{2 \ mol \ O_2}{2 \ mol \ H_2O})(\frac{32.00 \ g \ O_2}{1 \ mol \ O_2})$
Divide/Multiply [Cancel Units]: $\displaystyle 63.929 \ g \ O_2$