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Explain, in terms of particles, what happens when methane boils and why

Ronch [10]

Answer:Butane > ethane > methane, because between bigger molecules there are stronger van der Waals forces and also higher molar mass means they need to be given more energy to have enough kinetic energy to move quickly, freely in gas.

There are multiple butene isomers (Butene) and some (2-Butenes - cis and trans) actually have higher boiling point than n-Butane (there is also Isobutane, of course, with quite much lower boiling point than all of them) and some (1-Butene, Isobutylene) have lower, so this isn't really a fair or simple question. But on simplest level, it can again be said that 1-butene has lower boiling point because it has very similar shape but slightly lower molar mass (2H less) than n-butane.

Explanation:

3 0

2 years ago

Convert the following Grams: 0.200 moles of H2S

ANEK [815]

Answer:

6.82 g H₂S

General Formulas and Concepts:

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Chemistry

Atomic Structure

Reading a Periodic Table

Stoichiometry

Using Dimensional Analysis

Explanation:

Step 1: Define

0.200 mol H₂S

Step 2: Identify Conversions

Molar Mass of H - 1.01 g/mol

Molar Mass of S - 32.07 g/mol

Molar Mass of H₂S - 2(1.01) + 32.07 = 34.09 g/mol

Step 3: Convert

Set up: $\displaystyle 0.200 \ mol \ H_2S(\frac{34.09 \ g \ H_2S}{1 \ mol \ H_2S})$
Multiply: $\displaystyle 6.818 \ g \ H_2S$

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

6.818 g H₂S ≈ 6.82 g H₂S

7 0

3 years ago

A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a

Morgarella [4.7K]

Answer : The percent abundance of the heaviest isotope is, 78 %

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = $\frac{x}{100}$

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = $\frac{10}{100}$

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] % = (90 - x) %

Fractional abundance of middle-weight isotope = $\frac{(90-x)}{100}$

Now put all the given values in above formula, we get:

$48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]$

$x=78\%$

Therefore, the percent abundance of the heaviest isotope is, 78 %

5 0

3 years ago

Read 2 more answers

A 100.0-g sample of water at 27.0oC is poured into a 71.0-g sample of water at 89.0oC. What will be the final temperature of the

Sedbober [7]

Answer: The final temperature will be $52.74^oC$

Explanation:

Calculating the heat released or absorbed for the process:

$q=m\times C\times (T_2-T_1)$

In a system, the total amount of heat released is equal to the total amount of heat absorbed.

$q_1=-q_2$

OR

$m_1\times C\times (T_f-T_1)=-m_2\times C\times (T_f-T_2)$ ......(1)

where,

C = heat capacity of water = $4.184J/g^oC$

$m_1$ = mass of water of sample 1 = 100.0 g

$m_2$ = mass of water of sample 2 = 71.0 g

$T_f$ = final temperature of the system = ?

$T_1$ = initial temperature of water of sample 1 = $27^oC$

$T_2$ = initial temperature of the water of sample 2 = $89.0^oC$

Putting values in equation 1, we get:

$100.0\times 4.184\times (T_f-27)=-71.0\times 4.184\times (T_f-89)\\\\171T_f=9019\\\\T_f=\frac{9019}{171}=52.74^oC$

Hence, the final temperature will be $52.74^oC$

6 0

3 years ago

Temperature (°C)

amm1812

Answer:

heating point for water - 100°

8 0

3 years ago