C. Which molecule is not linear? a H₂O b. CO₂ c. N₂ d. Cl₂

Which of the following would remain a liquid longer as temperature continues to drop?

viktelen [127]

I just took a test with this question and got the answer wrong for saying ethane. The correct answer is propane.

3 0

2 years ago

Assuming the relative rate of secondary hydrogen atom abstraction for the chlorination of propane is 3.9 times faster than the r

Ghella [55]

Answer:

% of n-propyl chloride = 43.48 %

Explanation:

There are 2 secondary hydrogens and 6 primary hydrogens

The rate of abstraction of seondary hydrogen = 3.9 X rate of abstraction of primary hydrogen

probability of formation of isopropyl chloride = 3.9 X 1 (relative rate X relative number of secondary hydrogens)

Probability of formation of n-propyl chloride = 1 X 3 (relative rate X relative number of primary hydrogens)

Total probability = 3.9

% of n-propyl chloride = 3 X 100 / 6.9 = 43.48 %

7 0

3 years ago

Find the number of grams in 16.95 mol hydrogen peroxide (H2O2). Round your

Feliz [49]

Answer: There are 576.46 number of grams present in 16.95 mol hydrogen peroxide $(H_{2}O_{2})$ .

Explanation:

Number of moles is defined as the mass of substance divided by its molar mass.

The molar mass of $H_{2}O_{2}$ is 34.01 g/mol. Hence, mass of hydrogen peroxide present in 16.95 moles is calculated as follows.

$Moles = \frac{mass}{molarmass}\\16.95 mol = \frac{mass}{34.01 g/mol}\\mass = 576.46 g$

Thus, we can conclude that there are 576.46 number of grams present in 16.95 mol hydrogen peroxide $(H_{2}O_{2})$ .

3 0

2 years ago

Chlorine has two naturally occurring isotopes, 35C1 Gsotopic mass 34.9689 amu) and 370 (isotopic mass 36.9659 amu). If chlorine

Lilit [14]

Answer: The percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 75.77% and 24.23% respectively.

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:

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

Let the fractional abundance of $_{17}^{35}\textrm{Cl}$ isotope be 'x'. So, fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope will be '1 - x'

For $_{17}^{35}\textrm{Cl}$ isotope:

Mass of $_{17}^{35}\textrm{Cl}$ isotope = 34.9689 amu

Fractional abundance of $_{17}^{35}\textrm{Cl}$ isotope = x

For $_{17}^{37}\textrm{Cl}$ isotope:

Mass of $_{17}^{37}\textrm{Cl}$ isotope = 36.9659 amu

Fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope = 1 - x

Average atomic mass of chlorine = 35.4527 amu

Putting values in equation 1, we get:

$35.4527=[(34.9689\times x)+(36.9659\times (1-x))]\\\\x=0.7577$

Percentage abundance of $_{17}^{35}\textrm{Cl}$ isotope = $0.7577\times 100=75.77\%$

Percentage abundance of $_{17}^{37}\textrm{Cl}$ isotope = $(1-0.7577)=0.2423\times 100=24.23\%$

Hence, the percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 75.77% and 24.23% respectively.

6 0

2 years ago

What is the molarity of a solution that contains 9.63 grams of hcl in 1.5 liters of solution

Aleonysh [2.5K]

Molar mass HCl = 36.5 g/mol

number of moles:

mass of solute / molar mass:

9.63 / 36.5 => 0.263 moles

Volume = 1.5 L

M = n / V

M = 0.263 / 1.5

M = 0.1753 mol/L

espero ter ajudado!

8 0

2 years ago

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