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Nana76 [90]
4 years ago
13

Which of the following compounds is a gas at room temperature? A : CH3CH2OH B : CH3CH2CH2CH2CH2CH2CH3 C : CH3CH2CH3 D : HOCH2CH2

OH
Chemistry
1 answer:
attashe74 [19]4 years ago
8 0

Answer:

C. CH_3CH_2CH_3

Explanation:

Molecules with the stronger intermolecular forces are pulled tightly together  to form solid at higher temperatures and that's why the freezing point is higher.

Also, molecules with the stronger intermolecular force have greater interaction between the molecules and thus on heating do not boil easily and have high boiling point also.

Thus, melting point and boiling point increases with increase in number of carbon atoms and also increase in intermolecular forces (like hydrogen bonding, if present).

Thus, the compound which is gas at room temperature is CH_3CH_2CH_3 <u>because it has least number of carbon atoms and absence of hydrogen bonding.</u>

<u></u>

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2.3 Zinc has five naturally occurring isotopes: 48.63% of 64 Zn with an atomic weight of 63.929 amu; 27.90% of 66Zn with an atom
lakkis [162]

<u>Answer:</u> The average atomic mass of element Zinc is 65.40 amu.  

<u>Explanation:</u>

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)

  • <u>For _{30}^{64}\textrm{Zn} isotope:</u>

Mass of _{30}^{64}\textrm{Zn} isotope = 63.929 amu

Percentage abundance of _{30}^{64}\textrm{Zn} isotope = 48.63 %

Fractional abundance of _{30}^{64}\textrm{Zn} isotope = 0.4863

  • <u>For _{30}^{66}\textrm{Zn} isotope:</u>

Mass of _{30}^{66}\textrm{Zn} isotope = 65.926 amu

Percentage abundance of _{30}^{66}\textrm{Zn} isotope = 27.90 %

Fractional abundance of _{30}^{66}\textrm{Zn} isotope = 0.2790

  • <u>For _{30}^{67}\textrm{Zn} isotope:</u>

Mass of _{30}^{67}\textrm{Zn} isotope = 66.927 amu

Percentage abundance of _{30}^{67}\textrm{Zn} isotope = 4.10 %

Fractional abundance of _{30}^{67}\textrm{Zn} isotope = 0.0410

  • <u>For _{30}^{68}\textrm{Zn} isotope:</u>

Mass of _{30}^{68}\textrm{Zn} isotope = 67.925 amu

Percentage abundance of _{30}^{68}\textrm{Zn} isotope = 18.75 %

Fractional abundance of _{30}^{68}\textrm{Zn} isotope = 0.1875

  • <u>For _{30}^{70}\textrm{Zn} isotope:</u>

Mass of _{30}^{70}\textrm{Zn} isotope = 69.925 amu

Percentage abundance of _{30}^{70}\textrm{Zn} isotope = 0.62 %

Fractional abundance of _{30}^{70}\textrm{Zn} isotope = 0.0062

Putting values in equation 1, we get:

\text{Average atomic mass of Zinc}=[(63.929\times 0.4863)+(65.926\times 0.2790)+(66.927\times 0.0410)+(67.925\times 0.1875)+(69.925\times 0.0062)]

\text{Average atomic mass of Zinc}=65.40amu

Hence, the average atomic mass of element Zinc is 65.40 amu.

7 0
3 years ago
The empirical formula of a compound is CH3- The molecular formula of this compound could be C2H6 CH4 C3H6 C2H4
Liono4ka [1.6K]

Answer:

C2H6

Explanation:

7 0
3 years ago
in 1990, the men's singles winner of the us. open tennis tournament has his serves clocked at 127 mi/hr. How fast must a 56.6 g
Naddika [18.5K]

This problem is providing information about the mass of a tennis ball, 56.6 g (0.0566 kg) and asks for the velocity it will have to equal the wavelength of green light, which is 5400 A or 540 nm (5.4x10⁻⁷ m). Thus, after doing the math, the result is 2.17x10⁻²⁶ m/s.

<h3>Broglie's wavelength:</h3>

In this case, we recall the formula of the Broglie's wavelength as shown below:

\lambda =\frac{h}{mv}

Whereas lambda is the wavelength, h is the Planck's constant, m the mass and v the speed; thus, we solve for the speed according to the question:

v =\frac{h}{m\lambda}

<h3>Calculations:</h3>

Then, we just plug in the numbers we were given to get the answer:

v =\frac{6.626x10^{-34} kg*\frac{m^2}{s} }{0.056kg*5.4x10^{-7}m}\\\\v=2.17x10^{-26}m/s

Learn more about Broglie's wavelength: brainly.com/question/5440536

3 0
2 years ago
Calculate the solubility of copper(II) hydroxide, Cu(OH)2, in g/L​
Oduvanchick [21]

Answer:

Ksp = [ Cu+² ] [ OH-] ²

molar mass Cu(oH )2 ==> M= 63.546 (1) + 16 (2) + 1 (2) = 97.546 g/mol

Ksp = [ Cu+² ] [ OH-] ²

Ksp [ cu (OH)2 ] = 2.2 × 10-²⁰

|__________|___<u>Cu</u><u>+</u><u>²</u><u> </u>__|_<u>2</u><u>OH</u><u>-</u>____|

|<u>Initial concentration(M</u>)|___<u>0</u>__|_<u>0</u>______|

<u>|Change in concentration(M)</u>|_<u>+S</u><u> </u>|__<u>+2S</u>__|

|<u>Equilibrium concentration(M)|</u><u>_S</u><u> </u><u>_</u><u>|</u><u>2S___</u><u>|</u>

Ksp = [ Cu+² ] [ OH-] ²

2.2 ×10-²⁰ = (S)(2S)²= 4S³

s =  \sqrt[3]{ \frac{2.2 \times  {10}^{ - 20} }{4} }  = 1.8 \times  {10}^{ - 7}

S = 1.8 × 10-⁷ M

The molar solubility of Cu(OH)2 is 1.8 × 10-⁷ M

Solubility of Cu (OH)2 =

Cu (OH)2 =  \frac{1.8 \times  {10}^{ - 7} mol \:Cu (OH)2 }{1L}  \times  \frac{97.546 \: g \: Cu (OH)2}{1 \: mol \: Cu (OH)2}  \\  = 1.75428 \times 10 ^{ - 5}

<h3>Solubility of Cu (OH)2 = 1.75428 × 10 -⁵ g/ L</h3>

I hope I helped you^_^

8 0
3 years ago
I need help please ASAP
nalin [4]

Answer:

A

Explanation:

because CO2 is carbon dioxide and CO is carbon monoxide. mono meaning one which in this case is monoxide.

6 0
3 years ago
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