Answer:
Attraction between molecules of methane in liquid state is primarily due to "London dispersion force".
Explanation:
Methane is a non-polar and aprotic molecule. Hence there is no dipole moment in methane as well as no chance of hydrogen bonding formation by methane.
We know that all molecules contain electrons. Therefore transient dipole arises in every molecule due to revolution of electrons around nucleus in a non-circular orbit. Hence an weak intermolecular attraction force is always present in every molecule as a result of this which is termed as "London dispersion force".
So, attraction between molecules of methane in liquid state is primarily due to "London dispersion force".
Answer: (not sure)
T - A
T - A
T - A
A - T
C - G
G - C
C - G
C - G
A - T
im not sure about the amoni acid produced but
AAA produces -> Lys.
ACG produces -> Thr.
and i don't know about the last one
Answer:
C₂H₄O
Explanation:
In a compound that contains Cabon, hydrogen and oxygen, the combustion produce CO₂ from the carbon, and H₂O from the hydrogens. Using the mass of the products we can solve the moles of Carbon and hydrogen. The empirical formula is the simplest whole-number of atoms present in a molecule.
<em>Moles CO₂ = Moles C:</em>
11.8g CO₂ * (1mol / 44g) = 0.268 moles CO₂ = 0.268 moles C * (12g/mol) =
3.216g C
<em>Moles H₂O = 1/2 moles H:</em>
4.83g H₂O * (1mol / 18g) = 0.268 moles H₂O * (2 mol H / 1 mol H₂O) =
0.537 mol H * (1g/mol) = 0.537g H
<em>Mass O to find moles O:</em>
5.90g Sample - 3.216g C - 0.537g H = 2.147g O * (1mol / 16g) = 0.134 moles O
<em>Ratio of atoms -Dividing in 0.134 moles-:</em>
C = 0.268mol C / 0.134 mol O = 2
H = 0.537mol H / 0.134 mol O = 4
O = 0.134mol O / 0.134 mol O = 1
Empirical formula is:
<h3>C₂H₄O</h3>
Answer:
To obtain the grams of fat that the ground round has, knowing that it weighs 1.33 pounds we must pass this value to grams. Since 1 pound equals 453.59 grams, 1.33 pounds equals 603.27 (453.59 x 1.33).
Now, to obtain 29 percent of 603.27, we must make the following calculation: 603.27 / 100 x 29, which gives a total of 174.94 grams.
In this way, your reasoning is correct and it is probably a mistake in the book.
Data Given:
Moles = n = 3.91 mol
Pressure = P = 5.35 atm
Temperature = T = 323 K
Volume = V = ?
Formula used: Ideal Gas Equation is used,
P V = n R T
Solving for V,
V = n R T / P
Putting Values,
V = (3.91 mol × 0.0825 atm.L.mol⁻¹.K⁻¹ × 323 K) ÷ 5.35 atm
V = 19.36 L
