Answer:
d. 103.3
Explanation:
In the given question, the National Weather Service routinely supplies atmospheric pressure data to help pilots set their altimeters. And the units of atmospheric pressure used for reporting the atmospheric pressure data are inches of mercury. For a barometric pressure of 30.51 inches of mercury, we can calculate the pressure in kPa as follow:
In principle, 3.386 kPa is equivalent to the atmospheric pressure of 1 inch of mercury. Thus, 30.51 inches of mercury is equivalent to 30.51 in *(3.386 kPa/1 in) = 103.307 kPa.
Therefore, a barometric pressure of 30.51 inches of mercury corresponds to _____103.3_____ kPa.
Answer:
Explanation:
Hello there!
In this case, sine the solution of this problem require the application of the Raoult's law, assuming heptane is a nonvolatile solute, so we can write:
Thus, we first calculate the mole fraction of chloroform, by using the given masses and molar masses as shown below:
Therefore, the partial pressure of chloroform turns out to be:
Regards!
Answer:
B. They both contain three atoms around the central atom.
Explanation:
Do the unit test on edg20 and got it right!
Answer:
A. Intramolecular interactions are generally stronger.
B. a. Only intermolecular interactions are broken when a liquid is converted to a gas.
Explanation:
<em>A. Which is generally stronger, intermolecular interactions or intramolecular interactions?</em>
Intramolecular interactions, in which electrons are gained, lost or shared, constitute true bonds and are one or two orders of magnitude stronger than intermolecular interactions.
<em>B. Which of these kinds of interactions are broken when a liquid is converted to a gas?</em>
When a liquid vaporizes, the intermolecular attractions are broken, that is, molecules get more separated. However, true bonds are not broken which is why the molecules keep their chemical identity.
Answer:
see calculations in explanation
Explanation:
percent = part/total x 100%
part = ∑ atomic mass of element
- hydrogen = 1.008 amu (atomic mass units)
- carbon = 12.011 amu
- nitrogen = 14.007 amu
total = ∑ molecular mass of compound
= H amu + C amu + Namu
= 1.008 amu + 12.011 amu + 14.007 amu
= 27.026 amu
%H = (1.008amu/27.026amu)100% = 3.730%
%C = (12.011amu/27.026amu)100% = 44.442%
%N = (14.007amu/27.026amu)100% = 51.827%
Check results ∑%values = 100%
3.730% + 44.442% + 51.827% = 99.999% ≅ 100%
