Answer:
12.8 g of 
must be withdrawn from tank
Explanation:
Let's assume gas inside tank behaves ideally.
According to ideal gas equation- 
where P is pressure of , V is volume of , n is number of moles of , R is gas constant and T is temperature in kelvin scale.
We can also write, 
Here V, T and R are constants.
So, 
ratio will also be constant before and after removal of from tank
Hence, 
Here, 
and 
So, 
So, moles of must be withdrawn = (0.66 - 0.26) mol = 0.40 mol
Molar mass of = 32 g/mol
So, mass of must be withdrawn = 
"Compound" is the one among the following choices given in the question that is the <span>most specific classification of Ca(NO3)2. The correct option among all the options that are given in the question is the second option. It is actually an inorganic compound. I hope that this is the answer that has come to your help.</span>
Answer:
Dispersion forces.
Explanation:
CO2 contains dispersion forces, and covalent bonds. It is a linear molecule, and the bond angle of O-C-O is 180 degree. O is more electronegative than C, the C-O contains polar bond with the having negative end pointing towards the O.
CO contains two C-O bonds. They cancel each other out because of the dipoles point in opposite directions. Although, CO2 contains polar bonds, it is known as a nonpolar molecule. So, the only intramolecular forces which CO2 having are London dispersion forces.
Answers are:
2. It pushes on all objects that are on Earth’s surface.
3. It can be measured in atmospheres or kilopascals.
Barometric pressure (atmospheric pressure), is the pressure within the atmosphere of Earth
Atmospheric pressure decreases with increasing height, because there are fewer air molecules above a given object.
Barometer is an instrument used in meteorology to measure atmospheric pressure.
Atmospheric pressure (atm) is the force per unit area by the weight of air above that point.
Kilopascal (kPa) is a metric system pressure unit and equals to 1000 force of newton per square meter.
Atmospheric pressure results from molecular collisions of atmospheric gases.
Answer:
K = Ka/Kb
Explanation:
P(s) + (3/2) Cl₂(g) <-------> PCl₃(g) K = ?
P(s) + (5/2) Cl₂(g) <--------> PCl₅(g) Ka
PCl₃(g) + Cl₂(g) <---------> PCl₅(g) Kb
K = [PCl₃]/ ([P] [Cl₂]⁽³'²⁾)
Ka = [PCl₅]/ ([P] [Cl₂]⁽⁵'²⁾)
Kb = [PCl₅]/ ([PCl₃] [Cl₂])
Since [PCl₅] = [PCl₅]
From the Ka equation,
[PCl₅] = Ka ([P] [Cl₂]⁽⁵'²⁾)
From the Kb equation
[PCl₅] = Kb ([PCl₃] [Cl₂])
Equating them
Ka ([P] [Cl₂]⁽⁵'²⁾) = Kb ([PCl₃] [Cl₂])
(Ka/Kb) = ([PCl₃] [Cl₂]) / ([P] [Cl₂]⁽⁵'²⁾)
(Ka/Kb) = [PCl₃] / ([P] [Cl₂]⁽³'²⁾)
Comparing this with the equation for the overall equilibrium constant
K = Ka/Kb
