Answer:
0.07172 L = 7.172 mL.
Explanation:
- We can use the general law of ideal gas: <em>PV = nRT.
</em>
where, P is the pressure of the gas in atm (P = 1.0 atm, Standard P).
V is the volume of the gas in L (V = ??? L).
n is the no. of moles of the gas in mol (n = 3.2 x 10⁻³ mol).
R is the general gas constant (R = 0.0821 L.atm/mol.K),
T is the temperature of the gas in K (T = 273 K, Standard T).
<em>∴ V = nRT/P =</em> (3.2 x 10⁻³ mol)(0.0821 L.atm/mol.K)(273 K)/(1.0 atm) = <em>0.07172 L = 7.172 mL.</em>
Answer:
1.
Since both components of these solutions have the same molar mass, mole fractions would be the same as mass fractions.
0.110 atm = (2/3)(Pi) + (1/3)(Pn) [1]
0.089 atm = (1/3)(Pi) + (2/3)(Pn) [2]
2*[1] - [2]:
(2)(0.110) - 0.089 atm = Pi
Pi = 0.131 atm
2*[2] - [1]:
(2)(0.089) - 0.110 atm = Pn
Pn = 0.068 atm
2.
The hydroxyl (-OH) group on the end of a longer 1-propanol molecule makes it more polar than IPA. It follows that the intermolecular forces between 1-propanol are stronger than those of IPA and thus the vapor pressure of 1-propanol should be lower than IPA.
Explanation:
Answer:
Oxygen.
Explanation:
They take in oxygen from the air. This is the process of respiration.
Explanation:
Given parameters:
Number of molecules = 4.21 x 10²³ molecules
Unknown parameters:
Number of moles
Solution:
A mole can be defined as the amount of a substance that contains the avogadro's number of particles i.e 6.02 x 10²³
To find the number of moles:
Number of moles = ![\frac{number of molecules }{[tex]6.02 x 10^{23}](https://tex.z-dn.net/?f=%5Cfrac%7Bnumber%20of%20molecules%20%7D%7B%5Btex%5D6.02%20x%2010%5E%7B23%7D)
}[/tex]
Number of moles = ![\frac{4.21 x 10[tex]^{23}](https://tex.z-dn.net/?f=%5Cfrac%7B4.21%20x%2010%5Btex%5D%5E%7B23%7D)
}{
}[/tex]
Number of moles of CaCl₂ = 0.699moles
Learn more;
mole calculation brainly.com/question/13064292
#learnwithBrainly
<span>We know that protons gives positive charge, neutrons no charge and electrons negative, then +14 - 12 = +2
So the charge of the atom is sign positive and magnitude 2</span>
