Answer:
680 g/m is the molar mass for the unknown, non electrolyte, compound.
Explanation:
Let's apply the formula for osmotic pressure
π = Molarity . R . T
T = T° absolute (in K)
R = Universal constant gases
π = Pressure
Molarity = mol/L
As units of R are L.atm/mol.K, we have to convert the mmHg to atm
760 mmHg is 1 atm
28.1 mmHg is (28.1 .1)/760 = 0.0369 atm
0.0369 atm = M . 0.082 L.atm/mol.K . 293K
(0.0369 atm / 0.082 mol.K/L.atm . 293K) = M
0.0015 mol/L = Molarity
This data means the mol of solute in 1L, but we have 100mL so
Molarity . volume = mol
0.0015 mol/L . 0.1L = 1.5x10⁻⁴ mole
The molar mass will be: 0.102g / 1.5x10⁻⁴ m = 680 g/m
Answer:
The correct answer is: 1.035 x 10⁻³ M
Explanation:
The dissociation equilibrium for acetic acid (CH₃COOH) is the following:
CH₃COOH(aq) ↔ CH₃COO⁻(aq) + H⁺(aq) Kc = 1.8 x 10⁻⁵
The expression for the equilibrium constant (Kc) is the ratio of concentrations of products over reactants. The products are acetate ion (CH₃COO⁻) and hydrogen ion (H⁺) while the reactant is acetic acid (CH₃COOH):
![Kc=\frac{[CH_{3} COO^{-} ][H^{+} ]}{[CH_{3} COOH]}= 1.8 x 10^{-5}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BCH_%7B3%7D%20COO%5E%7B-%7D%20%5D%5BH%5E%7B%2B%7D%20%5D%7D%7B%5BCH_%7B3%7D%20COOH%5D%7D%3D%201.8%20x%2010%5E%7B-5%7D)
Given: [CH₃COOH]= 0.016 M and [CH₃COO⁻]= 0.92 M, we replace the concentrations in the equilibrium expression and we calculate [H⁺]:
![\frac{(0.016 M)[H^{+} ]}{(0.92M)}= 1.8 x 10^{-5}](https://tex.z-dn.net/?f=%5Cfrac%7B%280.016%20M%29%5BH%5E%7B%2B%7D%20%5D%7D%7B%280.92M%29%7D%3D%201.8%20x%2010%5E%7B-5%7D)
⇒[H⁺]= (1.8 x 10⁻⁵)(0.92 M)/(0.016 M)= 1.035 x 10⁻³ M
Hello!
<span>We have the following statement data:
</span>
Data:




<span>As the percentage is the mole fraction multiplied by 100:
</span>

<span>The mole fraction will be the percentage divided by 100, thus:
</span><span>What is the partial pressure of oxygen in this mixture?
</span>



<span>To calculate the partial pressure of the oxygen gas, it is enough to use the formula that involves the pressures (total and partial) and the fraction in quantity of matter:
</span>
In relation to

:




<span>
Answer:
</span><span>
b. 320.0 mm hg </span>
Answer:
Non-zero digits are always significant.
Any zeros between two significant digits are significant.
A final zero or trailing zeros in the decimal portion ONLY are significant. If a number ends in zeros to the right of the decimal point, those zeros are significant.
Explanation:
1.138 has 4 significant figures, which are 1, 1, 3 and 8. The numbers after the decimal point are decimals and are significant figures.