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Serjik [45]
2 years ago
6

Suppose that you add 26.7 g of an unknown molecular compound to 0.250 kg of benzene, which has a K f Kf of 5.12 oC/m. With the a

dded solute, you find that there is a freezing point depression of 3.89 oC compared to pure benzene. What is the molar mass (in g/mol) of the unknown compound
Chemistry
1 answer:
kiruha [24]2 years ago
5 0

From the calculation, the molar mass of the solution is 141 g/mol.

<h3>What is the molar mass?</h3>

We know that;

ΔT = K m i

K = the freezing constant

m = molality of the solution

i = the Van't Hoft factor

The molality of the solution is obtained from;

m = ΔT/K i

m = 3.89/5.12 * 1

m = 0.76 m

Now;

0.76 =  26.7 /MM/0.250

0.76 = 26.7 /0.250MM

0.76 * 0.250MM =  26.7

MM= 26.7/0.76 * 0.250

MM = 141 g/mol

Learn more about molar mass:brainly.com/question/12127540?

#SPJ12

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In which direction does the electric field point at a position directly east of positive charge.
butalik [34]

Answer:

The direction of the field is taken to be the direction of the force it would exert on a positive test charge. The electric field is radially outward from a positive charge and radially in toward a negative point charge.

Explanation:

4 0
3 years ago
El pH de una solución acuosa 10^-14 M de ácido acético, a 25°C, es igual a, teniendo en cuenta que Ke = 1.76x10^-5
frez [133]

Answer:

pH=14

Explanation:

Hola!

En este caso, consideramos que la disociación de ácido acético ocurre:

CH_3COOH\rightarrow CH_3COO^-+H^+

Así, mediante la solución del equilibrio ácido, podemos calcular la concentración de iones hidronio que posteriormente sirven para calcular el pH de la solución, por tal razón, debemos calcular el equilibrio dada la constante de equilibrio y por medio de la ley de acción de masas en términos del cambio x como cualquier problema de equilibrio:

Ke=\frac{CH_3COO^-][H^+]}{[CH_3COOH]}\\\\1.76x10^{-5}=\frac{x*x}{1x10^{-14}M-x}

Resolviendo para x, tenemos x=0.999x10^{-14}

Así, la concentración de hidrógeno es igual a x, por lo que el pH:

pH=-log([H^+])=-log(0.999x10^{-14})\\\\pH=14

Dicho valor tiene sentido desde que la concentración de hidrógeno es casi despreciable, por lo que se puede asumir que tiende a ser básica.

Saludos!

5 0
3 years ago
Read 2 more answers
An Erlenmeyer flask contains 25.00 mL of 0.500 M HCl before a titration is begun. How many moles of hydrogen ions are present in
Sliva [168]

Answer:

0.0125 mol.

Explanation:

<em>Molarity of a solution is the no. of moles of a solute per 1.0 L of the solution.</em>

<em></em>

M of HCl = 0.5 mol/L.

V = 25.0 mL = 0.025 L.

<em>∴ no. of moles of HCl present in the flask at this time = MV</em> = (0.5 mol/L)(0.025 L) = <em>0.0125 mol.</em>

6 0
3 years ago
A 3.28 L solution is prepared by dissolving 535 g CaCl2 in water. The molar mass of CaCl2 is 110.98 g. What is the morality of t
andrey2020 [161]

Answer:

              1.47 mol/L

Explanation:

Molarity is given as,

                           Molarity = Moles / Vol in L    ------- (1)

Moles of CaCl₂,

                           Moles = Mass / M.Mass

                           Moles = 535 g / 110.98 g/mol

                           Moles = 4.82 mol

Now, putting values in eq. 1.

                           Molarity = 4.82 mol / 3.28 L

                           Molarity = 1.47 mol/L

5 0
2 years ago
The pKb value for aqueous ammonia at 25
Alex17521 [72]

The value of pka for NH_3 in an aqueous solution is 9.2.

<h3>What is Kb?</h3>

Kb denotes the base dissociation constant.

pKa + pKb =14  at 25 degree celcius.

pKa + 4.8 =14

pKa = 9.2

pKb is the negative base-10 logarithm of the base dissociation constant (Kb) of a solution.

It is used to determine the strength of a base or alkaline solution.

The value of pka for NH_3 in an aqueous solution is 9.2.

Learn more about pka here:

brainly.com/question/14124805

#SPJ1

8 0
1 year ago
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