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Allushta [10]
2 years ago
12

how much potassium bromide will form a saturated solution with 100 grams of water at 40 degrees celsius

Chemistry
1 answer:
Citrus2011 [14]2 years ago
4 0

Answer:

75.4 g.

Explanation:

Hello there!

In this case, since the solubility stands for the maximum amount of a solute that is completely dissolved in a specific amount of solvent, typically 100 g of water as the universal solvent; we need to take into account that the higher the temperature the higher the solubility for the most of the solids.

In such a way, since the solubility of potassium bromide at 20 °C is 65.3 g per 100 g of water, at 40 °C is 75.4 g per 100 g of water, proving the aforesaid. Therefore, 75.4 g of potassium bromide will form a saturated solution with 100 g of water at 40 °C.

Regards!

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Answer:

-245.7°C es la temperatura del gas bajo 100 torr

Explanation:

Para resolver esta pregunta debemos hacer uso de la ley de Boyle que establece que la presión de un gas es directamente proporcional a la temperatura de este cuando el volumen permanece constante. La ecuación es:

P1T2 = P2T1

<em>Donde P es presión y T temperatura absoluta del estado inicial, 1, y final, 2.</em>

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Reemplazando:

P1 = 1000torr

T2 = ? -Incógnita-

P2 = 100torr

T1 = 273K -Temperatura del hielo fundido = 0°C = 273K

1000torrT2 = 100torr273

T2 = 27.3K

27.3K - 273 =

<h3>-245.7°C es la temperatura del gas bajo 100 torr</h3>

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2 years ago
I. Three different representations of glucose are<br> given below. Identify each type<br> of model.
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Some SbCl5 is allowed to dissociate into SbCl3 and Cl2 at 521 K. At equilibrium, [SbCl5] = 0.195 M, and [SbCl3] = [Cl2] = 6.98×1
Brilliant_brown [7]

Answer:

a) The equilibrium will shift in the right direction.

b) The new equilibrium concentrations after reestablishment of the equilibrium :

[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M

[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

Explanation:

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

a) Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

On increase in amount of reactant

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

If the reactant is increased, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where more product formation is taking place. As the number of moles of SbCl_5 is  increasing .So, the equilibrium will shift in the right direction.

b)

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

Concentration of SbCl_5  = 0.195 M

Concentration of SbCl_3  = 6.98\times 10^{-2} M

Concentration of Cl_2  = 6.98\times 10^{-2} M

On adding more [SbCl_5 to 0.370 M at equilibrium :

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

Initially

0.370 M         6.98\times 10^{-2}M    

At equilibrium:

(0.370-x)M   (6.98\times 10^{-2}+x)M  

The equilibrium constant of the reaction  = K_c

K_c=2.50\times 10^{-2}

The equilibrium expression is given as:

K_c=\frac{[SbCl_3][Cl_2]}{[SbCl_5]}

2.50\times 10^{-2}=\frac{(6.98\times 10^{-2}+x)M\times (6.98\times 10^{-2}+x)M}{(0.370-x) M}

On solving for x:

x = 0.0233 M

The new equilibrium concentrations after reestablishment of the equilibrium :

[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M

[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

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In the world of chemistry, a 'mole' is a number of something... a very large number of something.
kipiarov [429]

Given:

No of atoms present= 8.022 x 10^23 atoms

Now we know that 1 mole= 6.022 x 10^23 atoms


Hence number of moles present in 8.022 x 10^23 atoms is calculated as below.

Number of moles

= 8.022 x 10^23/6.022x 10^23

=1.3 moles.

Hence we have 1.3 moles present.

3 0
2 years ago
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