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GenaCL600 [577]
3 years ago
6

Question 8

Chemistry
1 answer:
nexus9112 [7]3 years ago
6 0
72 degrees C (minus 273)
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Please help me thanks
Digiron [165]

Answer:

your finger becomes negatively charged

Explanation:

5 0
2 years ago
Read 2 more answers
The chemical reaction involving an acid and a base or alkali is know as a ​
saul85 [17]

Ans

Neutralization reaction

Explanation:

The meaning is

acid + base = salt + water

5 0
3 years ago
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Calculate the energy required to heat 406.0mg of cyclohexane from 33.5°C to 38.9°C . Assume the specific heat capacity of cycloh
N76 [4]

Answer:

Q = 4.056 J

Explanation:

  • Q = m<em>C</em>ΔT

∴ m = 406.0 mg = 0.406 g

∴ <em>C </em>= 1.85 J/g.K

∴ T1 = 33.5°C ≅ 306.5 K

∴ T2 = 38.9°C = 311.9 K

⇒ ΔT = 311.9 - 306.5 = 5.4 K

⇒ Q = (0.406 g)(1.85 J/gK)(5.4 K)

⇒ Q = 4.056 J

6 0
2 years ago
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

3 0
3 years ago
What is the mass of a 1.71-l sample of a liquid that has a density of 0.921g/ml?
Temka [501]
Hey there!:

Volume in mL :

1.71 L * 1000 => 1710 mL

Density = 0.921 g/mL

Therefore:


Mass = Density * Volume

Mass = 0.921 * 1710

Mass = 1574.91 g
8 0
2 years ago
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