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Olenka [21]
3 years ago
15

Experiment on distillation

Chemistry
1 answer:
Temka [501]3 years ago
4 0

Answer:

ok.. what is your question??? confused

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What happens in a double-replacement reaction?
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Answer: D

Explanation: trust

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Consider this reaction: Upper K Upper O Upper H + Upper H Upper B 4 right arrow Upper K Upper B r + Upper H Subscript 2 Baseline
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Hbr is the answer when I work it out that’s what I got.
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Consider the reaction and its equilibrium constant at 25.0 ∘C :
olga55 [171]

Since Qp>Kp , the reaction is not at equilibrium.

<h3>What is the equilibrium constant?</h3>

The equilibrium constant shows the extent to which reactants are converted into products.

Now we have to obtain the Qp as follows;

Qp =[CH3OH]/[CO] [H2]^2

Qp = 0.265/(0.265) (0.265)^2

Qp = 14.2

Now we know that Kp = 6.09×10−3, Since Qp>Kp , the reaction is not at equilibrium.

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5 0
2 years ago
The solubility of nitrogen gas at 25 degrees C and 1 atm is 6.8 x 10^(-4) mol/L. If the partial pressure of nitrogen gas in air
Zolol [24]

Answer:

5.2 x 10⁻⁴ M.

Explanation:

  • The relationship between gas pressure and the  concentration of dissolved gas is given by  Henry’s law:

<em>P = kC</em>

where P is the partial pressure of the gaseous  solute above the solution.

k is a constant (Henry’s constant).

C is the concentration of the dissolved gas.

  • At two different pressures, there is two different concentrations of dissolved gases and is expressed in a relation as:

<em>P₁C₂ = P₂C₁,</em>

P₁ = 1.0 atm, C₁ = 6.8 x 10⁻⁴ mol/L.

P₂ = 0.76 atm, C₂ = ??? mol/L.

<em>∴ C₂ = (P₂C₁)/P₁ =</em> (0.76 atm)(6.8 x 10⁻⁴ mol/L)/(1.0 atm) = <em>5.168 x 10⁻⁴ mol/L ≅ 5.2 x 10⁻⁴ M.</em>

5 0
3 years ago
A hot lump of 39.9 g of iron at an initial temperature of 78.1 °C is placed in 50.0 mL H 2 O initially at 25.0 °C and allowed to
Drupady [299]

Answer : The final temperature of the mixture is 29.6^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of iron = 0.499J/g^oC

c_2 = specific heat of water = 4.18J/g^oC

m_1 = mass of iron = 39.9 g

m_2 = mass of water  = Density\times Volume=1g/mL\times 50.0mL=50.0g

T_f = final temperature of mixture = ?

T_1 = initial temperature of iron = 78.1^oC

T_2 = initial temperature of water = 25.0^oC

Now put all the given values in the above formula, we get

(39.9g)\times (0.499J/g^oC)\times (T_f-78.1)^oC=-(50.0g)\times 4.18J/g^oC\times (T_f-25.0)^oC

T_f=29.6^oC

Therefore, the final temperature of the mixture is 29.6^oC

8 0
3 years ago
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