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2 years ago

Which of the following indicates that a chemical reaction system has reached equilibrium?

1 answer:

4 0

D) if stuff is changing then the reaction is hardly in equilibrium is it? Everything is just chilling at equilibrium so there would be constant concentration

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Different isotopes of the same element must have a different

Naddika [18.5K]

Different isotopes of the same element must have a different mass number

5 0

3 years ago

Barium sulfate is made by the following reaction.

ipn [44]

Answer:

%age Yield = 96 %

Explanation:

The balance chemical equation for given double replacement reaction is,

Ba(NO₃)₂ + Na₂SO₄ → BaSO₄ + 2 NaNO₃

Step 1: Calculate moles of Ba(NO₃)₂:

Moles = Mass / M.Mass

Moles = 75.1 g / 261.33 g/mol

Moles = 0.2873 moles of Ba(NO₃)₂

Step 2: Find out moles of BaSO₄ formed:

According to balance chemical equation,

1 mole of Ba(NO₃)₂ produced = 1 mole of BaSO₄

So,

0.2873 moles of Ba(NO₃)₂ will produce = X moles of BaSO₄

Solving for X,

X = 0.2873 mol × 1 mol / 1 mol

X = 0.2873 moles of BaSO₄

Step 3: Calculate Theoretical Mass of BaSO₄:

Mass = Moles × M.Mass

Mass = 0.2873 mol × 233.38 g/mol

Mass = 67.07 g of BaSO₄

Step 4: Calculate %age Yield as:

Theoretical Yield = 67.07 g

Actual Yield = 64.4 g

%age Yield = ???

Formula Used:

%age Yield = (Actual Yield ÷ Theoretical Yield) × 100

Putting Values,

%age Yield = (64.4 g ÷ 67.07 g) × 100

%age Yield = 96.01 % ≈ 96 %

6 0

2 years ago

(01.01 LC)

Monica [59]

Paid sponsorship by a company

5 0

3 years ago

Read 2 more answers

The blank

Alik [6]

I think it’s “number” and “type”

5 0

3 years ago

2) A common "rule of thumb" -- for many reactions around room temperature is that the

babunello [35]

The question is incomplete. The complete question is :

A common "rule of thumb" for many reactions around room temperature is that the rate will double for each ten degree increase in temperature. Does the reaction you have studied seem to obey this rule? (Hint: Use your activation energy to calculate the ratio of rate constants at 300 and 310 Kelvin.)

Solutions :

If we consider the activation energy to be constant for the increase in 10 K temperature. (i.e. 300 K → 310 K), then the rate of the reaction will increase. This happens because of the change in the rate constant that leads to the change in overall rate of reaction.

Let's take :

$T_1=300 \ K$

$T_2=310 \ K$

The rate constant = $K_1 \text{ and } K_2$ respectively.

The activation energy and the Arhenius factor is same.

So by the arhenius equation,

$K_1 = Ae^{-\frac{E_a}{RT_1}}$ and $K_2 = Ae^{-\frac{E_a}{RT_2}}$

$\Rightarrow \frac{K_1}{K_2}= \frac{e^{-\frac{E_a}{RT_1}}}{e^{-\frac{E_a}{RT_2}}}$

$\Rightarrow \frac{K_1}{K_2}= e^{-\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)}$

$\Rightarrow \ln \frac{K_1}{K_2}= - \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{E_a}{R} \left(\frac{1}{T_1} -\frac{1}{T_2} \right)$

Given, $E_a = 0.269$ J/mol

R = 8.314 J/mol/K

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{0.269}{8.314} \left(\frac{1}{300} -\frac{1}{310} \right)$

$\Rightarrow \ln \frac{K_2}{K_1}= \frac{0.269}{8.314} \times \frac{10}{300 \times 310}$

$\Rightarrow \ln \frac{K_2}{K_1}= 3.479 \times 10^{-6}$

$\Rightarrow \frac{K_2}{K_1}= e^{3.479 \times 10^{-6}}$

$\Rightarrow \frac{K_2}{K_1}= 1$

∴ $K_2=K_1$

So, no this reaction does not seem to follow the thumb rule as its activation energy is very low.

8 0

2 years ago