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4 years ago

For a reaction system at equlibrium, le chateliers principle can be used to predict the

1 answer:

6 0

According to Le Chatelier's principle, adding additional reactant to a system will shift the equilibrium to the right, towards the side of the products. ... By Le Chatelier's principle, we can predict that the amount of methanol will increase, thereby decreasing the total change in CO.

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Several balls and sticks glued together to represent a molecule

rewona [7]

Answer:

The different elements are represented by their symbols one line shows a single bond and two lines show a double bond. This written representation is called a structural diagram

4 0

3 years ago

In the reaction: C6H6 + O2 = CO2 + H2O If 26.2 grams of C6H6 is reacted in a 0.250-liter container and this reaction is carried

Lelechka [254]

<h3>Answer:</h3>

43.33 atm

<h3>Explanation:</h3>

We are given;

Mass of C₆H₆ = 26.2 g

Volume of the container = 0.25 L

Temperature = 395 K

We are required to calculate the pressure inside the container;

First, we calculate the number of moles of C₆H₆

Molar mass of C₆H₆ = 78.1118 g/mol.

But; Moles = mass ÷ Molar mass

Moles of C₆H₆ = 26.2 g ÷ 78.1118 g/mol.

= 0.335 moles C₆H₆

Second, we calculate the pressure, using the ideal gas equation;

Using the ideal gas equation, PV = nRT , Where R is the ideal gas constant, 0.082057 L.atm/mol.K

Therefore;

P = nRT ÷ V

= (0.335 mol × 0.082057 × 395 K) ÷ 0.25 L

= 43.433 atm

Therefore, the pressure inside the container is 43.33 atm

6 0

3 years ago

If the block were to be cut in half, what would be the<br> density of the smaller block?

anyanavicka [17]

Answer:

Density stays the same

Explanation:

The density remains the same because cutting the object in half will divide the mass & volume by the same amount. Also, the density of a substance remains the same no matter what size it is.

6 0

3 years ago

Suppose 0.245 g of sodium chloride is dissolved in 50. mL of a 18.0 m M aqueous solution of silver nitrate.

Bezzdna [24]

Answer:

$\large \boxed{\text{ 0.066 mol/L}}$

Explanation:

We are given the amounts of two reactants, so this is a limiting reactant problem.

1. Assemble all the data in one place, with molar masses above the formulas and other information below them.

Mᵣ: 58.44

NaCl + AgNO₃ ⟶ NaNO₃ + AgCl

m/g: 0.245

V/mL: 50.

c/mmol·mL⁻¹: 0.0180

2. Calculate the moles of each reactant

$\text{Moles of NaCl} = \text{245 mg NaCl} \times \dfrac{\text{1 mmol NaCl}}{\text{58.44 mg NaCl}} = \text{4.192 mmol NaCl}\\\\\text{ Moles of AgNO}_{3}= \text{50. mL AgNO}_{3} \times \dfrac{\text{0.0180 mmol AgNO}_{3}}{\text{1 mL AgNO}_{3}} = \text{0.900 mmol AgNO}_{3}$

3. Identify the limiting reactant

Calculate the moles of AgCl we can obtain from each reactant.

From NaCl:

The molar ratio of NaCl to AgCl is 1:1.

$\text{Moles of AgCl} = \text{4.192 mmol NaCl} \times \dfrac{\text{1 mmol AgCl}}{\text{1 mmol NaCl}} = \text{4.192 mmol AgCl}$

From AgNO₃:

The molar ratio of AgNO₃ to AgCl is 1:1.

$\text{Moles of AgCl} = \text{0.900 mmol AgNO}_{3} \times \dfrac{\text{1 mmol AgCl}}{\text{1 mmol AgNO}_{3}} = \text{0.900 mmol AgCl}$

AgNO₃ is the limiting reactant because it gives the smaller amount of AgCl.

4. Calculate the moles of excess reactant

Ag⁺(aq) + Cl⁻(aq) ⟶ AgCl(s)

I/mmol: 0.900 4.192 0

C/mmol: -0.900 -0.900 +0.900

E/mmol: 0 3.292 0.900

So, we end up with 50. mL of a solution containing 3.292 mmol of Cl⁻.

5. Calculate the concentration of Cl⁻

$\text{[Cl$^{-}$] } = \dfrac{\text{3.292 mmol}}{\text{50. mL}} = \textbf{0.066 mol/L}\\\text{The concentration of chloride ion is $\large \boxed{\textbf{0.066 mol/L}}$}$

8 0

3 years ago

How may moles are in 88.4 grams of Al(OH)3?

mash [69]

Answer:

1.133mol Al(OH)3

Explanation:

88.4/78.003 = 1.133

5 0

3 years ago