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Vadim26 [7]
3 years ago
9

A chemist prepares a solution of potassium dichromate by measuring out of potassium dichromate into a volumetric flask and filli

ng the flask to the mark with water. Calculate the concentration in of the chemist's potassium dichromate solution. Be sure your answer has the correct number of significant digits.
Chemistry
1 answer:
riadik2000 [5.3K]3 years ago
4 0

Answer:

0.297 mol/L

Explanation:

<em>A chemist prepares a solution of potassium dichromate by measuring out 13.1 g of potassium dichromate into a 150 mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in mol/L of the chemist's potassium dichromate solution. Be sure your answer has the correct number of significant digits.</em>

<em />

Step 1: Calculate the moles corresponding to 13.1 g of potassium dichromate

The molar mass of potassium dichromate is 294.19 g/mol.

13.1 g × (1 mol/294.19 g) = 0.0445 mol

Step 2: Convert the volume of solution to L

We will use the relationship 1 L = 1000 mL.

150 mL × (1 L/1000 mL) = 0.150 L

Step 3: Calculate the concentration of the solution in mol/L

C = 0.0445 mol/0.150 L = 0.297 mol/L

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Answer:

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Explanation:

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Answer:

Carbon Cycle

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Steps of the Nitrogen Cycle

  • Atmospheric nitrogen (N2) is converted to ammonia (NH3) by nitrogen-fixing bacteria in aquatic and soil environments. These organisms use nitrogen to synthesize the biological molecules they need to survive.
  • NH3 is subsequently converted to nitrite and nitrate by bacteria known as nitrifying bacteria.
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Oxygen is an element that is essential to biological organisms. The vast majority of atmospheric oxygen (O2) is derived from photosynthesis. Plants and other photosynthetic organisms use CO2, water, and light energy to produce glucose and O2. Glucose is used to synthesize organic molecules, while O2 is released into the atmosphere. Oxygen is removed from the atmosphere through decomposition processes and respiration in living organisms.

Explanation:

6 0
3 years ago
The equilibrium constant is equal to 5.00 at 1300 K for the reaction:2 SO2(g) + O2(g) ⇌ 2 SO3(g). If initial concentrations are
oee [108]

This is an incomplete question, here is a complete question.

The equilibrium constant is equal to 5.00 at 1300 K for the reaction:

2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)

If initial concentrations are [SO₂] = 1.20 M, [O₂] = 0.45 M, and [SO₃] = 1.80 M, the system is

A) at equilibrium.

B) not at equilibrium and will remain in an unequilibrated state.

C) not at equilibrium and will shift to the left to achieve an equilibrium state.

D) not at equilibrium and will shift to the right to achieve an equilibrium state.

Answer : The correct option is, (A) at equilibrium.

Explanation :

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given balanced chemical reaction is,

2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)

The expression for reaction quotient will be :

Q=\frac{[SO_3]^2}{[SO_2]^2[O_2]}

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

Q=\frac{(1.80)^2}{(1.20)^2\times (0.45)}=5.0

The given equilibrium constant value is, K_c=5.00

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

There are 3 conditions:

When Q>K_c that means product > reactant. So, the reaction is reactant favored.

When Q that means reactant > product. So, the reaction is product favored.

When Q=K_c that means product = reactant. So, the reaction is in equilibrium.

From the above we conclude that, the Q=K_c that means product = reactant. So, the reaction is in equilibrium.

Hence, the correct option is, (A) at equilibrium.

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Aloiza [94]

Answer:

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Explanation:

Step 1: Data given

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Step 2: The unbalanced equation

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Step 3: Balancing the equation

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On the left we have 2x Cl (in Cl2) and on the right side we have 3x Cl (in FeCl3). To balance the amount of Cl we have to multiply Cl2 (on the left) by 3 and FeCl3 by 2.

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On the left side we have 1x Fe and on the right side we have 2x Fe (in 2FeCl3). To balance the amount of Fe, we have to multiply Fe on the left side by 2. Now the equation is balanced.

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Answer:

its a

Explanation:

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