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FromTheMoon [43]
2 years ago
13

there are several ways to express solution concentration dilute concentrated ppm molarity molality normality all of these have o

ne thing in common they describe the ​

Chemistry
2 answers:
klio [65]2 years ago
4 0

B: quantity of solute with a specific volome of solvent.

Have a Nice Day :)

Rina8888 [55]2 years ago
4 0

Answer: quantity of solute dissolved in a given quantity of solvent

Explanation:-

Dilute : is defined as a solution in which amount of solvent is more than the amount of solute.

Concentrated: is defined as a solution in which amount of solute is more than the amount of solvent.

ppm: ppm is defined as the concentration when 1 part of the solute is  dissolved per 1 million of the solution.

Molarity: Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

Molality : Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.

Normality: Normality of a solution is defined as the number of ram equivalents of solute dissolved per Liter of the solution.

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Find percent yield:
saveliy_v [14]

<u>Answer:</u> The percent yield of the reaction is 91.8 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For B_5H_9 :</u>

Given mass of B_5H_9 = 4.0 g

Molar mass of B_5H_9 = 63.12 g/mol

Putting values in equation 1, we get:

\text{Moles of }B_5H_9=\frac{4g}{63.12g/mol}=0.0634mol

  • <u>For oxygen gas:</u>

Given mass of oxygen gas = 10.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

\text{Moles of oxygen gas}=\frac{10g}{32g/mol}=0.3125mol

The chemical equation for the reaction of B_5H_9 and oxygen gas follows:

2B_5H_9+12O_2\rightarrow 5B_2O_3+9H_2O

By Stoichiometry of the reaction:

12 moles of oxygen gas reacts with 2 moles of B_2H_5

So, 0.3125 moles of oxygen gas will react with = \frac{2}{12}\times 0.3125=0.052mol of B_2H_5

As, given amount of B_2H_5 is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

12 moles of oxygen gas produces 5 moles of B_2O_3

So, 0.3125 moles of oxygen gas will produce = \frac{5}{12}\times 0.3125=0.130moles of water

Now, calculating the mass of B_2O_3 from equation 1, we get:

Molar mass of B_2O_3 = 69.93 g/mol

Moles of B_2O_3 = 0.130 moles

Putting values in equation 1, we get:

0.130mol=\frac{\text{Mass of }B_2O_3}{69.63g/mol}\\\\\text{Mass of }B_2O_3=(0.130mol\times 69.63g/mol)=9.052g

To calculate the percentage yield of B_2O_3, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of B_2O_3 = 8.32 g

Theoretical yield of B_2O_3 = 9.052 g

Putting values in above equation, we get:

\%\text{ yield of }B_2O_3=\frac{8.32g}{9.052g}\times 100\\\\\% \text{yield of }B_2O_3=91.8\%

Hence, the percent yield of the reaction is 91.8 %

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Adding O2 to the reaction 6CO2 + 6H2O C6H12O6 + 6O2. WHICH WAY WILL THE REACTION SHIFT?
Oxana [17]

Answer:

The reaction will shift to the left to produce more reactants.

Explanation:

According to the Le- Chatelier principle,

At equilibrium state when stress is applied to the system, the system will behave in such a way to nullify the stress.

The equilibrium can be disturb,

By changing the concentration

By changing the volume

By changing the pressure

By changing the temperature

Consider the following chemical reaction.

Chemical reaction:

6CO₂ +  6H₂O  ⇄  C₆H₁₂O₆ + 6O₂

In this reaction the equilibrium is disturb by increasing the concentration of Product.

When the concentration of product is increased the system will proceed in backward direction in order to regain the equilibrium. Because when product concentration is high it means reaction is not on equilibrium state. As the concentration of O₂  increased the reaction proceed in backward direction to regain the equilibrium state and more reactant is formed.

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