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jok3333 [9.3K]
2 years ago
6

Increasing the concentration of a reactant shifts the position of chemical equilibrium towards formation of more products. What

effect does adding a reactant have on the rates of the forward and reverse reactions?.
Chemistry
1 answer:
lakkis [162]2 years ago
4 0

Answer:

A.There is no effect on the forward and reverse reactions. For an equilibrium reaction, the forward and reverse rates are always equal.

B. Both the forward and reverse reactions speed up by the same amount.

C. The forward reaction speeds up immediately. As more product is made, the reverse reaction starts to speed up as the forward reaction starts to slow down until they are equal.

D. The forward reaction slows down initially. As the reaction proceeds, the reverse reaction slows down to meet the new forward reaction.

E. The forward reaction speeds up. Eventually, production of the product speeds up the reverse reaction rate to match the new forward rate.

F. It is impossible to say without more specific information.

Explanation:

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finlep [7]

Given parameters:

Heat of fusion of water  = 334j/g

Mass of ice  = 45g

Temperature of ice =  0.0°c

Unknown:

Amount of heat needed to melt = ?

Solution:

This is simply a phase change and a latent heat is required in this process.

  To solve this problem; use the mathematical expression below;

             H  = mL

where m is the mass

            L is the heat of fusion of water;

              H  = 45 x 334  = 15030J

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3 years ago
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3 years ago
Use wedge-bond perspective drawings (if necessary) to sketch the atom positions in a general molecule of formula (not shape clas
yuradex [85]

The atom positions in a general molecule of formula (not shape class) AXn that has shape square pyramidal at the corers of square and one at the above center of the square.

<h3>What is square pyramidal?</h3>

The square pyramidal is a shape geometry of the hybridization in which it consists of one lone pair and 5 bond pairs of electrons that repel each other and due to which the geometry changes from octahedral to square pyramidal.

As atoms are located at the four corners of the planer and one atom at the above center of the planner which is repelled by 4 atoms present at the corner of the planer.

Therefore, the atom positions in a general molecule of formula (not shape class) AXn that has a shape square pyramidal at the corners of the square and one at the above center of the square.

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7 0
1 year ago
What is chemical reaction ?​
butalik [34]

<u><em>Answer: Chemical reaction, a process in which one or more substances, the reactants, are converted to one or more different substances, the products.</em></u>

Explanation:

7 0
3 years ago
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 %

6 0
3 years ago
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