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Alona [7]
2 years ago
9

A solid reactant is placed into a beaker of a warm water. The liquid vigorously bubbles as the solid dissolves into the solution

. What will most likely happen if the temperature of the liquid is slightly reduced?
Chemistry
2 answers:
Andrej [43]2 years ago
3 0

Answer:

Fewer bubbles will be produced because of fewer collisions of reactant molecules

Explanation:

As the solid dissolves into the solution after the liquid has been vigorously bubbled, if the temperature of the liquid is reduced a little, what will happen is that fewer bubbles will be produced as a result of lesser amount of collisions occurring between the reactant molecules

ratelena [41]2 years ago
3 0

Answer:

b

Explanation:

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If there are 40 mol of NBr3 and 48 mol of NaOH, what is the excess reactant
Vanyuwa [196]
<h3>Answer:</h3>

              Excess Reagent =  NBr₃

<h3>Solution:</h3>

The Balance Chemical Equation for the reaction of NBr₃ and NaOH is as follow,

                       2 NBr₃ + 3 NaOH   →   N₂ + 3 NaBr + 3 HBrO

Calculating the Limiting Reagent,

According to Balance equation,

               2 moles NBr₃ reacts with  =  3 moles of NaOH

So,

           40 moles of NBr₃ will react with  =  X moles of NaOH

Solving for X,

                       X  =  (40 mol × 3 mol) ÷ 2 mol

                       X  =  60 mol of NaOH

It means 40 moles of NBr₃ requires 60 moles of NaOH, while we are provided with 48 moles of NaOH which is Limited. Therefore, NaOH is the limiting reagent and will control the yield of products. And NBr₃ is in excess as some of it is left due to complete consumption of NaOH.

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2 years ago
For each of the following reactions, identify the missing reactant(s) or products(s) and then balance the resulting equation. No
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Answer:

A. The reactants are= Li and O2

The balanced equation is:

4Li + O2 —> 2Li2O

B. The products are: MgCl2 and O2

The balanced equation is:

Mg(ClO3)2 —> MgCl2 + 3O2

C. The products are: Ca(NO3)2 and H2O

The balanced equation is:

2HNO3 + Ca(OH)2 —> Ca(NO3)2 +

2H2O

D. The products are: CO2 and H2O

The balanced equation is:

C5H12 + 8O2 —> 5CO2 + 6H2O

Explanation:

A. ____ —> Li2O

The reactants are Li and O2. Thus the equation is given below:

Li + O2 —> Li2O

Thus the equation is balanced as follow:

There are 2 atoms of O on the left side and 1 atom on the right side. It can be balance by putting 2 in front of Li2O as shown below:

Li + O2 —> 2Li2O

Now, we have 4 atoms of Li on the right side and 1 atom on the left. It can be balance by putting 4 in front of Li as shown below:

4Li + O2 —> 2Li2O

Now the equation is balanced

B. Mg(ClO3)2 —> __

The products are MgCl2 and O2

The equation is given below:

Mg(ClO3)2 —> MgCl2 + O2

The equation can be balance as follow:

There are 6 atoms of O on the left side and 2 atoms on the right side. It can be balance by putting 3 in front of O2 as shown below:

Mg(ClO3)2 —> MgCl2 + 3O2

Now the equation is balanced

C. HNO3 + Ca(OH)2 —>___

The products are: Ca(NO3)2 and H2O

The equation is given below:

HNO3 + Ca(OH)2 —> Ca(NO3)2 + H2O

The equation is balanced as follow:

There are 2 atoms of NO3 on the right side and 1 atom on the left. It can be balance by putting 2 in front of HNO3 as shown below:

2HNO3 + Ca(OH)2 —> Ca(NO3)2 + H2O

There are a total of 4 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 2 in front of H2O as shown below:

2HNO3 + Ca(OH)2 —> Ca(NO3)2 + 2H2O

Now the equation is balanced

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The products are: CO2 and H2O

The equation is given below:

C5H12 + O2 —> CO2 + H2O

The equation can be balance as follow:

There are 5 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 5 in front of CO2 as shown below:

C5H12 + O2 —> 5CO2 + H2O

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C5H12 + O2 —> 5CO2 + 6H2O

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C5H12 + 8O2 —> 5CO2 + 6H2O

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