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kolbaska11 [484]
2 years ago
15

A student is curious about the Ksp value for NaCl. The student looks up the value om the appendix of his textbook but cannot fin

d a value for NaCl. Why not?
Chemistry
1 answer:
Leno4ka [110]2 years ago
7 0

Given :

The student looks up the value of the appendix of his textbook but cannot find a value for NaCl.

To Find :

Why the Ksp of NaCl is not present .

Solution :

Ksp values are given for those compounds which are partially soluble or insoluble in water .

But , we know NaCl is strong electrolyte and it completely dissociated in water .

So , the reaction go forward and their is no need to calculate Ksp .

Therefore , the Ksp of NaCl is not present .

Hence , this is the required solution .

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1. Suppose 0.7542 g of magnesium reacts with excess oxygen to form magnesium oxide as the only product, what would be the theore
Alina [70]

Answer :

(1) The theoretical yield of product, MgO is, 1.257 grams.

(2) The percent yield of MgO is, 64.13 %

(3) If the percent yield is calculated to be over 100% then there might be some impurity present in the desired product.

Solution : Given,

Mass of Mg = 0.7542 g

Molar mass of Mg = 24 g/mole

Molar mass of MgO = 40 g/mole

First we have to calculate the moles of Mg.

\text{ Moles of }Mg=\frac{\text{ Mass of }Mg}{\text{ Molar mass of }Mg}=\frac{0.7542g}{24g/mole}=0.03142moles

Now we have to calculate the moles of MgO

The balanced chemical reaction is,

2Mg(s)+O_2(g)\rightarrow 2MgO(s)

From the reaction, we conclude that

As, 2 mole of Mg react to give 2 mole of MgO

So, 0.03142 moles of Mg react to give 0.03142 moles of MgO

Now we have to calculate the mass of MgO

\text{ Mass of }MgO=\text{ Moles of }MgO\times \text{ Molar mass of }MgO

\text{ Mass of }MgO=(0.03142moles)\times (40g/mole)=1.257g

Theoretical yield of MgO = 1.257 g

Experimental yield of MgO = 0.8922 g

Now we have to calculate the percent yield of MgO

\% \text{ yield of }MgO=\frac{\text{ Experimental yield of }MgO}{\text{ Theretical yield of }MgO}\times 100

\% \text{ yield of }MgO=\frac{0.8922g}{1.257g}\times 100=70.97\%

Therefore, the percent yield of MgO is, 70.97 %

If the percent yield is calculated to be over 100% then the product would be greater than 1.257 g which indicates that there might be some impurity present in the desired product.

4 0
2 years ago
. Calculate the mass of O2 produced if 3.450 g potassium chlorate is completely decomposed by heating in presence of a catalyst
Vesnalui [34]
 <span>2 KClO3(s) → 3 O2(g) + 2 KCl(s) 

</span><span>Note: MnO2 (Manganese Dioxide) is not part of the reaction. A catalyst lowers the activation energy and increases both forward and reverse reactions at equal rates. 
</span>
molar mass of KClO3 = 122.5
Moles of KClO3 =  3.45 / 122.55 = 0.028

Moles of O2 produce = \frac{3}{2} \times 0.028

= 0.042 moles

molar mass of O2 = 32

so, mass of O2 = 32 x 0.042  = 1.35 g



5 0
1 year ago
If the mass percentage composition of a compound is 72.1% Mn and 27.9% O, its empirical formula is
mojhsa [17]

Answer:

MnO- Manganese Oxide

Explanation:

Empirical formula: This is the formula that shows the ratio of elements

present in a  

compound.

   

How to determine Empirical formula

1. First arrange the symbols of the elements present in the compound

alphabetically to  determine the real empirical formula. Although, there

are exceptions to this rule, E.g H2So4

2. Divide the percentage composition by the mass number.

3. Then divide through by the smallest number.

4. The resulting answer is the ratio attached to the elements present in

a compound.

           

                                                                              Mn                         O    

                         

% composition                                                      72.1                      27.9    

                       

Divide by mass number                                       54.94                     16  

                                 

                                                                               1.31                      1.74    

                       

Divide by the smallest number                         1.31                      1.31                          

                                                                               1                    1.3

                                                 

The resulting ratio is 1:1

 

Hence the Empirical formula is MnO, Manganese oxide

8 0
2 years ago
Tag all the carbon atoms with pi bonds in this molecule. If there are none, please check the box.
snow_tiger [21]

Answer:

Pi bonds (π bonds) are covalent chemical bonds where two lobes of an orbital involved in the bond overlap with two lobes of the other orbital involved. These orbitals share a nodal plane that passes through the nuclei involved. Are generally weaker than sigma links, because their negatively charged electronic density is further from the positive charge of the atomic nucleus, which requires more energy.

They are frequent components of multiple bonds, as is the molecule indicated in our exercise.

The characteristics that distinguish pi bonds from other kinds of interactions between atomic species are described below, beginning with the fact that this union does not allow the free rotation movement of atoms, such as carbon. For this reason, if there is rotation of the atoms, the bond is broken.

Explanation:

In order to describe the formation of the pi bond, first we must talk about the hybridization process, as this is involved in some important links.

Hybridization is a process where hybrid electronic orbitals are formed; that is, where orbitals of atomic sub-levels s and p can get mixed. This causes the formation of sp, sp2 and sp3 orbitals, which are called hybrids.

In this sense, the formation of pi bonds occurs thanks to the overlapping of a pair of lobes belonging to an atomic orbital over another pair of lobes that are in an orbital that is part of another atom.

This orbital overlap occurs laterally, so the electronic distribution is mostly concentrated above and below the plane formed by the linked atomic nuclei, and causes the pi bonds to be weaker than the sigma bonds.

When talking about the orbital symmetry of this type of junction, it should be mentioned that it is equal to that of the p-type orbitals as long as it is observed through the axis formed by the bond. In addition, these junctions are mostly made up of p orbitals.

Since pi bonds are always accompanied by one or two more links (one sigma or another pi and one sigma), it is relevant to know that the double bond that is formed between two carbon atoms has less bond energy than that corresponding to two Sometimes the sigma link between them.

4 0
2 years ago
Show that the Newton has the units of mass times acceleration
Dimas [21]

F = ma = (kg)(m/s2) = kg ´ m/s2 N

hope this helps :D

7 0
2 years ago
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