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olga_2 [115]
3 years ago
7

Why does common table salt have a high melting point

Chemistry
1 answer:
tankabanditka [31]3 years ago
5 0

Sodium chloride has a high melting and boiling point. There are strong electrostatic attractions between the positive and negative ions, and it takes a lot of heat energy to overcome them. Ionic substances all have high melting and boiling points


You might be interested in
If you have 30.O g of hydrogen gas burned in excess oxygen how many moles of water can you make
vladimir1956 [14]

Answer:

15 moles

Explanation:

Data given:

mass of hydrogen (H₂) = 30.0 g

amount of oxygen (O₂) = excess

moles of water = ?

Solution:

First we look to the reaction in which hydrogen react with oxygen and make (H₂O)

Reaction:

              2H₂  + O₂  -----------> 2H₂O

Now look at the reaction for mole ratio

             2H₂  + O₂  -----------> 2H₂O

             2 mole                       2 mole

So it is 2:2 mole ratio of hydrogen to water

As we Know

molar mass of H₂  = 2(1) = 2 g/mol

molar mass of H₂O = 2(1) + 16 = 18 g/mol

Now convert moles to gram

                  2H₂         +       O₂        ----------->    2H₂O

          2 mole (2 g/mol)                                 2 mole (18 g/mol)

                    4 g                                                     36 g

So,

we come to know that 4 g of hydrogen gives 36 g of water then how many grams of water will be produce by 30 grams of hydrogen.

Apply unity formula

                       4 g of H₂ ≅ 36 g of H₂O

                        30 g of H₂ ≅ X of H₂O

Do cross multiplication

                  X of H₂O =  30 g x 36 g / 4 g

                  X of H₂O =  270 g

Now convert grams of H₂O into moles

               No. of moles = mass in grams/molar mass

Put values in above formula

               No. of moles = 270 g / 18 (g/mol)

               No. of moles = 15 mol

so 30 gram of hydrogen produce 15 mol of water.

5 0
3 years ago
Which best explains the relationship between parent rock and soil<br><br> composition?
irina1246 [14]
<span>Soil is partially the result of the physical and chemical weathering of its parent rock. The minerals found in the soil were either in that parent rock, or they were formed from the weathering products of the parent rock.</span>
8 0
3 years ago
Based on table N of the reference tables what is the number of hours required for 42k (potassium -42) to undergo three half-life
skelet666 [1.2K]

The number of hours required : 37.2 hours

<h3>Further explanation</h3>

Given

⁴²K (potassium -42)

Required

The number of hours

Solution

The atomic nucleus can experience decay into 2 particles or more due to the instability of its atomic nucleus.  

Usually, radioactive elements have an unstable atomic nucleus.  

Based on Table N(attached), the half-life for ⁴²K is 12.4 hours, which means half of a sample of ⁴²K will decay in 12.4 hours

For three half-life periods :

\tt 3\times 12.4=37.2~hours

3 0
3 years ago
Trans fats: Think about how the prefix trans- is used in naming alkenes.
wel
In an alkene, cis and trans isomers are possible because the double band is rigid, cannot rotate, has groups attached to the carbons of the double bond that are fixed relative to each other, and only occurs with double bonds-possibility that molecule will have different geometries; two different molecules with slightly different properties. 
-Trans-2 ends of chain across the double bond.
While naming Cis-Trans isomers the prefix cis or trans are placed in front of the alkene name when there are cis-trans isomers. 
4 0
3 years ago
Read 2 more answers
For the decomposition of A to B and C, A(s)⇌B(g)+C(g) how will the reaction respond to each of the following changes at equilibr
lys-0071 [83]

Answer:

a. No change.    

b. The equilibrium will shift to the right.

c. No change

d. No change

e.  The equilibrium will shift to the left

f.  The equilibrium will shift to the right      

Explanation:

We are going to solve this question by making use of Le Chatelier´s principle which states that any change in a system at equilibrium will react in such a way as to attain qeuilibrium again by changing the equilibrium concentrations attaining   Keq  again.

The equilibrium constant  for  A(s)⇌B(g)+C(g)  

Keq = Kp = pB x pC

where K is the equilibrium constant ( Kp in this case ) and pB and pC are the partial pressures of the gases. ( Note A is not in the expression since it is a solid )

We also use  Q which has the same form as Kp but denotes the system is not at equilibrium:

Q = p´B x p´C where pB´ and pC´ are the pressures not at equilibrium.

a.  double the concentrations of Q which has the same form as Kp but : products and then double the container volume

Effectively we have not change the equilibrium pressures since we know pressure is inversely proportional to volume.

Initially the system will decrease the partial pressures of B and C by a half:

Q = pB´x pC´     ( where pB´and pC´are the changed pressures )

Q = (2 pB ) x (2 pC) = 4 (pB x PC) = 4 Kp  ⇒ Kp = Q/4

But then when we double the volume ,the sistem will react to  double the pressures of A and B. Therefore there is no change.

b.  double the container volume

From part a we know the system will double the pressures of B and C by shifting to the right ( product ) side since the change  reduced the pressures by a half :

Q =  pB´x pC´  = (  1/2 pB ) x ( 1/2 pC )  =  1/4 pB x pC  = 1/4 Kp

c. add more A

There is no change in the partial pressures of B and C since the solid A does not influence the value of kp

d. doubling the  concentration of B and halve the concentration of C

Doubling the concentrantion doubles  the pressure which we can deduce from pV = n RT = c RT ( c= n/V ), and likewise halving the concentration halves the pressure. Thus, since we are doubling the concentration of B and halving that of C, there is no net change in the new equilibrium:

Q =  pB´x pC´  = ( 2 pB ) x ( 1/2 pC ) = K

e.  double the concentrations of both products

We learned that doubling the concentration doubles the pressure so:

Q =  pB´x pC´   = ( 2 pB ) x ( 2 pC ) = 4 Kp

Therefore, the system wil reduce by a half the pressures of B and C by producing more solid A to reach equilibrium again shifting it to the left.

f.  double the concentrations of both products and then quadruple the container volume

We saw from part e that doubling the concentration doubles the pressures, but here afterward we are going to quadruple the container volume thus reducing the pressure by a fourth:

Q =  pB´x pC´   = ( 2 pB/ 4 ) x (2 pC / 4) = 4/16  Kp = 1/4 Kp

So the system will increase the partial pressures of B and C by a factor of four, that is it will double the partial pressures of B and C shifting the equilibrium to the right.

If you do not see it think that double the concentration and then quadrupling the volume is the same net effect as halving the volume.

3 0
3 years ago
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