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TEA [102]
2 years ago
7

Write a balanced equation for the formation of one mole of nacl(aq) from its elements

Chemistry
1 answer:
Makovka662 [10]2 years ago
8 0
<span>2Na + Cl2 => 2NaCl 1 mol Na + 0.5 mol Cl2 => 1 mol NaCl</span>
You might be interested in
List four observations that suggest chemical change is occurring
allsm [11]

Answer:

Examples of Chemical Changes

Burning wood.

Souring milk.

Mixing acid and base.

Digesting food.

Cooking an egg.

Heating sugar to form caramel.

Baking a cake.

Rusting of iron.

8 0
3 years ago
Calculate the freezing point of a solution containing 5.0 grams of KCl and 550.0 grams of water. The molal-freezing-point-depres
yulyashka [42]

<u>Answer:</u> The freezing point of solution is -0.454°C

<u>Explanation:</u>

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

Or,

\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Freezing point of pure solution = 0°C

i = Vant hoff factor = 2

K_f = molal freezing point elevation constant = 1.86°C/m

m_{solute} = Given mass of solute (KCl) = 5.0 g

M_{solute} = Molar mass of solute (KCl) = 74.55 g/mol

W_{solvent} = Mass of solvent (water) = 550.0 g

Putting values in above equation, we get:

0-\text{Freezing point of solution}=2\times 1.86^oC/m\times \frac{5\times 1000}{74.55g/mol\times 550}\\\\\text{Freezing point of solution}=-0.454^oC

Hence, the freezing point of solution is -0.454°C

3 0
2 years ago
A polar covalent bond will form between which two atoms?
Alex_Xolod [135]

Types of Bonds can be predicted by calculating the difference in electronegativity.

If, Electronegativity difference is,

 

                Less than 0.4 then it is Non Polar Covalent

                

                Between 0.4 and 1.7 then it is Polar Covalent 

            

                Greater than 1.7 then it is Ionic

 

For Be and F,

                    E.N of Fluorine          =   3.98

                    E.N of Beryllium        =   1.57

                                                   ________

                    E.N Difference                2.41          (Ionic Bond)


For H and Cl,

                    E.N of Chorine           =   3.16

                    E.N of Hydrogen        =   2.20

                                                   ________

                    E.N Difference                0.96          (Polar Covalent Bond)


For Na and O,

                    E.N of Oxygen          =   3.44

                    E.N of Sodium          =   0.93

                                                       ________

                    E.N Difference                2.51          (Ionic Bond)


For F and F,

                    E.N of Fluorine          =   3.98

                    E.N of Fluorine          =   3.98

                                                        ________

                    E.N Difference                0.00         (Non-Polar Covalent Bond)

Result:

           A polar covalent bond is formed between Hydrogen and Chlorine atoms.

5 0
3 years ago
What is the concentration of hydronium ions and hydroxide ions
nikdorinn [45]
To find the concentration of hydronium ions, take 10 raised to the negative pH:
10^-9.56= 2.75 x10^-10M

To find the concentration of hydroxide ions, take 10 raised to the negative pOH: 10^-4.44 = 3.63 x10^-5M
6 0
3 years ago
What are essential and non essential proteins ?
katen-ka-za [31]
Nonessential amino acids can be made by the body, while essential amino acids cannot be made by the body so you must get them from your diet.
Nonessential acids : glutamine, proline, glycine etc
Essential acids: lysine, leucine etc.
6 0
3 years ago
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