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sergejj [24]
3 years ago
5

How do the interactions that are broken in water when it is boiled compare with those broken when water is electrolyzed? Boiling

water breaks intermolecular attractions and electrolysis breaks covalent bonds. Boiling water breaks covalent bonds and electrolysis breaks intermolecular attractions. Boiling water and electrolysis of water break covalent bonds. Boiling water and electrolysis of water break intermolecular forces.
Chemistry
1 answer:
nasty-shy [4]3 years ago
8 0

Answer:

Boiling water breaks intermolecular attractions and electrolysis breaks covalent bonds.

Explanation:

When water boils, hydrogen bonds are broken between adjacent water molecules. The hydrogen bond is an intermolecular bond between adjacent oxygen and hydrogen atoms of water molecules.  

During electrolysis, water dissociates in the presence of electric current. Here, ions are formed in the process. Therefore, covalent bonds are broken here.

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Name the following compound NiBr2?
Fantom [35]

Answer:

the Name of NiBr2 is

Nickel(II) bromide

5 0
3 years ago
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What does this symbol, 1/0N, mean?
Firdavs [7]

An element can be represented symbolically in terms of its atomic number and mass number as:

_{Z}^{A}\textrm{X}

where X = symbol of the element

A = atomic mass

Z = atomic number

In this case we have the representation as:

_{0}^{1}\textrm{n}

where n = symbol for neutron

mass number of neutron = 1

charge/atomic number of neutron = 0

Ans: D) A neutron of zero charge and one unit mass



5 0
4 years ago
what is the concentration of an NaOH solution that requires 50 mL of a 1.25 M H2SO4 solution to neutralize 78.0 ml of NaOH​
Dimas [21]

Answer:

  • The answer is the concentration of an NaOH = 1.6 M

Explanation:

The most common way to solve this kind of problem is to use the formula  

  • C₁ * V₁ = C₂ * V₂

In your problem,

For NaOH

C₁ =??     v₁= 78.0 mL = 0.078 L

For H₂SO₄

C₁ =1.25 M     v₁= 50.0 mL = 0.05 L

but you must note that for the reaction of NaOH with H₂SO₄

2 mol of NaOH raect with 1 mol H₂SO₄

So, by applying in above formula

  • C₁ * V₁ = 2 * C₂ * V₂
  • (C₁ * 0.078 L) = (2*  1.25 M * 0.05 L)
  • C₁ = (2*  1.25 M * 0.05 L) / (0.078 L) = 1.6 M  

<u>So, the answer is the concentration of an NaOH = 1.6 M</u>

3 0
3 years ago
Tetrahydrofuran (THF) is a common organic solvent with a boiling point of 339 K. Calculate the total energy (q) required to conv
poizon [28]

Explanation:

For the given reaction, the temperature of liquid will rise from 298 K to 339 K. Hence, heat energy required will be calculated as follows.

             Q_{1} = mC_{1} \Delta T_{1}

Putting the given values into the above equation as follows.

           Q_{1} = mC_{1} \Delta T_{1}

                      = 27.3 g \times 1.70 J/g K \times 41

                      = 1902.81 J

Now, conversion of liquid to vapor at the boiling point (339 K) is calculated as follows.

           Q_{2} = energy required = mL_{v}

    L_{v} = latent heat of vaporization

Therefor, calculate the value of energy required as follows.

             Q_{2} = mL_{v}

                         = 27.3 \times 444

                         = 12121.2 J

Therefore, rise in temperature of vapor from 339 K to 373 K is calculated as follows.

            Q_{3} = mC_{2} \Delta T_{2}

Value of C_{2} = 1.06 J/g,    \Delta T_{2} = (373 -339) K = 34 K

Hence, putting the given values into the above formula as follows.

             Q_{3} = mC_{2} \Delta T_{2}

                       = 27.3 g \times 1.06 J/g \times 34 K

                       = 983.892 J

Therefore, net heat required will be calculated as follows.

            Q = Q_{1} + Q_{2} + Q_{3}

                = 1902.81 J + 12121.2 J + 983.892 J

                = 15007.902 J

Thus, we can conclude that total energy (q) required to convert 27.3 g of THF at 298 K to a vapor at 373 K is 15007.902 J.

5 0
3 years ago
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La diferensia es que los volcanes caseros no existen solo existen los verdaderos
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