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VikaD [51]
3 years ago
5

When KNO3 dissolved in water, what is the intermolecular

Chemistry
1 answer:
tangare [24]3 years ago
5 0

Answer:Ion-dipole

Explanation:

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How would you prepare 250 mL of 0.125 M HCl from concentrated HCl (aq) that is 38.0% by mass with a density of 1.19 g/mL
hoa [83]
<h2>Step 1 : Identify the given </h2>

Volume = 250mL

Density = 1.19 g/ML

<h2>Step 2 . Calculate the mass of HCL </h2>

Density = mass/volume

∴Mass = Density * Volume

= 1.19g/mL* 250mL

= 297,5g

<h2>Step 3 : Calculate the total mass of the solution, given that concentration HCL is 38% </h2>

Mass of the total solution can be calculated by the following :

38% = Mc /297.5 * 100

Mc = 38/100 *297.5

= 113.05grams

• Finally, this means that mass of the total solution of 0.125M HCL i,s 113grams, ,you would use this mass to prepare 250 mL of 0.125 M HCl from concentrated HCl (aq) that is 38.0%

6 0
1 year ago
If you try to balance an equation by changing subscripts you change...
aksik [14]

Answer:

kwkrofofoxosowoqoaododpdprofpcoxozoskawkdjdn

Explanation:

sklwlrlfclxoskkekrdododosoekekrkrododowoekekfkdodkwkeororkdkdkwejrjrkfidiwi3jr

8 0
3 years ago
Consider the following equilibrium: 2SO^2(g) + O2(9) = 2 SO3^(g)
saul85 [17]

Answer:

At equilibrium, the forward and backward reaction rates are equal.

The forward reaction rate would decrease if \rm O_2 is removed from the mixture. The reason is that collisions between \rm SO_2 molecules and \rm O_2\! molecules would become less frequent.

The reaction would not be at equilibrium for a while after \rm O_2 was taken out of the mixture.

Explanation:

<h3>Equilibrium</h3>

Neither the forward reaction nor the backward reaction would stop when this reversible reaction is at an equilibrium. Rather, the rate of these two reactions would become equal.

Whenever the forward reaction adds one mole of \rm SO_3\, (g) to the system, the backward reaction would have broken down the same amount of \rm SO_3\, (g)\!. So is the case for \rm SO_2\, (g) and \rm O_2\, (g).

Therefore, the concentration of each species would stay the same. There would be no macroscopic change to the mixture when it is at an an equilibrium.

<h3>Collision Theory</h3>

In the collision theory, an elementary reaction between two reactants particles takes place whenever two reactant particles collide with the correct orientation and a sufficient amount of energy.

Assume that \rm SO_2\, (g) and \rm O_2\, (g) molecules are the two particles that collide in the forward reaction. Because the collision has to be sufficiently energetic to yield \rm SO_3\, (g), only a fraction of the reactions will be fruitful.

Assume that \rm O_2\, (g) molecules were taken out while keeping the temperature of the mixture stays unchanged. The likelihood that a collision would be fruitful should stay mostly the same.

Because fewer \!\rm O_2\, (g) molecules would be present in the mixture, there would be fewer collisions (fruitful or not) between \rm SO_2\, (g) and \rm O_2\, (g)\! molecules in unit time. Even if the percentage of fruitful collisions stays the same, there would fewer fruitful collisions in unit time. It would thus appear that the forward reaction has become slower.

<h3>Equilibrium after Change</h3>

The backward reaction rate is likely going to stay the same right after \rm O_2\, (g) was taken out of the mixture without changing the temperature or pressure.

The forward and backward reaction rates used to be the same. However, right after the change, the forward reaction would become slower while the backward reaction would proceed at the same rate. Thus, the forward reaction would become slower than the backward reaction in response to the change.

Therefore, this reaction would not be at equilibrium immediately after the change.

As more and more \rm SO_3\, (g) gets converted to \rm SO_2\, (g) and \rm O_2\, (g), the backward reaction would slow down while the forward reaction would pick up speed. The mixture would once again achieve equilibrium when the two reaction rates become equal again.

5 0
2 years ago
What do atoms form when they bond chemically?
V125BC [204]
They form molecules which can be in solution form if diluted in water, but some do form solutions on exposure to the atmosphere i.e they are deliquescent like pellets of sodium hydroxide
7 0
3 years ago
What is kind of property is denisty?
Virty [35]

Answer:

INTENSIVE PROPERTY OF MATTER

Explanation:

DENSITY IS AN INTENSIVE PROPERTY OF MATTER THAT ILLUSTRATES HOW MUCH MASS A SUBSTANCE HAS IN A GIVEN AMOUNT OF VALUE,

7 0
2 years ago
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