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Tems11 [23]
3 years ago
14

What is the theoretical yield of calcium phosphate?

Chemistry
1 answer:
BARSIC [14]3 years ago
5 0
  <span>I'm going to assume that the other reactant is sodium phosphate(you did not indicate this) 
so i'm guessing that the reaction did this 
CaCL2 + Na3(PO4) ------------> Ca3(PO4)2 + NaCl 
(since this reaction was done in water(aqueous solutions involved) the NaCl would be in the solution and thus the filtrate of the product so it won't influence the mass of the calcium phosphate product) 

so then you need to write the balanced equation for this reaction(lets assume you can do this) 
3 CaCL2 + 2 Na3(PO4) ------------> Ca3(PO4)2 + 6 NaCl 

so in theoretical analysis 3 moles of calcium chloride is involved in the production of 1 mole of calcium phosphate 

in terms of molar mass equivalents 3 molesw of calcium chloride weigh = 3 x (40 + 71)g/mol = 333g 
1 mole of calcium phosphate = 310 g 
so if you only have 0.513 g of calcium chloride then this would calculate to get the mass of calcium phosphate  
mass = [0.513 / 333] x 310 g = 0.477grams  

your actual yield is what you massed when you performed the lab. hope that helped.
</span>
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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.

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3 years ago
Which cell part contains direction for cell growth and reprodution?
DedPeter [7]

Answer:

Nucleus

Explanation:

Nucleus is your blueprint for the cell. It has all the directions packaged in tiny DNA molecules. The nucleus of a cell is an organelle that stores the cell's hereditary material, or DNA, and it coordinates the cell's activities, which include growth, intermediary metabolism, protein synthesis, and reproduction.

7 0
3 years ago
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Two sound waves are emitted from identical sources at the same time. They each travel a distance of 10 m, but one wave reaches t
kompoz [17]

Answer:

<em>Varying frequency</em> between both waves accounts for difference in speed.

Explanation:

The speed of a wave is dependent on four major factors:

  1. wavelength
  2. frequency
  3. medium, and
  4. temperature 

Assuming equal temperature and medium of travel of these sound waves, and given that the wavelength (that is distance of travel) is equal, the only varying factor would be their frequency.

Wave speed is calculated by multiplying the wavelength times the frequency

⇒ Speed = λ * <em>f</em>

4 0
3 years ago
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At 1000 K, Kp=19.9 for the reaction Fe2O3(s)+3CO(g)&lt;---&gt;2Fe(s)+3CO2(g) What are the equilibrium partial pressures of CO an
SOVA2 [1]

<u>Answer:</u> The equilibrium concentration of CO is 0.243 atm

<u>Explanation:</u>

We are given:

Initial partial pressure of carbon dioxide = 0.902 atm

As, carbon dioxide is present initially. This means that the reaction is proceeding backwards.

For the given chemical equation:

                      Fe_2O_3(s)+3CO(g)\rightleftharpoons 2Fe(s)+3CO_2(g)

<u>Initial:</u>                                                                  0.902

<u>At eqllm:</u>                            3x                           (0.902-3x)

The expression of K_p for above equation follows:

K_p=\frac{(p_{CO_2})^3}{(p_{CO})^3}

We are given:

K_p=19.9

Putting values in above equation, we get:

19.9=\frac{(0.902-3x)^3}{(3x)^3}\\\\x=0.0810

So, equilibrium concentration of CO = 3x = (3 × 0.0810) = 0.243atm[/tex]

Hence, the equilibrium concentration of CO is 0.243 atm

6 0
3 years ago
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