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Novay_Z [31]
2 years ago
11

The chemical formula for sodium citrate is Nag Cs H5O7

Chemistry
1 answer:
Yuliya22 [10]2 years ago
3 0

Answer:

C. Sodium citrate is a compound with a total of 21 atoms

Explanation:

Here's to help reassure you that the answer is correct, look at the picture

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A bag of fertilizer is labeled 10-20-20. What is the actual percentage of phosphorus in the fertilizer?
aleksandr82 [10.1K]

Answer:

20% of phosphorus

Explanation:

A fertilizer is used to improve the fertility of soils. Most fertilizers contains the element nitrogen, phosphorus and potassium.

They are often designated NPK fertilizers.

  Now we know that the numbers 10-20-20 depicts the nitrogen-phosphorus and potassium content of the fertilizer.

From the designation,

    The actual percentage is 20% of phosphorus.

                                               10%  of nitrogen

                                                20% of potassium

8 0
3 years ago
PLZ HELP I HAVE 5 MIN!! how do waves transport energy?
nataly862011 [7]

Answer:

'Wave' is a common term for a number of different ways in which energy is transferred: In electromagnetic waves, energy is transferred through vibrations of electric and magnetic fields. In sound waves, energy is transferred through vibration of air particles or particles of a solid through which the sound travels.

Explanation:

5 0
2 years ago
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Which statements correctly describe thermal energy in a block of ice, a glass of liquid water, and a container of steam?
DanielleElmas [232]

Answer:

Steam always has more thermal energy than ice.

Explanation:

6 0
3 years ago
Dichotomous Keys are used to identify organisms based on their observed characteristics. Dichotomous means..
BartSMP [9]

Dichotomous keys consist of a series of statements with two choices in each step that will lead users to the correct identification. So I believe your answer would be A.

7 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
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