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yuradex [85]
2 years ago
13

The ksp for silver sulfate (ag2so4) is 1. 2 ✕ 10-5. Calculate the solubility of silver sulfate in each of the following

Chemistry
2 answers:
Allisa [31]2 years ago
8 0

For ksp of silver sulfate (ag2so4) at 1. 2 ✕ 10-5, the solubility of silver sulfate is mathematically given as

x = 5.26*10^{-4}

<h3>What is the solubility of silver sulfate in 0.15 M AgNO3?</h3>

Generally, the equation for the  Chemical reaction is mathematically given as

Ag2SO4 ⇄ 2Ag+  + SO4 2-

Therefore

Ksp = [Ag+ ]2 *[SO4 2-]

1.2*10-5 = ( 0.15 +2x)2*(x)

x = 5.26*10^{-4}

In conclusion, the solubility of silver sulfate is

x = 5.26*10^{-4}

Read more about Chemical reaction

brainly.com/question/16416932

Y_Kistochka [10]2 years ago
3 0

If the ksp of silver sulfate (ag2so4) at 1.2 × 10⁻⁵, the solubility of silver sulfate is  5.26 × 10⁽⁻⁴⁾.

<h3>What is the ksp value?</h3>

ksp value determines the equilibrium between the solids and its respective ions in a solution.

The reaction is \rm Ag_2SO_4 \leftrightharpoons 2Ag+  + SO_4^2-

The ksp pf silver sulfate is  1.2 × 10⁻⁵

Tho calculate the solubility?

Putting the values in the equation

\rm Ksp = [Ag^+ ]^2 \times [SO_4^ 2^-]\\\\1.2\times 10^-^5 = ( 0.15 +2x)2\times (x)\\x = 5.26\times 10^{-4}

Thus, the solubility of silver sulfate is  5.26 × 10⁽⁻⁴⁾.

Learn more about silver sulfate

brainly.com/question/17094293

#SPJ4

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STALIN [3.7K]

Answer:

73.4% is the percent yield

Explanation:

2KClO₃ →  2KCl  + 3O₂

This is a decomposition reaction, where 2 moles of potassium chlorate decompose to 2 moles of potassium chloride and 3 moles of oxygen.

We determine the moles of salt: 400 g . 1. mol /122.5g= 3.26 moles of KClO₃

In the theoretical yield of the reaction we say:

2 moles of potassium chlorate can produce 3 moles of oxygen

Therefore, 3.26 moles of salt, may produce (3.26 . 3) /2 = 4.89 moles of O₂

The mass of produced oxygen is: 4.89 mol . 32 g /1mol = 156.6g

But, we have produced 115 g. Let's determine the percent yield of reaction

Percent yield = (Produced yield/Theoretical yield) . 100

(115g / 156.6g) . 100 = 73.4 %

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The number of glyceraldehyde-3-phosphate molecules that would be produced from 24 turns of the calvin cycle would be
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One product of the Calvin cycle is the glyceraldehyde-3-phosphate (G3P), which is later on used in the production of glucose and in the regeneration of <span>Ribulose 1,5-bisphosphate (RuBP), which is an organic compound</span> essential to the reactions in the cycle. 

One turn of Calvin Cycle produces 2 G3P molecules, each comprising of 3 carbons. This gives a total of 6 carbons. Five (5) of these carbons will be used to regenerate RuBP and only 1 will be available to form a surplus G3P later on. This surplus G3P will be used for the production of glucose (a 6-carbon sugar). 

Thus, 3 turns of the carbon cycle will produce 1 surplus G3P. There are 8 sets of 3-turns in 24 cycles, therefore, 

                       1 net G3P molecule * 8 sets of 3-turns  = 8 G3P molecules

Therefore, there are 8 net or surplus G3P molecules produced for 24 cycles of the Calvin Cycle. The total G3P molecules produced, including the ones that participated in the regeneration of RuBP would be 48 G3Ps. 

For every 3 turns, 6 G3P molecules are produced, 5 of which will be used in the regeneration of RuBP and 1 will be the net or surplus, to be used for the production of glucose. The 48 G3Ps then come from the calculation, 

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The figure below shows the products of the cycle after 3 turns (Source: https://ka-perseus-images.s3.amazonaws.com/2f4bdc8f8275834d3f5ef434d93bf16b991b2357.png). 

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Explanation:

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Answer:

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Oxidation is simply the loss in electrons. A chemical specie that undergoes oxidation is called the reducing agent.

Let us look at the species.

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