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Tanzania [10]
3 years ago
10

The atomic weight of iodine is less than the atomic weight of tellurium. However, Mendeleev listed iodine after tellurium in his

original periodic table because he suspected the atomic weights for these elements were inaccurate. What was Mendeleev's reason for this suspicion?
Chemistry
1 answer:
QveST [7]3 years ago
3 0

Explanation :

As we know that Mendeleev arranged the elements in horizontal rows and vertical columns of a table in order of their increasing relative atomic weights.

He placed the elements with similar nature in the same group.

According to the question, the atomic weight of iodine is less than the atomic weight of tellurium. So according to this, iodine should be placed before tellurium in Mendeleev's tables. But Mendeleev placed iodine after tellurium in his original periodic table.

However, iodine has similar chemical properties to chlorine and bromine. So, in order to make iodine queue up with chlorine and bromine in his periodic table, Mendeleev exchanged the positions of iodine and tellurium.

As we know that the positions of iodine and tellurium were reversed in Mendeleev's table because iodine has one naturally occurring isotope that is iodine-127  and tellurium isotopes are tellurium-128 and tellurium-130.

Due to high relative abundance of tellurium isotopes gives tellurium the greater relative atomic mass.

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The terms Q and K refer to reaction components at non-equilibrium and equilibrium conditions, respectively. For a forward reacti
daser333 [38]

Answer:

The value of Q must be less than that of K.

Explanation:

The difference of K and Q can be understood with the help of an example as follows

         A ⇄ B

In this reaction A is converted into B but after some A is converted , forward reaction stops At this point , let equilibrium concentration of B be [B] and let equilibrium concentration of A be [A]

In this case ratio of  [B] and  [A]  that is

K =  [B] / [A] which is called equilibrium constant.

But if we measure the concentration of A and B ,before equilibrium is reached , then the ratio of the concentration of A and B will be called Q. As reaction continues concentration of A increases and concentration of B decreases. Hence Q tends to be equal to K.

Q = [B] / [A] . It is clear that Q < K before equilibrium.

If Q < K , reaction will proceed towards equilibrium or forward reaction will

proceed .

8 0
3 years ago
3 A soil has 70% sand, 20% silt
Kisachek [45]

Today, as part of the series of posts on soils, we are going to look at ‘soil texture’. Soil forms the basis for all life but it’s important to know about its mineral constitution as well as its biological profile.

Texture refers to the ‘feel’ of the soil. This is affected by the constituent materials found within it, specifically sand, silt and clay particles. A coarse sand will feel gritty but a wet clay will feel heavy and sticky. The texture of a soil has a direct impact on the way the soil reacts to certain environmental conditions – for example, towards drought or heavy rain (with sandy soils more freely draining).

There is a big difference in the size of the different particles.

Coarse sand = diameter 2-0.2mm

Fine sand = diameter 0.2-0.02mm

Silt = diameter 0.02-0.002mm

Clay = diameter less than 0.002mm

Note how the clay particles are much smaller than the sand particles – this is important as it means the total surface area of a clay soil is much greater and so the capacity to hold water is also much greater.

Between the sand, silt and clay particles there are lots of pores. In fact a soil as a whole is generally 45% mineral, 5% organic matter (depending on the soil) and 50% pore space through which air and water can pass.

Sand –

Made up of weathered primary rock minerals.

The particles are irregular in outline.

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Large pore spaces in between.

Air gets in very easily and water flows rapidly through it.

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7 0
3 years ago
Why do different molecules require different masses to make 1 mol
Wewaii [24]

Answer:

Different substances have different molecular masses. Thus, equal masses have different numbers of atoms, molecules, or moles. On the other hand, equal numbers of moles of different substances have different masses.

3 0
3 years ago
A student needs to prepare 50.0 mL of a 1.20 M aqueous H2O2 solution. Calculate the volume of 4.9 M H2O2 stock solution that sho
Nonamiya [84]

Answer : The volume of 4.9 M H_2O_2 stock solution used to prepare the solution is, 12.24 ml

Solution : Given,

Molarity of aqueous H_2O_2 solution = 1.20 M = 1.20 mole/L

Volume of aqueous H_2O_2 solution = 50.0 ml = 0.05 L

(1 L = 1000 ml)

Molarity of H_2O_2 stock solution = 4.9 M = 4.9 mole/L

Formula used :

M_1V_1=M_2V_2

where,

M_1 = Molarity of aqueous H_2O_2 solution

M_2 = Molarity of H_2O_2 stock solution

V_1 = Volume of aqueous H_2O_2 solution

V_2 = Volume of H_2O_2 stock solution

Now put all the given values in this formula, we get the volume of H_2O_2 stock solution.

(1.20mole/L)\times (0.05L)=(4.9mole/L)\times V_2

By rearranging the term, we get

V_2=0.01224L=12.24ml

Therefore, the volume of 4.9 M H_2O_2 stock solution used to prepare the solution is, 12.24 ml

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