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3 years ago

Of the elements fe li te u and he which are considered group b elements

Chemistry

2 answers:

kramer3 years ago

7 0

Only fe (iron) is in the group b of elements

givi [52]3 years ago

5 0

Answer: Iron is considered as a Group B element.

Explanation:

Mendeleev divided the periodic table into 8 groups, each divided into two sub-groups named as A and B. He arranged the elements according to their atomic masses.

Group A elements are considered as the elements which are present in s-block and p-block.

Group B elements are the transition elements. Transition elements are the elements that belong to d-block.

From the given elements:

Lithium (Group 1-A) and helium (Group VIII-A) are s-block elements.

Tellurium is a p-block (Group VI-A)element.

Uranium is a f-block element.

Iron is the d-block (Group VIII-B)element (transition element)

Hence, iron is considered as a Group B element.

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Only some particles split up into smaller particles

Makovka662 [10]

Answer:

Everything around you can be broken down into smaller particles called atoms. The particles of one substance are all the same and different substances are made up of different particles.

Explanation:

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2 years ago

How do I solve the following word problem: The half-life of Phosphorus - 32 is 14.3 days. It is used to study a plant's use of f

BaLLatris [955]

Half life is the time that it takes for half of the original value of some amount of a radioactive element to decay.

We have the following equation representing the half-life decay:

$A=A_o\times2^{(-\frac{t}{t_{half}})_{}_{}}$

A is the resulting amount after t time

Ao is the initial amount = 50 mg

t= Elapsed time

t half is the half-life of the substance = 14.3 days

We replace the know values into the equation to have an exponential decay function for a 50mg sample

$A=\text{ 50 }\times2^{\frac{-t}{14.3}}$

That would be the answer for a)

To know the P-32 remaining after 84 days we have to replace this value in the equation:

$\begin{gathered} A=\text{ 50 }\times2^{\frac{-84}{14.3}} \\ A=0.85\text{ mg} \end{gathered}$

So, after 84 days the P-32 remaining will be 0.85 mg

4 0

11 months ago

Chemoautotrophs use ___ as an energy source and ___ as a carbon source. select one:

Agata [3.3K]

The answer is E.
Chemoautotrophs are organisms that use inorganic compounds as an energy source and CO2 as a carbon source. The organisms live in hostile environments like dark depths of the sea. Inorganic energy sources that they use are mostly hydrogen sulfide, elemental sulfur, ferrous iron, molecular hydrogen, and ammonia.

3 0

2 years ago

Read 2 more answers

Which of the following is the correctly balanced chemical equation for the reaction of Ca(OH)2 and HNO3?

sattari [20]

Explanation:

The reaction between calcium hydroxide and nitric acid is as follows.

$Ca(OH)_{2} + HNO_{3} \rightarrow Ca(NO_{3})_{2} + H_{2}O$

Number of reactant atoms are as follows.

Ca = 1
O = 4
H = 3
N = 1

Number of product atoms are as follows.

Ca = 1
O = 4
H = 2
N = 3

To balance the given chemical equation, multiply $HNO_{3}$ by 2 on reactant side and multiply $H_{2}O$ by 2 on the product side.

Therefore, the balanced chemical equation will be as follows.

$Ca(OH)_{2} + 2HNO_{3} \rightarrow Ca(NO_{3})_{2} + 2H_{2}O$

4 0

3 years ago

Noble gas compounds like KrF, XeCl, and XeBr are used in excimer lasers. Draw an approximate molecular orbital diagram appropria

Aneli [31]

Answer:

Here's what I get.

Explanation:

The MO diagrams of KrBr, XeCl, and XeBr are shown below.

They are similar, except for the numbering of the valence shell orbitals.

Also, I have drawn the s and p orbitals at the same energy levels for both atoms in the compounds. That is obviously not the case.

However, the MO diagrams are approximately correct.

The ground state electron configuration of KrF is

$(1\sigma_{g})^{2}\, (1\sigma_{u}^{*})^{2} \, (2\sigma_{g})^{2} \, (2\sigma_{u}^{*})^{2} \, (3\sigma_{g})^{2} \, (1\pi_{u})^{4} \, (1\pi_{g}^{*})^{4} \, (3\sigma_{g}^{*})^{1}$

KrF⁺ will have one less electron than KrF.

You remove the antibonding electron from the highest energy orbital, so the bond order increases.

The KrF bond will be stronger.

6 0

3 years ago