Ask question

Ask question

All categories

3 years ago

14

Arrange the following elements in order of decreasing atomic radius: Ba, Sn, S, Pb, and As. Rank elements from largest to smalle

st.

1 answer:

vodka [1.7K]3 years ago

8 0

Answer:

Ba>Sn>Pb>As>S

Explanation:

Hello,

In this case, we can write the atomic radius for the given elements:

$r_{Ba}=268pm\\r_{Sn}=225pm\\r_{S}=100pm\\r_{Pb}=202pm\\r_{As}=185pm$

In such a way, the required order is:

Ba>Sn>Pb>As>S

This is also in agreement as a periodic trend which states that the higher the period, the higher the radius.

Regards.

You might be interested in

What is the mole ratio of H2O to C8H8?

zepelin [54]

Answer:

Mole ratio for a compound

The chemical formula tells us the mole ratio.

CO2 = 1 CO2 molecule : 1 C atom : 2 O atoms.

Mole ratio for a reaction

The balanced chemical reactions tells us.

C12H22O11 + 12 O2  12 CO2 + 11 H2O

1 C12H22O11 molecule: 12 O2 molecules : 12 CO2

molecules : 11 H2O molecules.

Applications of the mole ratio concept

grams <--> moles <--> moles <--> grams

Explanation:

4 0

3 years ago

To learn more about the genetic material of plant and animal cells, where would a person look?

Agata [3.3K]

The person would look B) in the nucleus.

Hope this helps!

-Payshence xoxo

6 0

3 years ago

How many moles of carbon, hydrogen, and oxygen are present in a 100-g sample of ascorbic acid?

Y_Kistochka [10]

There are:

3.41 moles of C

4.54 moles of H

3.40 moles of O.

Why?

To solve the problem, the first thing that we need to do is to write the chemical formula of the ascorbic acid.

$C_{6}H_{8}O_{6}$

Now, we know that there are 100 grams of the compound, so, the masses of each element will represent the percent in the compound.

We have that:

$C_{6}=12.0107g*6=72.08g\\\\H_{8}=1.008g*8=8.064g\\\\O_{6}=15.999g*6=95.994g\\\\C_{6}H_{8}O_{6}=72.08g+8.064g+95.994g=176.138g$

To know the percent of each element, we need to to the following:

$C=\frac{72.08g}{176.138g}*100=0.409*100=40.92(percent)\\\\H=\frac{8.064g}{176.138g}*100=4.58(percent)\\\\O=\frac{95.994}{176.138g}*100=54.49(percent)$

So, we know that for the 100 grams of the compound, there are:

40.92 grams of C

4.58 grams of H

54.49 grams of O

We know the molecular masses of each element:

$C=12.0107\frac{g}{mol}\\\\H=1.008\frac{g}{mol}\\\\O=15.999\frac{g}{mol}{mol}$

Now, to calculate the number of moles of each element, we need to divide the mass of each element by the molecular mass of each element:

$C=\frac{40.92g}{12.010\frac{g}{mol}}=3.41mol\\\\H=\frac{4.58g}{1.008\frac{g}{mol}}=4.54mol\\\\O=\frac{54.49g}{15.999\frac{g}{mol}}=3.40mol$

Hence, we have that there are 3.41 moles of C, 4.54 moles of H, and 3.40 moles of O.

Have a nice day!

5 0

3 years ago

Identify the balanced chemical equation that represents a single displacement reaction.

UNO [17]

3H₂SO₄ + 2Al₂(SO₄)₃ → Al₂(SO₄)₃ + 3H₂ is the balanced chemical equation that represents a single displacement reaction.

<h3>What is a balanced chemical equation?</h3>

The equation in which the number of atoms of all the molecules is equal on both sides of the equation is known as a balanced chemical equation.

3H₂SO₄ + 2Al₂(SO₄)₃ → Al₂(SO₄)₃ + 3H₂

In this type of reaction, one substance is replacing another:

A + BC → AC + B

In a single displacement reaction, atoms replace one another based on the activity series. Elements that are higher in the activity series. Also, if the element that is to replace the other in a compound is more reactive the reaction will occur. If it is less reactive, there will be no reation.

In the first equation, fluorine is more reactive than bromine. Therefore, bromine cannot replace bromine.

In the second equation, the displacement is between hydrogen and aluminium. Hydrogen is lower in the activity series, this implies that aluminium will replace it.

Hence, 3H₂SO₄ + 2Al₂(SO₄)₃ → Al₂(SO₄)₃ + 3H₂ is the balanced chemical equation that represents a single displacement reaction.

Learn more about the displacement reaction here:

brainly.com/question/15052184

#SPJ1

8 0

2 years ago

g Ammonia has been studied as an alternative "clean" fuel for internal combustion engines, since its reaction with oxygen produc

Gala2k [10]

$\text{Ammonia has been studied as an alternative "clean" fuel for internal combustion}$

$\text{engines, since its reaction with oxygen produces only nitrogen and water vapor,}$

$\text{and in the liquid form it is easily transported. An industrial chemist studying this}$

$\text{reaction fills a} \ \mathbf{100 \ L }\ \text{tank with} \ \mathbf{8.6 \ mol} \ \text{of ammonia gas and} \ \mathbf{28 \ mol} \ \ \text{of oxygen gas, }$

$\text{to be} \ \mathbf{2.6\ mol} \ .\ \text{Calculate the concentration equilibrium constant for the combustion of}$

$\text{ammonia at the final temperature of the mixture. Round your answer to 2 significant digits.}$

Answer:

Explanation:

From the correct question above:

The reaction can be represented as:

$\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }$

From the above reaction; the ICE table can be represented as:

$\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }$

I (mol/L) 0.086 0.28 0 0

C -4x -3x +2x +6x

E 0.086 - 4x 0.28 - 3x +2x +6x

At equilibrium;

The water vapor = $\dfrac{2.6 \ mol}{100 \ L} = 6x$

$x = \dfrac{2.6}{100} \times \dfrac{1}{6}$

$x = 0.00433$

$\text{equilibrium constant} ({k_c}) = \dfrac{ [N_2]^2 [H_2O]^6 }{ [[NH_3]^4] [O_2]^3 }$

$\implies \dfrac{(2x)^2 (6x)^6}{(0.086-4x)^4\times (0.28-3x)^3} \\ \\$

Replacing the value of x, we have:

$K_c = \dfrac{4 \times 46,656 \times x^8}{(0.086-4x)^4\times (0.28 -3x)^3} \\ \\ K_c = \dfrac{4 \times 46656 \times (0.00433)^8}{(0.06868)^4(0.26701)^3} \\ \\ K_c = \mathbf{5.4446 \times 10^{-8}}$

$K_c = \mathbf{5.5 \times 10^{-8} \ to \ 2 \ significant \ figures}$

5 0

3 years ago