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4 years ago

The symbol for xenon (xe) would be a part of the noble gas notation for the element

Chemistry

1 answer:

daser333 [38]4 years ago

6 0

Answer:

Explanation:

To write the electronic configuration of an element using the preceding noble gas configuration, we simply use the noble gas in the previous period of that particular element. We do not use the noble gas of that period in which the element belongs to but the one in the preceding group.

A is wrong

This is because the noble gas in the preceding period is krypton.

In fact, the electronic configuration is [Kr] 4d105s25p3

C is wrong

The noble gas in the last period here is Radon. The electronic configuration of Radium in fact is [Rn] 7s2

D is wrong

The noble gas in the last period before the period of uranium is Radon also. In fact, the electronic configuration of the element is [Rn] 5f36d17s2

B is correct, the actual electronic configuration of the element Cesium is [Xe] 6s1

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What happens to the electrons in an ionic bond

asambeis [7]

Answer:

Result of the formation of positive and negative ions.

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3 years ago

Read 2 more answers

1.15 g of a metallic element needs 300 cm3 of oxygen for complete reaction, at 298 K and 1 atm

sashaice [31]

1) Calculate the number of moles of O2 (g) in 300 cm^3 of gas at 298 k and 1 atm

Ideal gas equation: pV = nRT => n = pV / RT

R = 0.0821 atm*liter/K*mol

V = 300 cm^3 = 0.300 liter

T = 298 K

p = 1 atm

=> n = 1 atm * 0.300 liter / [ (0.0821 atm*liter /K*mol) * 298K] = 0.01226 mol

2) The reaction of a metal with O2(g) to form an ionic compound (with O2- ions) is of the type

X (+) + O2 (g) ---> X2O or

2 X(2+) + O2(g) ----> X2O2 = 2XO or

4X(3+) + 3O2(g) ---> 2X2O3

In the first case, 1 mol of metal react with 1 mol of O2(g); in the second case, 2 moles of metal react with 1 mol of O2(g); in the third, 4 moles of X react with 3 moles of O2(g)

So, lets probe those 3 cases.

3) Case 1: 1 mol of metal X / 1 mol O2(g) = x moles / 0.01226 mol

=> x = 0.01226 moles of metal X

Now you can calculate the atomic mass of the hypotethical metal:

1.15 grams / 0.01226 mol = 93.8 g / mol

That does not correspond to any of the metal with valence 1+

So, now probe the case 2.

4) Case 2:

2moles X metal / 1 mol O2(g) = x / 0.01226 mol

=> x = 2 * 0.01226 = 0.02452 mol

And the atomic mass of the metal is: 1.15 g / 0.02452 mol = 46.9 g/mol

That is similar to the atomic mass of titanium which is 47.9 g / mol and whose valece is 2+.

4) Case 3

4 mol meta X / 3 mol O2 = x / 0.01226 => x = 0.01226 * 4 / 3 = 0.01635

atomic mass = 1.15 g / 0.01635 mol = 70.33 g/mol

That does not correspond to any metal.

Conclusion: the identity of the metallic element could be titanium.

5 0

3 years ago

True or false natural elements don't reflect light.

Burka [1]

I think the statement is false. Natural elements do reflect light. Almost everything in this world reflect light. Reflection is when light bounces off an object. Hope this answers the question. Have a nice day.

3 0

3 years ago

How many moles in 30.0 grams of h3po4

jok3333 [9.3K]

Molar mass H₃PO₄ = 98.0 g/mol

1 mole ----- 98.0 g
? mole ------ 30.0 g

moles = 30.0 * 1 / 98.0

= 0.306 moles

hope this helps!

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3 years ago

7. Suppose 1.01 g of iron (III) chloride is placed in a 10.00-mL volumetric flask with a bit of water in it. The flask is shaken

Nana76 [90]

Answer: The molarity of Iron (III) chloride is 0.622 M.

Explanation:

Molarity is defined as the number of moles present in one liter of solution. The equation used to calculate molarity of the solution is:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Or,

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Mass of iron (III) chloride = 1.01 g

Molar mass of iron (III) chloride = 162.2 g/mol

Volume of the solution = 10 mL

Putting values in above equation, we get:

$\text{Molarity of Iron (III) chloride}=\frac{1.01g\times 1000}{162.2g/mol\times 10mL}\\\\\text{Molarity of Iron (III) chloride}=0.622M$

Hence, the molarity of Iron (III) chloride is 0.622 M.

3 0

3 years ago