How many molecules are in 3.40L of H2O vapor at STP?*

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

trapecia [35]

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

For writing the electronic configuration of any element by using the preceding noble gas configuration, we simply use the symbols of noble gas belongs to the previous period of that particular elements. We can't use the symbol of noble gas of same period from which the element belong.

A is the wrong option because the noble gas in the preceding period to the period from which antimony belongs is krypton.

The actual electronic configuration of antimony is as follow:

[Kr] 4d10 5s2 5p3

B is correct option because the noble gas in the preceding period to the period from which Cesium belongs is Xenon.

The actual electronic configuration of Cesium is as follow:

[Xe] 6s1

Thus, we concluded that the symbol for xenon (xe) would be a part of the noble gas notation for the element cesium.

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8 0

1 year ago

PLEASE HELP!!!!! Calcium hydroxide is _____. an acid a base a neutral

Bingel [31]

I believe it is a base!

4 0

2 years ago

Read 2 more answers

What volume will 0.405 g of krypton gas occupy at STP?

Rufina [12.5K]

Answer:

The answer to your question is V = 0.108 L or 108 ml

Explanation:

Data

Volume = ?

mass = 0.405 g

Temperature = 273°K

Pressure = 1 atm

Process

1.- Convert mass of Kr to moles

83.8 g of Kr -------------------- 1 mol

0.405 g ------------------- x

x = (0.405 x 1) / 83.8

x = 0.0048 moles

2.- Use the Ideal gas law to solve this problem

PV = nRT

- Solve for V

V = nRT / P

- Substitution

V = (0.0048)(0.082)(273) / 1

- Simplification

V = 0.108 / 1

- Result

V = 0.108 L

8 0

2 years ago

Read 2 more answers

If 100 mg of ferrocene is reacted with 75 mg of anhydrous aluminum chloride and 40 microliters of acetyl chloride and 100 mg of

Alex_Xolod [135]

Answer:

81.3 %

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

For ferrocene:-

Mass of ferrocene = 100 mg = 0.1 g

Molar mass of ferrocene = 186.04 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{0.1\ g}{186.04\ g/mol}$

$Moles\ of\ ferrocene= 0.0005375\ mol$

For acetyl chloride:-

Volume = 40 microliters = 0.04 mL

Density = 1.1 g / mL

Density is defined as:-

$\rho=\frac{Mass}{Volume}$

or,

$Mass={\rho}\times Volume=1.1\times 0.04\ g=0.044 g$

Mass of acetyl chloride = 0.044 g

Molar mass of acetyl chloride = 78.49 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{0.044\ g}{78.49\ g/mol}$

$Moles\ of\ acetyl\ chloride= 0.0005606\ mol$

As per the reaction stoichiometry, one mole of ferrocene reacts with one mole of acetyl chloride to give one mole of monoacetylferrocene

Limiting reagent is the one which is present in small amount. Thus, ferrocene is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

one mole of ferrocene on reaction forms one mole of monoacetylferrocene

0.0005375 mole of ferrocene on reaction forms 0.0005375 mole of monoacetylferrocene

Moles of product formed = 0.0005375 moles

Molar mass of monoacetylferrocene = 228.07 g/mole

Mass of monoacetylferrocene produced = Moles*molecular weight = 0.0005375*228.07 g = 0.123 grams = 123 mg

Given experimental yield = 100 mg

% yield = (Experimental yield / Theoretical yield) × 100 = (100/ 123) × 100 = 81.3 %

5 0

3 years ago

Use the chemical reaction: 2Mg + O2 > 2MgO to answer the question below.

Arturiano [62]

10 mols of MgO will form

5 0

2 years ago