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3 years ago

11. What are transition metals?

Chemistry

1 answer:

kolezko [41]3 years ago

8 0

Answer:

Elements that fall between those on the left and right sides of the periodic table

Explanation:

Transition metals:

These are present at the center of periodic table.

These are d-block elements.

They include the elements of group 3 to 12 in periodic table.

They have large charge to radius ratio.

They mostly form paramagnetic compounds.

They shoes more than one oxidation state.

They form colored compounds.

They all have high melting and boiling point.

They have high densities.

They form stable complexes.

The elements of f-block are also transition but they are called inner transition.These are consist of two series lanthanide and actinides.

You might be interested in

What is the mass of 0.55 mole<br> of magnesium chloride?

Gala2k [10]

Answer:

SYMBOLS, FORMULAS AND MOLAR MASSES

OBJECTIVES

1. To correctly write and interpret chemical formulas

2. To calculate molecular weights from chemical formulas

3. To calculate moles from grams using chemical formulas

INTRODUCTION

Part I. Symbols and formulas

An element is a homogeneous pure substance made up of identical atoms. All matter is made

up of elements and, since chemistry is the study of matter, it is convenient to use symbols to represent

the elements rather than using the entire name.

By international agreement, specific symbols are assigned to each element (Note: This means

that while names of the elements vary with language, symbols are constant throughout the world.) Each

element is assigned a one- or two-letter symbol. The first letter is capitalized, the second (if there is

one) is not. While this often seems trivial, it is in fact a very important point. For example, in chemical

language Co represents cobalt, which is a metal and an element, while CO represents carbon monoxide,

a compound which is a colorless, odorless gas! Even when there is not an obvious correspondence,

for instance "MN", it can cause confusion. Do you mean the element manganese? Did you forget a

letter and mean something else? Are you using "M" to represent something else entirely? Chemists

sometimes use "M" to represent any metal. It is well worth the trouble to memorize the symbols for

common elements.

Since compounds consist of elements, the chemical formulas of compounds also consist of

elements with subscripts used to denote the number of atoms per molecule. If there is no subscript, it is

implied that there is one of that kind of atom. Ones never appear in chemical formulas. Not only do

subscripts denote ratios of atoms, they also denote the ratio of moles of element to one mole of

compound. Parentheses can be used to show groups of atoms, with the subscripts showing how many

groups there are. Parentheses are not used if there is only one group.

Examples: For one mole of the following compounds, how many moles of each element are

present?

MgCl2 1 mole Mg, 2 moles Cl

Mg(NO3)2 1 mole Mg, 2 moles N, 6 moles O

NaNO3 1 mole Na, 1 mole N, 3 mole O

AgCl 1 mole Ag, 1 mole ClPart II. Molar Masses

Each atom has a different size and therefore a different mass. The relative masses of each

element can be found on the periodic table. For example, one atom of magnesium weighs 24.31 amu

(atomic mass units). However, one mole of magnesium weighs 24.31 g. (Moles were planned that

way!) Since one mole of MgCl2 consists of one mole of magnesium and two moles of chlorine, the

mass of one mole of MgCl2 must be the sum of the masses of the elements. The mass of one mole of a

substance is called the molar mass or molecular weight.

Examples: What is the molar mass of the following compounds?

MgCl2 24.31 + 2(35.45) = 95.21 g/mol

Mg(NO3)2 24.31 + 2(14.01) + 6(16.00) = 148.33 g/mol

NaNO3 23.00 + 14.01 + 3(16.00) = 85.01 g/mol

AgCl 107.9 + 35.45 = 143.4 g/mol

(Note: Yes! You DO have to count significant figures when calculating molecular weight/molar

mass. However, the number of significant figures may vary depending on which periodic table you use.)

Chemists are generally interested in number of moles. Unfortunately, it is impossible to measure

moles directly. However, masses are easily measured, and if the chemical formula of the compound is

known, the molar mass can be used to determine the number of moles. The molar mass is defined as:

molar mass = grams/moles = g/mol (1)

Moles may be calculated by using molar mass as a conversion factor in dimensional analysis where

molar mass in grams = 1 (exactly) mole of compound (2)

This method is used in multi-step calculations. For example, if 0.873 g of MgCl2 is weighed out, it

is 9.17 x 10-3

moles.

1 mole

0.873g x 95.21 g = 9.17 x 10-3

mol MgCl2 (3)

However, 0.873 g of AgCl is only 6.09 x 10-3

mol.

1 mole

0.873g x 143.4 g = 6.09 x 10-3

mol AgCl (4)Molar mass may also be used to relate moles to grams. For example, 0.158 mol of MgCl2 is 15.2 g.

0.158 mol x 95.21 g = 15.2 g MgCl2 (5)

1 mol

Percent is used to express parts per one hundred. Usually in chemistry, it refers to

g of species of interest x 100 = % (6)

g of whole thing

Example: For the % Mg in MgCl2: In one mole of MgCl2, there are 24.31 g of Mg (molar mass of Mg,

the part we are interested in) and 95.21 g of MgCl2 (the whole thing), so %Mg in MgCl2 is

(24.31/95.21) x 100 = 25.53% Mg (7)

PROCEDURE

Work individually.

The formula for calcium phosphate is Ca3(PO4)2. Weigh about 2 g of calcium phosphate to the

nearest 0.001 g. In other words, you do not have to have exactly 2.000g, but you must know the

weight you have exactly. Acceptable results include but are not limited to: 1.985g , 2.035g, 2.314g

etc.

Be sure to report all results with the correct number of significant figures and appropriate units!

5 0

3 years ago

1. What is an isomer? How many possible isomers of hexane are there? What are the structural differences between these isomers?

VARVARA [1.3K]

Answer 1:
Isomers are compounds with same molecular formula but different structure formula. Isomers are classified into two types
a) Structural/configurational isomers
b) Stereo isomers
In structural/configurational isomers atom and functional groups are attached in different fashion. Structural isomers may have different functional groups. Structural isomers are further classified as chain isomers, position isomers and functional isomers. In case of stereo-isomers, compounds have same functional group, but different orientation in space. They also have difference activity towards polarized light.

Answer 2:
Hexane has a molecular formula of C6H14. It exhibits following structural isomers
a) hexane<span>,
b) 2-methylpentane
c)3-methylpentane
d) 2,2-dimethylbutane
e) 2,3-dimethylbutane
Thus, in all there are 5 isomers of hexane

Answer 3:
</span><span>Butane has two possible isomers but that decane has 75 possible isomers. This can be attributed to the fact that butane has 4 carbon atoms, while decane has 10 carbon atom. As the number of carbon atom increases, there are higher possible sites of linkage, in different fashion. Therefore, as number 69 of carbon atoms increases, number of different possible isomers increases.

Answer 4:
It has been observed that, though isomers have same molecular formula, but the have different boiling points. Infact, branched isomers have lower boiling point as compared to linear isomers. For example, hexane has boiling point = 69 oC, 2 methyl pentane has boiling point = 60 oC, 2,4, dimethyl butane has boiling point = 58 oC and 2,2 dimethyl butane has boiling point = 50 oC. Thus, it can be observed that branched isomers have lower boiling points as compared to linear isomers. This can be attributed to lower van der Waal's forces of interaction in branched isomers as compared to linear isomers.
</span><span>
</span>

5 0

3 years ago

if you were to engineer an everyday solution for conserving water which activity do you think would be the most impactful ?

Ugo [173]

If you were to engineer an everyday solution for conserving water which activity do you think would be the most impactful ?

7 0

2 years ago

How many moles of ethanol are produced starting with 500.g glucose?

Monica [59]

<h3>Answer:</h3>

5.55 mol C₂H₅OH

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Tables
Moles

<u>Stoichiometry</u>

Using Dimensional Analysis
Analyzing Reactions RxN

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂

[Given] 500. g C₆H₁₂O₆ (Glucose)

[Solve] moles C₂H₅OH (Ethanol)

<u>Step 2: Identify Conversions</u>

[RxN] 1 mol C₆H₁₂O₆ → 2 mol C₂H₅OH

[PT] Molar mass of C - 12.01 g/mol

[PT] Molar Mass of H - 1.01 g/mol

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of C₆H₁₂O₆ - 6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol

<u>Step 3: Stoichiometry</u>

[DA] Set up conversion: $\displaystyle 500 \ g \ C_6H_{12}O_6(\frac{1 \ mol \ C_6H_{12}O_6}{180.18 \ g \ C_6H_{12}O_6})(\frac{2 \ mol \ C_2H_5OH}{1 \ mol \ C_6H_{12}O_6})$
[DA} Multiply/Divide [Cancel out units]: $\displaystyle 5.55001 \ mol \ C_2H_5OH$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

5.55001 mol C₂H₅OH ≈ 5.55 mol C₂H₅OH

8 0

3 years ago

Calculate the electric double layer thickness of a alumina colloid in a dilute (0.1 mol/dm3) CsCI electrolyte solution at 30 °C.

Ad libitum [116K]

Explanation:

The given data is as follows.

Concentration = 0.1 $mol/dm^{3}$

= 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions

= $6.022 \times 10^{25} ions/m^{3}$

T = $30^{o}C$ = (30 + 273) K = 303 K

Formula for electric double layer thickness ( $\lambda_{D}$ ) is as follows.

$\lambda_{D}$ = $\frac{1}{k} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}$

where, $n^{o}$ = concentration = $6.022 \times 10^{25} ions/m^{3}$

Hence, putting the given values into the above equation as follows.

$\lambda_{D}$ = $\sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}$

= $\sqrt \frac{78 \times 8.854 \times 10^{-12} c^{2}/Jm \times 1.38 \times 10^{-23}J/K \times 303 K}{2 \times 6.022 \times 10^{25} ions/m^{3} \times (1)^{2} \times (1.6 \times 10^{-19}C)^{2}}$

= $9.669 \times 10^{-10}$ m

or, = $9.7 A^{o}$

= 1 nm (approx)

Also, it is known that $\lambda_{D}$ = $\sqrt \frac{1}{n^{o}}$

Hence, we can conclude that addition of 0.1 $mol/dm^{3}$ of KCl in 0.1 $mol/dm^{3}$ of NaBr " $\lambda_{D}$ " will decrease but not significantly.

7 0

3 years ago