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3 years ago

Which option correctly orders the elements by increasing reactivity? Hint: Review your periodic table.

Chemistry

1 answer:

myrzilka [38]3 years ago

5 0

Answer is: Li < Na < K.

Lithium (Li), sodium (Na) and potassium (K) are alkaline metals.

Alkaline metals (far left in main group) have lowest ionizations energy and easy remove valence electrons (one electron).

Potassium has lowest ionization energy and it is most reactive, because valcense electrons is farthest from nucleus.

Electron configuration of lithium atom: ₃Li 1s² 2s¹.

Electron configuration of sodium atom: ₁₁Na 1s² 2s² 2p⁶ 3s¹.

Electron configuration of potassium is: ₁₉K 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹.

The ionization energy (Ei) is the minimum amount of energy required to remove the valence electron, when element lose electrons, oxidation number of element grows (oxidation process).

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Use VSEPR theory to predict the molecular

sergey [27]

Answer:

1. trigonal-planar

Explanation:

NO3- (nitrate) has three domains around it and each oxygen pushes against each other creating a planar structure.

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2 years ago

Suppose that 7.25 x 10^22 atoms of a hypothetical element have a mass of 3.88 g. What would be the molar mass (g/mol) of this el

S_A_V [24]

Answer:

32.23 to 4 significant figures.

Explanation:

The molar mass of the element is the mass of 6.022 * 10^23 atoms (Avogadro's number).

So by proportion it is 6.022 * 10^23 * 3.88 / 7.25 * 10^22

= 32.23 to 4 significant figures.

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4 years ago

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strojnjashka [21]

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3 years ago

Read 2 more answers

A) Combustion analysis of toluene, a common organic solvent, gives 5.86 mg of CO2 and 1.37 mg of H2O. If the compound contains o

Mumz [18]

Answer:

For a: The empirical formula for the given compound is $CH$

For b: The empirical and molecular formula for the given organic compound are $C_{10}H_{20}O$

Explanation:

For a:

The chemical equation for the combustion of hydrocarbon follows:

$C_xH_y+O_2\rightarrow CO_2+H_2O$

where, 'x', and 'y' are the subscripts of Carbon and hydrogen respectively.

We are given:

Conversion factor used: 1 g = 1000 mg

Mass of $CO_2=5.86mg=5.86\times 10^{-3}g$

Mass of $H_2O=1.37mg=1.37\times 10^{-3}g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in $5.86\times 10^{-3}g$ of carbon dioxide, $\frac{12}{44}\times 5.86\times 10^{-3}=1.60\times 10^{-3}g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in $1.37\times 10^{-3}g$ of water, $\frac{2}{18}\times 1.37\times 10^{-3}=0.152\times 10^{-3}g$ of hydrogen will be contained.

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.60\times 10^{-3}g}{12g/mole}=0.133\times 10^{-3}moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.152\times 10^{-3}g}{1g/mole}=0.152\times 10^{-3}moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $0.133\times 10^{-3}$ moles.

For Carbon = $\frac{0.133\times 10^{-3}}{0.133\times 10^{-3}}=1$

For Hydrogen = $\frac{0.152\times 10^{-3}}{0.133\times 10^{-3}}=1.14\approx 1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H = 1 : 1

Hence, the empirical formula for the given compound is $CH$

For b:

The chemical equation for the combustion of menthol follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2$ = 0.2829 g

Mass of $H_2O$ = 0.1159 g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 0.2829 g of carbon dioxide, $\frac{12}{44}\times 0.2829=0.077g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 0.1159 g of water, $\frac{2}{18}\times 0.1159=0.013g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.1005) - (0.077 + 0.013) = 0.105 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.077g}{12g/mole}=0.0064moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.013g}{1g/mole}=0.013moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.0105g}{16g/mole}=0.00065moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00065 moles.

For Carbon = $\frac{0.0064}{0.00065}=9.84\approx 10$

For Hydrogen = $\frac{0.013}{0.00065}=20$

For Oxygen = $\frac{0.00065}{0.00065}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 10 : 20 : 1

The empirical formula for the given compound is $C_{10}H_{20}O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 156 g/mol

Mass of empirical formula = 156 g/mol

Putting values in above equation, we get:

$n=\frac{156g/mol}{156g/mol}=1$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(1\times 10)}H_{(1\times 20)}O_{(1\times 1)}=C_{10}H_{20}O$

Hence, the empirical and molecular formula for the given organic compound are $C_{10}H_{20}O$

3 0

4 years ago

In salt, what is the nature of the bond between sodium and chlorine? in salt, what is the nature of the bond between sodium and

kifflom [539]

A chemical bond that involves the transfer of electrons from a metal to a nonmetal and thus forms ions and the atoms in the molecule are attracted towards each other through electrostatic force of attraction are said be ionic bond. In short, the bond formed between the two oppositely charged ions.

In salt, the compound formed between sodium and chlorine is sodium chloride having molecular formula $NaCl$ where sodium is a metal and chlorine is the non-metal. The formation of the compound takes place by transfer of an electron from sodium thus forming a cation of formula $Na^{+}$ to chlorine which gains electron and thus results in an anion formation having formula $Cl^{-}$ . The force of attraction between these oppositely charged ions that helds them together and results in formation of $NaCl$ compound is electrostatic force of attraction.

Hence in salt, the nature of bond between sodium and chlorine is ionic.

4 0

4 years ago