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3 years ago

24. Na or K-Which is bigger in atomic size? The metallic character of an element depend on the

Chemistry

1 answer:

Vinvika [58]3 years ago

3 0

Answer: K is bigger in atomic size. The metallic character increases as the size increases.

Explanation:

Reactivity of a metal is defined as the tendency of an element to lose electrons. The metallic character increases as we move down a group and it decreases as we move across a period.

When the size of an element increases, the valence electrons gets away from the nucleus and the tendency of an element to loose electrons increases.

In a group, the size of an element increases because there is an addition of new shell and electron is added in that shell.

Thus for alkali metals, the metallic character increases from Lithium to sodium to potassium and so on.

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See the questions on the sheet

Sergio [31]

Answer:

Question 1. Nonpolar covalent

Question 2. The fluorine atom is able to attract the shared electrons more strongly than a hydrogen atom

Question 3. True

Explanation:

Question 1

First of all, oxygen is a molecular compound, as it consists of non-metal atoms only (oxygen). This means we wouldn't expect to have any ionic bonding in it, as it doesn't contain a metal ion. A molecular compound has covalent bonding.

Whenever a diatomic molecule contains the same two atoms bonded by a bond, we expect to have a non-polar bond. This is due to the fact that the two atoms are identical and have the same values of electronegativity, meaning the difference in their electronegativity values is 0 and we have no net polarity within the bond.

For a bond between two different atoms, the molecule would be polar, as one atom would have a greater electronegativity (electron withdrawing force) compared to the other atom.

Question 2

Based on the principles of polarity, whenever we have a diatomic molecule, it's only non-polar when the two atoms are the same. In case of HF, we have two different atoms: hydrogen and fluorine. Since the two atoms are not identical, the molecule would be polar overall, as fluorine is more electronegative than hydrogen. Simply speaking, it means that fluorine attracts the shared electrons within the H-F bond stronger than hydrogen does. This makes a difference in electronegativity values between H and F non-zero and an overall polar bond.

Question 3

We may recall the Coulombic force equation. It states that the attraction force is directly proportional to the charge and inversely proportional to the square of a distance between the two charges.

A bond formed between two atoms or ions is the closest distance the two species can approach each other. Intermolecular forces, in contrast, are the forces that atoms experience within a distance greater than the bond length. We may conclude that for a greater distance, the Coulombic force is lower and, hence, the strength of intermolecular forces are significantly lower compared to covalent or ionic bonds.

7 0

3 years ago

Which is an example of a heterogeneous mixture? A coffee B soil C gelatin D air

mariarad [96]

Heterogeneous mixtures are made of different substances that remain physically seperate. An example would be mixing sand and sugar together.
the answer would be soil
The answer is B soil

6 0

3 years ago

Sulfur dioxide has a vapor pressure of 462.7 mm Hg at –21.0 °C and a vapor pressure of 140.5 mm Hg at –44.0 °C. What is the enth

stealth61 [152]

Answer : The value of $\Delta H_{vap}$ is 28.97 kJ/mol

Explanation :

To calculate $\Delta H_{vap}$ of the reaction, we use clausius claypron equation, which is:

$\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$P_1$ = vapor pressure at temperature $-21.0^oC$ = 462.7 mmHg

$P_2$ = vapor pressure at temperature $-44.0^oC$ = 140.5 mmHg

$\Delta H_{vap}$ = Enthalpy of vaporization = ?

R = Gas constant = 8.314 J/mol K

$T_1$ = initial temperature = $-21.0^oC=[-21.0+273]K=252K$

$T_2$ = final temperature = $45^oC=[-41.0+273]K=232K$

Putting values in above equation, we get:

$\ln(\frac{140.5mmHg}{462.7mmHg})=\frac{\Delta H_{vap}}{8.314J/mol.K}[\frac{1}{252}-\frac{1}{232}]\\\\\Delta H_{vap}=28966.6J/mol=28.97kJ/mol$

Therefore, the value of $\Delta H_{vap}$ is 28.97 kJ/mol

4 0

4 years ago

A mixture of 75 mole% methane and 25 mole% hydrogen is burned with 25% excess air. Fractional conversions of 90% of the methane

son4ous [18]

Solution :

Consider a mixture of methane and hydrogen.

Take the basis as 100 moles of the mixture.

The mixture contains 75% of methane and 25% of hydrogen by mole and it is burned with 25% in excess air.

Moles of methane = 0.75 x 100

Moles of hydrogen = 0.25 x 100

The chemical reactions involved during the reaction are :

$CH_4+2O_2 \rightarrow CO_2 + 2H_2O$

$CH_4+1.5O_2 \rightarrow CO+2H_2O$

$H_2+0.5O_2 \rightarrow H_2O$

The fractional conversion of methane is 90%

Number of moles of methane burned during the reaction is = 0.9 x 75

= 67.5

Moles of methane leaving = initial moles of methane - moles of methane burned

= 75 - 67.5

= 7.5 moles

Fractional conversion of hydrogen is 85%

The number of moles of hydrogen burned during the reaction is = 0.85 x 25

= 21.25

Moles of hydrogen leaving = initial moles of hydrogen - moles of hydrogen burned

= 25 - 21.25

= 3.75 moles

Methane undergoing complete combustion is 95%.

$CO_2$ formed is = 0.95 x 67.5

= 64.125 moles

$CO$ formed is = 0.05 x 67.5

= 3.375 moles

Oxygen required for the reaction is as follows :

From reaction 1, 1 mole of the methane requires 2 moles of oxygen for the complete combustion.

Hence, oxygen required is = 2 x 75

= 150 moles

From reaction 3, 1 mole of the hydrogen requires 0.5 moles of oxygen for the complete combustion.

Hence, oxygen required is = 0.5 x 25

= 12.5 moles

Therefore, total oxygen is = 150 + 12.5 = 162.5 moles

Air is 25% excess.

SO, total oxygen supply = 162.5 x 1.25 = 203.125 moles

Amount of nitrogen = $203.125 \times \frac{0.79}{0.21}$

= 764.136 moles

Total oxygen consumed = oxygen consumed in reaction 1 + oxygen consumed in reaction 2 + oxygen consumed in reaction 3

Oxygen consumed in reaction 1 :

1 mole of methane requires 2 moles of oxygen for complete combustion

= 2 x 64.125

= 128.25 moles

1 mole of methane requires 1.5 moles of oxygen for partial combustion

= 1.5 x 3.375

= 5.0625 moles

From reaction 3, 1 mole of hydrogen requires 0.5 moles of oxygen

= 0.5 x 21.25

= 10.625 moles.

Total oxygen consumed = 128.25 + 5.0625 + 10.625

= 143.9375 moles

Total amount of steam = amount of steam in reaction 1 + amount of steam in reaction 2 + amount of steam in reaction 3

Amount of steam in reaction 1 = 2 x 64.125 = 128.25 moles

Amount of steam in reaction 2 = 2 x 3.375 = 6.75 moles

Amount of steam in reaction 3 = 21.25 moles

Total amount of steam = 128.25 + 6.75 + 21.25

= 156.25 moles

The composition of stack gases are as follows :

Number of moles of carbon dioxide = 64.125 moles

Number of moles of carbon dioxide = 3.375 moles

Number of moles of methane = 7.5 moles

Number of moles of steam = 156.25 moles

Number of moles of nitrogen = 764.136 moles

Number of moles of unused oxygen = 59.1875 moles

Number of moles of unused hydrogen = 3.75 moles

Total number of moles of stack gas

= 64.125+3.375+7.5+156.25+764.136+59.1875+3.75

= 1058.32 moles

Concentration of carbon monoxide in the stack gases is

$=\frac{3.375}{1058.32} \times 10^6$

= 3189 ppm

b). The amount of carbon monoxide in the stack gas can be decreased by increasing the amount of the excess air. As the amount of the excess air increases, the amount of the unused oxygen and nitrogen in the stack gases will increase and the concentration of CO will decrease in the stack gas.

6 0

3 years ago

How many chlorine atoms are in each set? six calcium chloride formula units.

Aleonysh [2.5K]

Answer:

tae ko malutong

Explanation:

dahil utong ko maitim

3 0

3 years ago