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Shkiper50 [21]
2 years ago
15

Oxygen and nitrogen are both gases at room temperature. Explain why oxygen has a greater force of attraction between its particl

es than nitrogen. (Room temperature is 20°C) please help
Chemistry
1 answer:
RideAnS [48]2 years ago
8 0

Answer:

Explanation:

Oxygen molecules and Nitrogen molecules forms in a very similar way. The attraction between particles of oxygen is great due to its very high electronegativity value. Oxygen has a higher electronegative value compared to nitrogen.

Electronegativity of an atom is the relative tendency with which atoms of an element attracts valence electrons in a chemical bond. Valence electrons are used in forming chemical bonds. They can be transferred from one atom to the other or they can be shared.

Oxygen is the second most electronegative atom on the periodic table. To form a bond, it shares the valence electrons in order for its octet to be complete. Pull for the valence electrons between the contributing atoms is very strong due to their large electronegative values. This pull is stronger compared to that between nitrogen atoms.

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What's the while compound called knowing all those lines are carbon​
Kisachek [45]

Answer:

<u>2-chlorohexane</u>

Explanation:

<u>In this figure</u> :

  • There are 6 carbon atoms
  • The Cl atom is bonded to the 2nd carbon atom

⇒ The Cl is a substituent group, termed as -chloro

⇒ Based on IUPAC nomenclature, the 6 atom chain starts with hex

⇒ There are only single bonds present, so it is an alkane

<u>The name is</u> :

  • <u>2-chlorohexane</u>
6 0
2 years ago
For question numbers 1 and 2, two statements are given - one labelled
Svetllana [295]

Answer:

(iv) (A) is false, but (R) is true.

Explanation:

It is not true that carbon has a strong tendency to either lose or gain electrons to attain noble gas configuration. Carbon is a member of group 14, it is the first member of the group and carbon is purely a non metal. Only metals metals can loose electrons to attain the noble gas configuration. Moreover, carbon does not participate in ionic bonding so it does not gain electrons to attain the noble gas configuration.

However, carbon participates in covalent bonding where it is covalently bonded to four other chemical species using its four outermost electrons. Carbon forms covalent bonds in which four electrons are shared with other chemical species.

5 0
3 years ago
What is the concentration of a solution containing 1.11 g sugar (sucrose, C12H22O11, MW = 342.3 g/mol, d = 1.587 g/cm3) in 432 m
djyliett [7]

Answer:

0.0075 M

0.0060 m

Explanation:

Our strategy here is to use the definition of molarity and molality to solve this question.

The molarity is the number of moles of solute, sucrose in this case, per liter of solution.

The molality is the number of moles of solute per kilogram of solvent.

So the molarity of the  solution is

M = moles of solute/ V solution

As we see we need the volume of solution since we are only given the volume of solvent, but this will be easy to compute since we have the density of  sucrose.

So determine the moles of sucrose , and the volume of solution:

Moles sucrose = 1.11 g/342.3 g/mol = 3.24 x 10⁻³ M

Volume of solution = Vol Sucrose + Vol glycerine

d = m/V ⇒ Vsucrose = m / d = 1.11 g/ 1.587 g/cm³ = 0.70 cm³

Vol solution = 432 mL + 0.70 mL = 432.7 mL  (1cm³  = 1 mL)

Vol solution = 432.7 mL x 1 L / 1000 mL = 0.4327 L

⇒ M = 3.2 x 10⁻³  mol / 0.4327 L = 0.0075  M

For the molarity what we need is to first calculate the kilograms of glycerine from the given density:

d = m/v ⇒ m = d x v = 1.261 g/cm³ x  432 cm³ = 544.75 g

Converting to Kg:

544.75 g x 1 Kg/ 1000 g = 0.544 kg

Now the molality is

m = mol sucrose/ kg solvent = 3.24 x 10⁻³ mol / 0.544 Kg = 0.0060 m

Note: In the calculation for  volume of solution we could have approximated it to that of just glycerine, but since the density of sucrose was given we calculated the total volume of solution to be more rigorous.

8 0
3 years ago
Please help me with these two
Anna [14]
They exist in the outer orbitals
3 0
3 years ago
In Part A, we saw that the theoretical yield of aluminum oxide is 1.60 mol . Calculate the percent yield if the actual yield of
garri49 [273]

Taking into account definition of percent yield, the percent yield for the reaction is 76.25%.

<h3>Percent yield</h3>

The percent yield is the ratio of the actual return to the theoretical return expressed as a percentage.

The percent yield is calculated as the experimental yield divided by the theoretical yield multiplied by 100%:

percent yield=\frac{actual yield}{theorical yield}x100

where the theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.

<h3>Percent yield in this case</h3>

In this case, you know:

  • actual yield= 1.22 mol
  • theorical yield= 1.60 mol

Replacing in the definition of percent yields:

percent yield=\frac{1.22 mol}{1.60 mol}x100

Solving:

<u><em>percent yield= 76.25%</em></u>

Finally, the percent yield for the reaction is 76.25%.

Learn more about percent yield:

brainly.com/question/14408642

#SPJ1

4 0
2 years ago
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