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stiks02 [169]
4 years ago
11

The Lewis structure of an oxygen atom should have _____________ dots drawn around the symbol .The Lewis structure of a calcium a

tom should have ____________ dots drawn around the symbol Ca. The Lewis structure of a nitrogen atom should have ___________ dots drawn around the symbol N. The Lewis structure of an aluminum atom should have ___________ dots drawn around the symbol Al. The Lewis structure of a fluorine atom should have _________ dots drawn around the symbol F.
Chemistry
1 answer:
nydimaria [60]4 years ago
4 0

Answer:

Oxygen- 6

Calcium-2

Nitrogen-5

Aluminum-3

Flourine-7

Explanation:

A Lewis structure or dot electron structure is a representation of an atom in which the valence electrons on the outermost shell of the atom are represented as dots.

These dots show the number of valence electrons as well as the group to which the atom belongs in the periodic table.

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Which substance contains bonds that involve a transfer of electrons from one atom to another
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An ionic bond is a type of chemical bond formed through an electrostatic attraction between two oppositely charged ions. Ionic bonds are formed between a cation, which is usually a metal, and an anion, which is usually a nonmetal. A covalent bond involves a pair of electrons being shared between atoms.
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3 years ago
How many moles of water h2o are present in 75.0 g h2o?
nikklg [1K]
4.17 moles. Good luck! :)
7 0
3 years ago
A student titrates 10.00 milliliters of hydrochloric acid of unknown molarity with 1.000 m naoh. it takes 21.17 milliliters of b
Dima020 [189]
Mole ratio for the reaction is 1:1
no of moles in NaOH that reacted= 1*21.17/1000=0.02117mols
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4 0
3 years ago
What is the mass of 5.84 × 10^21 atoms of xenon?
podryga [215]

Answer:

5.84e+21

Explanation:

8 0
3 years ago
A cubic piece of platinum metal (specific heat capacity = 0.1256 J/°C・g) at 200.0°C is dropped into 1.00 L of deuterium oxide ('
polet [3.4K]

Answer:

a=5.65cm

Explanation:

Hello,

In this case, for this heat transfer process in which the heat lost by the hot platinum is gained by the cold deuterium oxide based on the equation:

Q_{Pt}=-Q_{Deu}

We can represent the heats in terms of mass, heat capacities and temperatures:

m_{Pt}Cp_{Pt}(T_f-T_{Pt})=-m_{Deu}Cp_{Deu}(T_f-T_{Deu})

Thus, we solve for the mass of platinum:

m_{Pt}=\frac{-m_{Deu}Cp_{Deu}(T_f-T_{Deu})}{Cp_{Pt}(T_f-T_{Pt})} \\\\m_{Pt}=\frac{-1.00L*1110g/L*4.211J/(g\°C)*(41.9-25.5)\°C}{0.1256J/(g\°C)*(41.9-200.0)\°C} \\\\m_{Pt}=3860.4g

Next, by using the density of platinum we compute the volume:

V_{Pt}=\frac{3860.4g}{21.45g/cm^3}\\ \\V_{Pt}=180cm^3

Which computed in terms of the edge length is:

V=a^3

Therefore, the edge length turns out:

a=\sqrt[3]{180cm^3}\\ \\a=5.65cm

Best regards.

6 0
3 years ago
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