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mash [69]
3 years ago
14

1.Build or draw the Lewis structure for each of the molecules listed below.

Chemistry
1 answer:
bagirrra123 [75]3 years ago
5 0
This may help you

Allright for <span><span>H2</span>O:</span> - The central atom is? --> the oxygen atom - How many atoms are bonded to the central atom? --> 2 hydrogen atoms - How many lone pairs of electrons are on the central atom? --> O has 6 electrons and has 2 single bonds, so 2 pairs - How many single bonds are there in this molecule? --> 2 - How many multiple bonds (double and/or triple) are there in this molecule? --> none For each of your molecules, answer the following questions: 1. Determine the electronegativity between the atoms of each molecule. Electronegativity O = 3.44 Electronegativity H = 2.20 3.44-2.20=1.24, so the electronegativity between O and H = 1.24 2. Identify the bond as either ionic or covalent. Electronegativity of 0.0-1.7 = covalent Electronegativity of 1.7-3.3 = ionic So it's a covalent bond 3. State whether the molecule is polar or non polar. Electronegativity of 0.5-1.7= polar covalent 4. Identify the structure as having hydrogen bonding, dipole-dipole moments or London dispersion forces (LDF). <span><span>H2</span>O</span><span> = hydrogen bonding</span>
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Please help me do this question.
masya89 [10]

Answer:

In chemistry, there are three definitions in common use of the word base, known as Arrhenius bases, Brønsted bases and Lewis bases. All definitions agree that bases are substances which react with acids as ... A ...

Acids are species with pH less than 7, while bases those with pH greater than 7. ... Using Arrhenius' definition, acids are those that produce the hydrogen ion when dissolved in water ( ) while bases produce the hydroxide ion ( ). These ions, when in water, act as charge carriers and can hence conduct electricity.

Acid–base reactions require both an acid and a base. ... For example, according to the Arrhenius definition, the reaction of B If inorganic, determine whether the compound is acidic.

Explanation:

I HAVE GIVE DIFINITIO FROM THAT YOU WRITE TRUE OR FALSE

8 0
3 years ago
What volume of air at 25°C and 1.00 atm can he stored in a 10.0 L high-pressure air tank if compressed to 25°C and 175 atm?
DaniilM [7]

Answer:

1750L

Explanation:

Given

Initial Temperature = 25°C

Initial Pressure = 175 atm

Initial Volume = 10.0L

Final Temperature = 25°C

Final Pressure = 1 atm

Final Volume = ?

This question is an illustration of ideal gas law.

From the given parameters, the initial temperature and final temperature are the same; this implies that the system has a constant temperature.

As such, we'll make use of Boyle's Law to solve this;

Boyle's Law States that:

P₁V₁ = P₂V₂

Where P₁ and P₂ represent Initial and Final Pressure, respectively

While V₁ and V₂ represent Initial and final volume

The equation becomes

175 atm * 10L = 1 atm * V₂

1750 atm L = 1 atm * V₂

1750 L = V₂

Hence, the final volume that can be stored is 1750L

5 0
3 years ago
Please help. 1.600x10^8 / 8.000x10^11
sweet-ann [11.9K]

Answer:

2x10^-4 = 2E-4

Explanation:

3 0
2 years ago
Read 2 more answers
In a solution, the solvent is the substance in greater supply. True or False
Dmitriy789 [7]

Answer:

true

Explanation:

in a solution solvent is the one in which solute is been dissolved that is solvent is in greater amount.

7 0
3 years ago
what is the volume of the air in a balloon that occupies 0.730 L at 28.0 c if the temperature is lowered to 0.00 C
svetoff [14.1K]

Answer:

The volume of the air is 0.662 L

Explanation:

Charles's Law is a gas law that relates the volume and temperature of a certain amount of gas at constant pressure. This law says that for a given sum of gas at a constant pressure, as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases because the temperature is directly related to the energy of the movement they have. the gas molecules. This is represented by the quotient that exists between volume and temperature will always have the same value:

\frac{V}{T}=k

If you have a certain volume of gas V1 that is at a temperature T1 at the beginning of the experiment and several the volume of gas to a new value V2, then the temperature will change to T2, and it will be true:

\frac{V1}{T1}=\frac{V2}{T2}

In this case:

  • V1= 0.730 L
  • T1= 28 °C= 301 °K (0°C= 273°K)
  • V2= ?
  • T2= 0°C= 273 °K

Replacing:

\frac{0.730 L}{301K}=\frac{V2}{273K}

Solving:

V2=273K*\frac{0.730L}{301K}

V2=0.662 L

<u><em>The volume of the air is 0.662 L</em></u>

6 0
3 years ago
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