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Arte-miy333 [17]
3 years ago
13

At 298 K and 1 atm, bromine is a liquid with a high vapor pressure, whereas chlorine is a gas. This provides evidence that, unde

r these conditions, the forces among Br2 molecules are _______ than those among Cl2 molecules
Chemistry
1 answer:
Galina-37 [17]3 years ago
3 0

Explanation:

In liquids, the molecules are held by less strong intermolecular forces of attraction as compared to solids. Due to which they are able to slide past each other. Hence, they have medium kinetic energy.

In gases, the molecules are held by weak Vander waal forces. Hence, they have high kinetic energy due to which they move rapidly from one place to another leading to more number of collisions.

So, when at 298 K and 1 atm Br_{2} exists in liquid state and Cl_{2} exists as a gas then it means there occurs strong force of attraction between the molecules of Br_{2} due to which it exists in liquid form.

Thus, we can conclude that at 298 K and 1 atm, bromine is a liquid with a high vapor pressure, whereas chlorine is a gas. This provides evidence that, under these conditions, the forces among Br_2 molecules are greater than those among Cl_2 molecules.

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What amount (moles) of compound is present in 1.00 g of each of the compounds in Exercise 54?
Anika [276]

Answer:

1. Caffeine, C₈H₁₀N₄O₂

Amount = 1.00/194 = 0.00515 moles

2. Ethanol, C₂H₅OH

Amount = 0.0217 moles

3. Dry Ice, CO₂

amount = 0.0227 moles

<em>Note: The question is incomplete. The compound are as follows:</em>

<em> 1. Caffeine, C₈H₁₀N₄O₂;</em>

<em>2. Ethanol, C₂H₅OH;</em>

<em>3. Dry Ice, CO₂</em>

Explanation:

Amount (moles) = mass in grams /molar mass in grams per mole

1. Caffeine, C₈H₁₀N₄O₂

molar mass of caffeine = 194 g/mol

Amount = 1.00 g/194 g/mol = 0.00515 moles

2. Ethanol, C₂H₅OH

molar mass of ethanol = 46 g/mol

Amount = 1.00 g/46 g/mol = 0.0217 moles

3. Dry Ice, CO₂

molar mass of dry ice = 44 g/mol

amount = 1.00 g/44 g/mol = 0.0227 moles

5 0
2 years ago
Write the empirical formula of at least four binary iconic compounds that could be formed from the following ions:
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Explanation:

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7 0
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Answer:

2nd one In my opinion....

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Sophie [7]

Answer:

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b) the mass number is 15+16 = 31

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d)Group 15 period 3

6 0
2 years ago
A glass holds 6 ounces of 60-proof rum (30 alcohol). how much fruit juice must be added to the rum so that the mixture is dilute
Elenna [48]

We need an equation that will relate the concentrated mixture and the diluted one. To solve this we use the equation, 

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where M1 is the concentration of the stock solution, V1 is the volume of the stock solution, M2 is the concentration of the new solution and V2 is its volume.

M1V1 = M2V2

30 % x 6 oz = 20 %  x V2

V2 = 9 oz 


The volume of the diluted mixture would be 9 oz. Therefore, you will need to add 9 oz - 6 oz = 3 oz of fruit juice to dilute the 30 percent alcohol to 20 percent alcohol.

7 0
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