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Aneli [31]
3 years ago
14

Indicate whether each of the statements below is true or false. Match the words in the left column to the appropriate blanks in

the sentences on the right. Make certain each sentence is complete before submitting your answer. ResetHelp False True blank: C B r 4 has a higher boiling point than C C l 4.: CBr4 has a higher boiling point than CCl4. blank: C B r 4 has weaker intermolecular forces than C C l 4.: CBr4 has weaker intermolecular forces than CCl4. blank: C B r 4 has a higher vapor pressure at the same temperature than C C l 4.: CBr4 has a higher vapor pressure at the same temperature than CCl4. blank: C B r 4 is more volatile than C C l 4.: CBr4 is more volatile than CCl4.
Chemistry
1 answer:
otez555 [7]3 years ago
5 0

Answer:

1) ) CBr₄ has a higher boiling point than CCl₄: True

2 CBr₄ has weaker intermolecular forces than CCl₄: False

3) CBr₄ has a higher vapor pressure at the same temperature than CCl₄: False

4) CBr₄ is more volatile than CCl₄: False

Explanation:

1) ) CBr₄ has a higher boiling point than CCl₄: True :

Due to higher molecular weight CBr₄ has more london disperion forces thus making the intermolecular interactions stronger and thus it need more temperature to boil it off.

2 CBr₄ has weaker intermolecular forces than CCl₄: False

Due to higher molecular weight CBr₄ has more london disperion forces thus making the intermolecular interactions stronger.

3) CBr₄ has a higher vapor pressure at the same temperature than CCl₄: False

Due to higher molecular weight CBr₄ has more london disperion forces thus making the intermolecular interactions stronger. Thus the vapor pressure of it will be less than CCl₄ at the same temperature.

4) CBr₄ is more volatile than CCl₄: False

Due to higher molecular weight CBr₄ has more london disperion forces thus making the intermolecular interactions stronger. Thus CCl₄ is more volatile.

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8. What volume (ml) of a 5.45 M lead nitrate solution must be diluted to 820.7 ml to make a
Ganezh [65]

212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.

Answer:

Option A.

Explanation:

Similar to Avagadro's law, there is another law termed as dilution law. As the product of volume and normality of the reactant is equal to the product of volume and normality of the product from the Avagadro's law. In dilution law, it will be as product of volume and concentration of the solute of the reactant is equal to the product of volume and concentration of solution.

C_{1} V_{1} =C_{2} V_{2}

So, as per the given question C1 = 5.45 M of lead nitrate and V1 has to be found. While C2 is 1.41 M of lead nitrate and V2 is 820.7 ml.

Then, (5.45*V_{1} ) = (820.7*1.41)

V_{1}=\frac{820.7*1.41}{5.45}=  212.33 ml

So nearly 212 ml of lead nitrate is required to prepare a dilute solution of 820.7 ml of lead nitrate.

3 0
3 years ago
Choose all the answers that apply.
max2010maxim [7]

Answer: has properties similar to other elements in group 18, does not react readily with other elements, is part of the noble gas group

Explanation: I’ve done on edg before

6 0
3 years ago
A chemical reaction in which compounds break up into simpler constituents is a _______ reaction.
sweet [91]
The awnser to ur question is B
7 0
3 years ago
Read 2 more answers
If you have 10,000 grams of a substance that decays with a half-life of 14 days, then how much will you have after 56 days?
Crank

Answer: 625 grams

Explanation:

14 goes into 56, 4 times.

10,000÷2=5,000 - first half life

5,000÷2=2,500 - second half life

2,500÷2=1,250 - 3rd half life

1,250÷2= 625 - 4th half life

Note: don't type "grams" after 625, just type "625".

5 0
2 years ago
at a particulat temperature, a smaple of pure water has a Kw of 7.7x10-14. what is the hydroxide concentration of this sample
ycow [4]

Answer: The hydroxide concentration of this sample is 3.85\times 10^{-7}

Explanation:

When an expression is formed by taking the product of concentration of ions raised to the power of their stoichiometric coefficients in the solution of a salt is known as ionic product.

The ionic product for water is written as:

K_w=[H^+]\times [OH^-]

7.7\times 10^{-14}=[H^+]\times [OH^-]

As [H^+]=[OH^-]

2[OH^-]=7.7\times 10^{-14}

[OH^-]=3.85\times 10^{-7}

Thus hydroxide concentration of this sample is 3.85\times 10^{-7}

4 0
3 years ago
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