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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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3 years ago
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VA 19.75-g sample was heated by 12.35 calories. The specific heat of the sample is 0.125 cal/g°C. What was the initial temperatu
SOVA2 [1]

Answer:

31.9 °C  

Explanation:

The formula for the heat q absorbed by an object is

q = mCΔT where ΔT = (T₂ - T₁)

Data:

q = 12.35 cal

m = 19.75 g

C = 0.125 cal°C⁻¹g⁻¹

T₂ = 37.0 °C

Calculations

(a) Calculate ΔT

q = mCΔT

12.35 cal = 19.25 g × 0.125 cal°C⁻¹g⁻¹ × ΔT

12.35 = 2.406ΔT °C⁻¹  

ΔT  = 12.35/(2.406 °C⁻¹) = 5.13 °C

(b) Calculate T₂

ΔT = T₂ - T₁

T₁ = T₂ - ΔT = 37.0 °C - 5.13 °C = 31.9 °C

The original temperature was 31.9 °C.

 

6 0
3 years ago
A glass flask whose volume is 1000 cm3 at a temperature of 0.800 C is completely filled with mercury at the same temperature. Wh
ira [324]

Answer: the coefficient of volume expansion of glass = 0.86/(1000 * 52) = 0.00001654 per degree.

Explanation:

Original volume of mercury = 1000 cm3.

The final volume of mercury considering its volume expansion quotient = 1000 + 1000*(1.8*10^-4 *52) = 1000 + 9.36 = 1009.36 cm^3

Considering the glass as a non expanding substance, the complete excess volume of 9.36 cm3 of mercury should have overflown the container, but due to the expansion of glass, the capacity of mercury containment increases and so a lesser amount of mercury flows out.

The amount of mercury that actually flowed out = 8.50 cm3.

So, the expansion of the glass container = 9.36-8.50 = 0.86 cm3.

Using the formula for coefficient of expansion,

coefficient of volume expansion of glass = 0.86/(1000 * 52) = 0.00001654 per degree.

5 0
3 years ago
If a lab requires each lab group to have 25ml of a solution and it takes 15 grams of CuNO3 to make 1 liter of solution how many
tino4ka555 [31]

We need to do some general algebra here.

We will find that you need 8.25 grams of CuNO₃ to make enough solution for the 22 labs.

<em>We know that:</em>

  • Each lab group needs 25 ml of solution.
  • it takes 15 g of CuNO₃ to make one L of that solution.
  • There are 22 labs.

Because each lab needs 25 ml of solution, 22 labs will need that amount 22 times, so the <u>total amount of solution needed</u> is:

22*25ml = 550 ml

Now we know that we need 15 grams to make one liter of solution, and:

1 L = 1000ml

Then you need 15g to make 1000ml

and x (we want to find this amount) to make 550ml

Then we can write two equations (not actual equations, as these are different units) like:

x = 550ml

15g = 1000ml

Now we can take the quotient between these two equations:

x/15 g = (550ml/1000ml)

And now we can solve this for x:

x = (550ml/1000ml)*15g = 8.25g

So you need 8.25 grams of CuNO₃ to make enough solution for the 22 labs.

If you want to learn more, you can read:

brainly.com/question/8743486

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What’s IV DV and Controls
zavuch27 [327]

Answer:

Iv-Independent Variable the part that doesnt get affected by the Dependent. DV- Dependent Variable- the part that gets affected by the independent. Controls is what everything is gettong compared to

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