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Alla [95]
3 years ago
13

Explain why pentane has a higher boiling point than butane

Chemistry
1 answer:
creativ13 [48]3 years ago
8 0

Answer: Larger molecules have stronger London forces.

Explanation:

Pentane therefore has a stronger force than butane.

You might be interested in
A buffer consists of 0.120 M HNO2 and 0.150 M NaNO2 at 25°C. pka of HNO2 is 3.40. a. What is the pH of the buffer? b. What is th
Mashcka [7]

Explanation:

It is known that K_{a} of HNO_{2} = 4.5 \times 10^{-4}.

(a)  Relation between K_{a} and pK_{a} is as follows.

                       pK_{a} = -log (K_{a})

Putting the values into the above formula as follows.

                      pK_{a} = -log (K_{a})

                                    = -log(4.5 \times 10^{-4})

                                     = 3.347

Also, relation between pH and  pK_{a} is as follows.

              pH = pK_{a} + log\frac{[conjugate base]}{[acid]}

                     = 3.347+ log \frac{0.15}{0.12}

                    = 3.44

Therefore, pH of the buffer is 3.44.

(b)   No. of moles of HCl added = Molarity \times volume

                                            = 11.6 M \times 0.001 L

                                             = 0.0116 mol

In the given reaction, NO^{-}_{2} will react with H^{+} to form HNO_{2}

Hence, before the reaction:

No. of moles of NO^{-}_{2} = 0.15 M \times 1.0 L

                                           = 0.15 mol

And, no. of moles of HNO_{2} = 0.12 M \times 1.0 L

                                               = 0.12 mol

On the other hand, after the reaction :  

No. of moles of NO^{-}_{2} = moles present initially - moles added

                                          = (0.15 - 0.0116) mol

                                          = 0.1384 mol

Moles of HNO_{2} = moles present initially + moles added

                               = (0.12 + 0.0116) mol

                                = 0.1316 mol

As, K_{a} = 4.5 \times 10^{-4}

           pK_{a} = -log (K_{a})

                         = -log(4.5 \times 10^{-4})

                         = 3.347

Since, volume is both in numerator and denominator, we can use mol instead of concentration.

As, pH = pK_{a} + log \frac{[conjugate base]}{[acid]}

            = 3.347+ log {0.1384/0.1316}

            = 3.369

            = 3.37 (approx)

Thus, we can conclude that pH after the addition of 1.00 mL of 11.6 M HCl to 1.00 L of the buffer solution is 3.37.

6 0
3 years ago
HELP PLEASE<br>Explain why the elements of group 1 and 7 are quite reactive
Sauron [17]
Group 17 is the most readily reduced elements on the periodic table, meaning that they are so close to being a stable elements, only missing 1 electron to complete their valance electron shell. Thus they will essentially react with anything to get that last electron! 

Group 1 elements are extremely reactive because they are the most readily oxidized, they are very close to reaching stability by giving up only 1 electron. Thus they will react with almost anything to give up their electron. 
8 0
3 years ago
n a coffee cup calorimeter, 100.0 mL of 1.0M NaOH and 100.0 mL of 1.0M HCl are mixed. Both solutions were originally at 24.6 deg
RoseWind [281]

Answer:

Explanation:

mass of the solution = volume x density = 200 x 1 = 200 gm

heat absorbed = m x s x Δ t , s is specific heat , Δt is rise in temperature

= 200 x 4.18 x ( 31.3 - 24.6 )

= 5601 J .

This is the enthalpy change required.

3 0
3 years ago
You determine that Lauren's EER is 2200 kcal/day. You recommend and intake of 60% CHO, 15% protein, and 25% fat. Approximately h
const2013 [10]

Answer:

The answer to your question is 330 g of CHO

Explanation:

Data

Calories needed = 2200 kcal/day

CHO = 60%

Proteins = 15%

Fats = 25%

Grams of carbohydrates needed = ?

Process

1.- Calculate the number of calories  in 60% of 2200 kcal

                      2200 kcal ----------------  100%

                          x            ---------------      60%

                       x = (60 x 2200) / 100

                       x = 1320 kcal

2.- Calculate the grams of CHO

                       1 g of CHO ---------------- 4 kcal

                       x                 ----------------  1320 kcal

                       x = (1320 x 1) / 4

                       x = 1320/4

                       x = 330 g of CHO

6 0
3 years ago
The c—c—c bond angle in propane, c3h8, is closest to
Archy [21]
Propane has a molecular formula C3H8.

It is of the form CnH2n+2.
where n = number of C atoms. In present case n = 3.

Each carbon atom in propane is sp3 hydribized. Thus, there are 4 hydrid orbital associated with each carbon atom

The central C atom is sigma bonded to two other carbon atoms and two hydrogen atoms.

Such, sp3 hydridized orbitals are spacial orientated at an  angle of 109^{0}28^{'}, in order to minimize repulsion between the electrons.

Hence, the c—c—c bond angle in propane, c3h8, is closest to 109^{0}28^{'}
3 0
3 years ago
Read 2 more answers
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