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Marta_Voda [28]
3 years ago
8

Why is paraffin oil constantly stirred to determine the melting point of benzene acid?

Chemistry
1 answer:
Sidana [21]3 years ago
7 0

Why is paraffin oil constantly stirred to determine the melting point of benzene acid?

  • Paraffin oil is used for determination of boiling point and melting point for the following reasons: It has a very high boiling point and so it can be used to maintain high temperatures in the boiling and melting point apparatus without loss of the substance.

#CarryOnLearning

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A compound distributes between benzene (solvent 1) and water (solvent 2) with a distribution coefficient, K = 2.7. If 1.0g of th
mote1985 [20]

Explanation:

The given data is as follows.

Solvent 1 = benzene,          Solvent 2 = water

 K_{p} = 2.7,         V_{S_{2}} = 100 mL

V_{S_{1}} = 10 mL,       weight of compound = 1 g

       Extract = 3

Therefore, calculate the fraction remaining as follows.

                  f_{n} = [1 + K_{p}(\frac{V_{S_{2}}}{V_{S_{1}}})]^{-n}

                                  = [1 + 2.7(\frac{100}{10})]^{-3}

                                  = (28)^{-3}

                                  = 4.55 \times 10^{-5}

Hence, weight of compound to be extracted = weight of compound - fraction remaining

                                  = 1 - 4.55 \times 10^{-5}

                                  = 0.00001

or,                               = 1 \times 10^{-5}

Thus, we can conclude that weight of compound that could be extracted is 1 \times 10^{-5}.

7 0
3 years ago
Read 2 more answers
A sheet of gold weighing 8.8 g and at a temperature of 10.5°C is placed flat on a sheet of iron weighing 19.5 g and at a tempera
timofeeve [1]

Answer:

T = 36.393\,^{\textdegree}C

Explanation:

The contact between the sheet of gold and the sheet of iron allows a heat transfer until thermal equilibrium is done, which means that both sheets have the same temperature:

-Q_{iron} = Q_{gold}

-(0.008\,kg)\cdot (452\,\frac{J}{kg\cdot ^{\textdegree}C} )\cdot (T-54.4\,^{\textdegree}C) = (0.0195\,kg)\cdot (129\,\frac{J}{kg\cdot ^{\textdegree}C} )\cdot (T-10.5\,^{\textdegree}C)

-(3.616\,\frac{J}{^{\textdegree}C})\cdot (T-54.4\,^{\textdegree}C) = (2.515\,\frac{J}{^{\textdegree}C})\cdot (T-10.5^{\textdegree}C)

-1.438\cdot (T - 54.4^{\textdegree}C) = T-10.5^{\textdegree}C

-1.438\cdot T +78.227^{\textdegree}C = T - 10.5^{\textdegree}C

2.438\cdot T = 88.727\,^{\textdegree}C

The final temperature is:

T = 36.393\,^{\textdegree}C

8 0
3 years ago
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Write 2 typical properties that are only common to transition metals​
Vinil7 [7]

Answer:

Properties of transition elements

they are all metals and that most of them are hard, strong, and lustrous, have high melting and boiling points, and are good conductors of heat and electricity.

6 0
3 years ago
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Which phrase best describes a scientists?
larisa86 [58]

Answer:

They are helpers of the world  who find out about the natural world and try to explain what they have observed.

Explanation:

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The radioactivity of U-235 is of low intensity. Why then were the people of Hiroshima exposed to high intensity radiation in the
Marina86 [1]

Answer:

Explanation:

What occurred then is as a result of nuclear fission. This occurs as the Uranium-235 split into two smaller nuclei while releasing high energy neutrons. These neutrons bombard existing U-235 in the atmosphere and this reaction continue in a spontaneous manner until a chain reaction is formed of U-235, whose fall out fills the environment. This process was what led to people been exposed to high intensity radiation in the days and months after the atomic bomb was dropped.

7 0
3 years ago
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