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3 years ago

The plastic which do not remould against on the heating are called ___..

Chemistry

1 answer:

andre [41]3 years ago

5 0

Answer:

Explanation:

Thermoseeting is termed to the plastic which caanot be remouled by heating, hence reausing them is not possible effeciently hence must be avoided to use in day to day applciations.

bakelite is an an examle

The reverse of the "thermosettin" are "thermoplastic" which can be remolded and are good for long time use.

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Phthalates used as plasticizers in rubber and plastic products are believed to act as hormone mimics in humans. The value of ΔHc

jeka94

Answer:

$T_f$ = 25.05°C

Explanation:

Given:

the value of ΔHcomb (heat of combustion) for dimethylphthalate (C10H10O4) is = 4685 kJ/mol.

mass = 0.905g of dimethylphthalate

molar mass = 194.18g dimethylphthalate

number of moles of dimethylphthalate = ???

$T_i$ = 21.5°C

$C_{calorimeter}$ = 6.15 kJ/°C

$T_f$ = ???

since we have our molar mass and mass of dimethylphthalate ;we can determine the number of moles as;

0.905g of dimethylphthalate × $\frac{1 mole (dimethylphthalate)}{194.184g(dimethylphthalate)}$

number of moles of dimethylphthalate = 0.000466 moles

Heat released = moles of dimethylphthalate × heat of combustion

= 0.000466 moles × 4685 kJ

= 21.84 kJ

∴ Heat absorbed by the calorimeter = $C_{calorimeter}$ $(T_f-T_i} )$

21.84 kJ =6.15 kJ/°C $* (T_f-21,5^0C)$

21.84 KJ = $(6.15 kJ/^0C * T_f) - (6.15 kJ/^0C*21.5^0C)$

21.84 KJ = $(6.15 kJ/^0C * T_f)$ - 132.225 kJ

21.84 KJ + 132.225 kJ = $(6.15 kJ/^0C * T_f)$

154.065 kJ = $(6.15 kJ/^0C * T_f)$

$T_f$ = $\frac{154.065kJ}{6.15kJ/^0C}$

$T_f$ =25.05°C

4 0

3 years ago

The energy received by Earth from the sun is what type of energy

DedPeter [7]

Answer:

Solar Energy

Explanation:

3 0

3 years ago

Read 2 more answers

Why do health care professionals limit the number of radiation treatments that a person can be given

katovenus [111]

Radiation is s very powerful substance and can cause major health problems do the health care professionals limit the use to help not create sickness

6 0

3 years ago

If the initial temperature of a movable cylinder was 50 degrees Celsius

slega [8]

Answer:

8.45 L

Explanation:

From the question given above, the following data were obtained:

Initial temperature (T₁) = 50 °C

Initial pressure (P₁) = 2 atm

Initial volume (V₁) = 5 L

Final temperature (T₂) = 0 °C

Final pressure (P₂) = 1 atm

Final volume (V₂) =?

Next, we shall convert celsius temperature to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T₁) = 50 °C

Initial temperature (T₁) = 50 °C + 273

Initial temperature (T₁) = 323 K

Final temperature (T₂) = 0 °C

Final temperature (T₂) = 0 °C + 273

Final temperature (T₂) = 273 K

Finally, we shall determine the new volume. This can be obtained as follow:

Initial temperature (T₁) = 323 K

Initial pressure (P₁) = 2 atm

Initial volume (V₁) = 5 L

Final temperature (T₂) = 273 k

Final pressure (P₂) = 1 atm

Final volume (V₂) =?

P₁V₁ / T₁ = P₂V₂ / T₂

2 × 5 / 323 = 1 × V₂ / 273

10 / 323 = V₂ / 273

Cross multiply

323 × V₂ = 10 × 273

323 × V₂ = 2730

Divide both side by 323

V₂ = 2730 / 323

V₂ = 8.45 L

Thus, the new volume is 8.45 L

5 0

3 years ago

After completing an experiment to determine gravimetrically the percentage of water in a hydrate, a student reported a value of

SIZIF [17.4K]

Answer:

b) The dehydrated sample absorbed moisture after heating

Explanation:

a) Strong initial heating caused some of the hydrate sample to splatter out.

This will result in a higher percent of water than the real one, because you assume in the calculation that the splattered sample was only water (which in not true).

b) The dehydrated sample absorbed moisture after heating.

Usually inorganic salts may absorbed moisture from the atmosphere so this will explain the 13% difference between calculated water percent the real content of water in the hydrate.

c) The amount of the hydrate sample used was too small.

It will create some errors but they do not create a difference of 13% difference as stated in the problem.

d) The crucible was not heated to constant mass before use.

Here the error is small.

e) Excess heating caused the dehydrated sample to decompose.

Usually the inorganic compounds are stable in the temperature range of this kind of experiments. If you have an organic compound which retain water molecules you may decompose the sample forming volatile compounds which will leave crucible so the error will be quite high.

6 0

3 years ago

The plastic which do not remould against on the heating are called ___..​

The plastic which do not remould against on the heating are called ___..