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irinina [24]
2 years ago
8

Draw a diagram to show how you will separate the salt from sugar using the solvent ethanol​

Chemistry
1 answer:
SIZIF [17.4K]2 years ago
5 0
I can’t draw sorry but tbh i’m just trying to get sum points so i can do my work u feel me ;)
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What is a stream of charged protons and electrons
Law Incorporation [45]

Answer:

It is called the solar wind thx for asking brainly.com

Explanation:

5 0
3 years ago
In an electrically heated boiler, water is boiled at 140°C by a 90 cm long, 8 mm diameter horizontal heating element immersed in
RideAnS [48]

Explanation:

The given data is as follows.

Volume of water = 0.25 m^{3}

Density of water = 1000 kg/m^{3}

Therefore,  mass of water = Density × Volume

                       = 1000 kg/m^{3} \times 0.25 m^{3}

                       = 250 kg  

Initial Temperature of water (T_{1}) = 20^{o}C

Final temperature of water = 140^{o}C

Heat of vaporization of water (dH_{v}) at 140^{o}C  is 2133 kJ/kg

Specific heat capacity of water = 4.184 kJ/kg/K

As 25% of water got evaporated at its boiling point (140^{o}C) in 60 min.

Therefore, amount of water evaporated = 0.25 × 250 (kg) = 62.5 kg

Heat required to evaporate = Amount of water evapotaed × Heat of vaporization

                           = 62.5 (kg) × 2133 (kJ/kg)

                           = 133.3 \times 10^{3} kJ

All this heat was supplied in 60 min = 60(min)  × 60(sec/min) = 3600 sec

Therefore, heat supplied per unit time = Heat required/time = \frac{133.3 \times 10^{3}kJ}{3600 s} = 37 kJ/s or kW

The power rating of electric heating element is 37 kW.

Hence, heat required to raise the temperature from 20^{o}C to 140^{o}C of 250 kg of water = Mass of water × specific heat capacity × (140 - 20)

                      = 250 (kg) × 40184 (kJ/kg/K) × (140 - 20) (K)

                     = 125520 kJ  

Time required = Heat required / Power rating

                       = \frac{125520}{37}

                       = 3392 sec

Time required to raise the temperature from 20^{o}C to 140^{o}C of 0.25 m^{3} water is calculated as follows.

                    \frac{3392 sec}{60 sec/min}

                     = 56 min

Thus, we can conclude that the time required to raise the temperature is 56 min.

4 0
3 years ago
Question
andreev551 [17]

Answer:

Sln

n=m/mr

n=25/100

n=0.25mole of Caco3

Malality =number of moles/volume (divided by number of moles both sides)

volume =Malality /number of moles

v=0.125/0.25

v=0.500L

I hope this help

5 0
3 years ago
The specific heat for liquid ethanol is 2.46 J/(g•°C). When 210 g of ethanol is cooled from
avanturin [10]

Answer:

a) heat it from 23.0 to 78.3

q = (50.0 g) (55.3 °C) (2.46 J/g·°C) =  

b) boil it at 78.3

(39.3 kJ/mol) (50.0 g / 46.0684 g/mol) =  

c) sum up the answers from the two calculations above. Be sure to change the J from the first calc into kJ

Explanation:

3 0
3 years ago
You go to the same school as me BAHAHHAHA
Reptile [31]
Oh my i really hope no one from my school sees my comments
4 0
3 years ago
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