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choli [55]
3 years ago
9

How many grams of sodium chloride are in 250mL of a 2.5M NaCI solution

Chemistry
1 answer:
Elden [556K]3 years ago
4 0
I think I did this right.

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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
Heating curve shows temperature verses energy gain. Which parts of the curve represent a gain in potential energy?
Brilliant_brown [7]

Answer:

Those two horizontal lines.

Explanation:

Hello there!

In this case, when focusing on these heating curves, it is important to say they tend to have two constant-temperature sections and three variable-temperature sections. Thus, from lower to higher temperature, the first constant-temperature section corresponds to melting and the second one vaporization, whereas the three variable-temperature sections correspond to the heating of the solid until melting, the liquid until vaporization and the gas until the critical point.

In such a way, we infer that the boxes referred to constant temperature are referred to a gain in potential energy, that is, the two horizontal lines.

Regards!

6 0
3 years ago
Who made polio vaccine
yulyashka [42]
The creator of the Polio Vaccine was Jonas Salk.
5 0
3 years ago
Read 2 more answers
A student measures the pressure and volume of an empty water bottle to be 1. 4 atm and 2. 3 l
rewona [7]

The new volume of the bottle is 5.07 L.

<h3>What is Boyle's Law?</h3>

It is a gas law that states the pressure decrease with the increase in the pressure.

By the formula

\rm P_1V_1= P_1V_2

Given that, volume1 is 2.3l

Pressure1 is 4 atm.

Pressure2 to 0.65 atm.

V2 is to find?

Putting the values in the equation

\rm 1.4\times 2.3 = 0.65 \times V_2\\\\V_2 = 5.07 L

Thus, the new volume of the bottle is 5.07 L.

Learn more about Boyle's Law

brainly.com/question/1437490

#SPJ4

6 0
2 years ago
What is a factor that affects the outcome of an experiment?
Anni [7]

Answer:

Observation affects the outcome

7 0
3 years ago
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