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anastassius [24]
2 years ago
5

The ___ is the number of protons in the nucleus of an atom of an element. Fill in the blank.

Chemistry
1 answer:
Maurinko [17]2 years ago
8 0
The atomic number is the number of protons in the nucleus
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How many Joules must be removed to condense 150.0g of steam at 100.0 degrees Celsius to water at 100.0 degrees celcius
Elina [12.6K]

Answer:

339kJ

Explanation:

Given parameters:

Mass of steam  = 150g  = 0.15kg

Initial temperature of steam  = 100°C

Final temperature of water  = 100°C  

Unknown:

Quantity of heat that must be removed to condense the steam = ?

Solution:

The heat involved here is a latent heat because there is no change temperature. The process is just a phase change.

  H  = mL

m is the mass

L is the latent heat of vaporization  = 2,260 kJ/kg

Insert the parameters and solve;

 H = 0.15kg x 2,260 kJ/kg

 H  = 339kJ

7 0
3 years ago
After the solution reaches equilibrium, what concentration of zn2 (aq remains?
Fantom [35]
According to sources, the most probable answer to this query is that when solutions reaches equilibrium, the amount of concentration of two or more matter combined in this solution becomes equal. 

Thank you for your question. Please don't hesitate to ask in Brainly your queries. 
8 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
Which question cannot be answered using scientific methods?
vfiekz [6]
The question that cannot be answered using scientific method is "Which is the most interesting acid?" 
7 0
3 years ago
Read 2 more answers
Explain why the benzene molecule usually reacts with electrophiles<br>​
Genrish500 [490]

Answer:

Explanation:Because of the delocalised electrons exposed above and below the plane of the rest of the molecule, benzene is obviously going to be highly attractive to electrophiles - species which seek after electron rich areas in other molecules.

3 0
3 years ago
Read 2 more answers
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