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Nata [24]
3 years ago
15

I didn’t study for this lol if your answer correct I’ll mark ur answer brainliest

Chemistry
1 answer:
kotykmax [81]3 years ago
4 0

Answer:

3

Explanation:

3 is the answer is did this a little while ago

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Which statement best describes how an existing theory is often affected by the development of new technology?
grigory [225]

Answer: An existing theory is modified so that it can explain both the old and new observations.

Explanation:

4 0
3 years ago
Choose the letter of the choice that best completes the statement or answers the question. Elements in the same group have the s
pav-90 [236]
Elements in the same group have D. Same number of valence electrons.
6 0
3 years ago
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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
A certain sample of a liquid has a mass of 42 grams and a volume of 35 centimeters3. What is the density of the liquid?
BabaBlast [244]

Answer:

            1.2 g.cm⁻³

Solution:

Data Given:

                             Mass  =  42 g

                             Volume  =  35 cm³

Formula Used:

                             Density  =  Mass ÷ Volume

Putting values,

                             Density  =  42 g ÷ 35 cm³

                             Density  =  1.2 g.cm⁻³

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3 years ago
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alukav5142 [94]
There are many safety precautions and rules you MUST follow during labs. 
for this incident here is what you should do:
1)Notify your Instructor and partner 
2) if the liquid is toxic (like not water or vinegar) then let your Instructor handle it properly, or follow your instructors orders (like if they say to put a towel over it or something like that)
3) Broken glassware, minus mercury thermometer, must be immediately cleaned up, do not use your bare hand, always wear gloves.
4)dispose of the broken glass properly and clean the liquid up (unless it is harmful, then let your instructor do it)

6 0
3 years ago
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