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Marrrta [24]
3 years ago
9

Which equation is balanced in an acidic solution?

Chemistry
1 answer:
liberstina [14]3 years ago
5 0

Answer:

I think C but it could be D

Explanation:

I would say go for C

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This planet is about the size of Earth. It has a very thick atmosphere with
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Answer:

Venus

Explanation:

Mars has a thin atmosphere

Saturn is really cold (-288F)

Mercury has no atmosphere

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A scale model of the solar system
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There you go BRAINLIEST would be appreciated

3 0
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Which horizon has the least weathered rocks? How do you know?
ollegr [7]

Answer:

C: The C horizon is a subsurface horizon. It is the least weathered horizon. Also known as the saprolite, it is unconsolidated, loose parent material. The master horizons may be followed by a subscript to make further distinctions between differences within one master horizon.

Explanation:

4 0
3 years ago
What are the functions of ammonia in the solvay process​
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The ammonia gas is absorbed in the concentrated brine to produce aqueous sodium chloride and aqueous ammonia. This ammoniation process is exothermic, so energy is released as heat. The ammonia tower eventually needs to be cooled.

7 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
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