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kvasek [131]
3 years ago
11

1-Witch compound is composed of the most atoms?

Chemistry
1 answer:
PilotLPTM [1.2K]3 years ago
6 0

Answer:

good luck with these problems I've never seen anything like it

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How many moles are in 23.8 g of sulfur dioxide
Yuki888 [10]

Answer:

0.371

Explanation:

number of moles = mass / Ar

= 23.8g / 32.1 + (16.0 × 2)

= 0.37129485179 = 0.371 (3 s.f).

hope it helps :)

4 0
2 years ago
How many methods of heat transfer are there<br>​
Eva8 [605]
There are three methods of heat transfer
3 0
3 years ago
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What decay process occurs in step one? A) beta decay B) alpha decay C) thorium decay D) positron decay
forsale [732]

Answer:

Alpha i think

Explanation:

5 0
3 years ago
Read 2 more answers
Please explain your reasoning and give explanation of why you chose that answer
natita [175]

Answer:

a. Cellular Respiration produces more ATP than Anaerobic Respiration.

Explanation:

Cellular Respiration creates a total of 36-38 ATP per round. Anaerobic processes only produce 2 ATP per cycle.

5 0
3 years ago
The halpy of vaporization of H2O at 1 atm and 100 C is 2259 kJ/kg. The heat capacity of liquid water is 4.19 kJ/kg.C, and the he
Naddik [55]

Answer: Option (a) is the correct answer.

Explanation:

The given data is as follows.

        C_{p}_{liquid} = 4.19 kJ/kg ^{o}C

        C_{p}_{vaporization} = 1.9 kJ/kg ^{o}C

Heat of vaporization (\DeltaH^{o}_{vap}) at 1 atm and 100^{o}C is 2259 kJ/kg

        H^{o}_{liquid} = 0

Therefore, calculate the enthalpy of water vapor at 1 atm and 100^{o}C as follows.

            H^{o}_{vap} = H^{o}_{liquid} + \DeltaH^{o}_{vap}        

                                   = 0 + 2259 kJ/kg

                                   = 2259 kJ/kg

As the desired temperature is given 179.9^{o}C and effect of pressure is not considered. Hence, enthalpy of liquid water at 10 bar and 179.9^{o}C is calculated as follows.

             H^{D}_{liq} = H^{o}_{liquid} + C_{p}_{liquid}(T_{D} - T_{o})

                             = 0 + 4.19 kJ/kg ^{o}C \times (179.9^{o}C - 100^{o}C)

                              = 334.781 kJ/kg

Hence, enthalpy of water vapor at 10 bar and 179.9^{o}C is calculated as follows.

               H^{D}_{vap} = H^{o}_{vap} + C_{p}_{vap} \times (T_{D} - T_{o})

             H^{D}_{vap} = 2259 kJ/kg + 1.9 \times (179.9 - 100)            

                              = 2410.81 kJ/kg

Therefore, calculate the latent heat of vaporization at 10 bar and 179.9^{o}C as follows.

       \Delta H^{D}_{vap} = H^{D}_{vap} - H^{D}_{liq}              

                         = 2410.81 kJ/kg - 334.781 kJ/kg

                         = 2076.029 kJ/kg

or,                      = 2076 kJ/kg

Thus, we can conclude that at 10 bar and 179.9^{o}C latent heat of vaporization is 2076 kJ/kg.

3 0
4 years ago
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