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lora16 [44]
3 years ago
7

Please help!!! I will mark Brainliest !!!!

Chemistry
2 answers:
shepuryov [24]3 years ago
5 0

Answer:

c

Explanation:

Andrej [43]3 years ago
4 0

Answer: C. C6H12O6 + 6 O2 → 6 CO2 + 6 H2O + Energy

Explanation: We know that 'Aerobic Cellular Respiration' is the process to which all organisms use C6H12O6, Glucose, and 6O2 to create 6CO2 and 36 ATP. This is called 'Aerobic' because it has to require oxygen. In the reaction, C6H12O6+6O2, are all the reactants and 6CO2+6H20+36ATP, are all the products.

I hope this helps you! ☺

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Explanation:

The given data is as follows.

     V_{1} = 50 ml,      T_{1} = 345 K

     T_{2} = 298 K,      T_{f} = 317 K,

    V_{2} = 50 ml

Now, we will calculate the heat energy as follows.

        Q_{hot} = m \times C_{p} \times (T_{1} - T_{f})

                     = 50 g \times 4.184 \times (345 - 317)

                     = 5857.6 J

       Q_{cold} = m \times C_{p} \times (T_{f} - T_{2})

                     = 50 g \times 4.184 \times (317 - 298)

                     = 3974.8 J

As,   Q_{hot} = -Q_{cold} so there will be loss of heat. And, some heat will go to the calorimeter.

Hence,     Q_{hot} = Q_{cold} + Q_{cal}

                 5857.6 = 3974.8 + Q_{cal}

               Q_{cal} = 1882.8 J

We assume that the temperature of (calorimeter + water) is 298 K. Hence,

                  dT = T_{f} - T_{2}

                        = (317 - 298) K

                        = 19 K

Hence, we will calculate the specific heat as follows.

               C = \frac{Q}{dT}

                   = \frac{1882.8 J}{19}

                   = 99.1 J/K

Thus, we can conclude that the value of C_{cal} for the calorimeter is 99.1 J/K.

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