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Hunter-Best [27]
3 years ago
5

Please answer the question

Chemistry
1 answer:
Tom [10]3 years ago
7 0

Answer:

Stamens

Explanation:

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A resting adult requires about 240 ml of pure oxygen/mind and breathes about 12 times every minute. if inhaled air contains 20 p
Dmitry_Shevchenko [17]

12 times breathe give 240 ml of pure O_{2}. Each breathe gives 20 ml of O_{2}.

Let us consider, volume of air per breathe= x ml.

Pure O_{2} from inhaled air= \frac{20}{100}x ml and Pure O_{2} from exhaled air= \frac{16}{100}x ml.

Pure O_{2} from inhaled and exhaled air= 20 ml

So, \frac{20}{100}x + \frac{16}{100}x = 20

Therefore, x = 55.5 ml

So, volume of air per breath= 55.5 ml.


7 0
3 years ago
Through which conversion is energy released? adp to atp atp to adp nadp to nadph co2 and h2o to c6h12o6 and o2
aleksandrvk [35]
  • Energy transformation includes ATP and ADP. at the point when energy is delivered, the response will in general separate a bigger particle to a more modest structure.
  • For this situation, the bigger particle is ATP comprised of three phosphates bunches while ADP is just made out of two phosphate gatherings.
<h3>What is the energy source for the conversion of ADP and ATP?</h3>
  • The energy expected for the change of ADP into ATP is acquired from light during photosynthesis and from exothermic responses during cell breath in the two plants and creatures.
  • ADP is produced on hydrolysis of ATP and the energy released in the process is utilised to carry out various cellular processes.

To learn more about energy release from the given link

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3 0
2 years ago
What part of the cell is a cop
Tanya [424]

Answer: This is from a wiki i found. Approximately one third of a cell’s proteins are destined to function outside the cell’s boundaries or while embedded within cellular membranes. Ensuring these proteins reach their diverse final destinations with temporal and spatial accuracy is essential for cellular physiology. In eukaryotes, a set of interconnected organelles form the secretory pathway, which encompasses the terrain that these proteins must navigate on their journey from their site of synthesis on the ribosome to their final destinations. Traffic of proteins within the secretory pathway is directed by cargo-bearing vesicles that transport proteins from one compartment to another. Key steps in vesicle-mediated trafficking include recruitment of specific cargo proteins, which must collect locally where a vesicle forms, and release of an appropriate cargo-containing vessel from the donor organelle (Figure 1). The newly formed vesicle can passively diffuse across the cytoplasm, or can catch a ride on the cytoskeleton to travel directionally. Once the vesicle arrives at its precise destination, the membrane of the carrier merges with the destination membrane to deliver its cargo. Have a nice day.

Explanation: Plz make brainliest

6 0
2 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
Through which material do sound waves tend to move fastest
ira [324]

Answer:

METAL

Explanation:

3 0
3 years ago
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