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Alekssandra [29.7K]
3 years ago
6

Contrast How are photosynthesis and cellular respiration different?

Chemistry
1 answer:
vovikov84 [41]3 years ago
3 0

Answer:

Photosynthesis converts carbon dioxide and water into oxygen and glucose. Cellular respiration converts oxygen and glucose into water and carbon dioxide.

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i have a balloon that can hold 100 liters of air. if i blow up this balloon with 3 moles of oxygen gas at a pressure of 1.0 atmo
Tcecarenko [31]

Baloon with 3 moles og oxygen at 1 atm.The temperature of the balloon is <u>4 Kelvin</u>.

An ideal gas is a theoretical gas composed of many randomly transferring factor particles that aren't difficult to interparticle interactions. the best gasoline idea is beneficial because it obeys the precise gas law, a simplified equation of country, and is amenable to evaluation under statistical mechanics.

An ideal gas is described as one for which both the extent of molecules and forces between the molecules are so small that they have got no effect at the behavior of the gas. The real gas that acts almost like a really perfect gasoline is helium. that is due to the fact helium, in contrast to maximum gases, exists as an unmarried atom, which makes the van der Waals dispersion forces as low as viable

Using the ideal gas equation:-

Given;

P₁ = 1 atm

V₁ = 100  L

n = 3

r = 8.314

T = PV/nR

  = 1 × 100 / 3 × 8.314

 = 4 K

Learn more about ideal gas here:-brainly.com/question/20348074

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4 0
1 year ago
Pressure gauge at the top of a vertical oil well registers 140 bars. The oil well is 6000 m deep and filled with natural gas dow
andreyandreev [35.5K]

Explanation:

(a)  The given data is as follows.

              Pressure on top (P_{o}) = 140 bar = 1.4 \times 10^{7} Pa       (as 1 bar = 10^{5})

              Temperature = 15^{o}C = (15 + 273) K = 288 K

         Density of gas = \frac{PM}{ZRT}

                \frac{dP}{dZ} = \rho \times g

               \frac{dP}{dZ} = \int \frac{PM}{ZRT}

                \int_{P_{o}}^{P_{1}} \frac{dP}{dZ} = \frac{Mg}{ZRT} \int_{0}^{4700} dZ

           ln (\frac{P_{1}}{P_{o}}) = \frac{18.9 \times 10^{-3} \times 9.81 \times 4700 m}{0.80 \times 8.314 J/mol K \times 288 K}

                              = 0.4548

                     P_{1} = P_{o} \times e^{0.4548}

                                 = 1.4 \times 10^{7} Pa \times 1.5797

                                 = 2.206 \times 10^{7} Pa

Hence, pressure at the natural gas-oil interface is 2.206 \times 10^{7} Pa.

(b)   At the bottom of the tank,

                 P_{2} = P_{1}  + \rho \times g \times h

                             = 2.206 \times 10^{7} Pa + 700 \times 9.81 \times (6000 - 4700)[/tex]

                             = 309.8 \times 10^{5} Pa

                             = 309.8 bar

Hence, at the bottom of the well at 15^{o}C pressure is 309.8 bar.

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They all are correct , so with that being said anyone of them can be right
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Answ burh i dont even know

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