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muminat
3 years ago
8

The following conditions exist in a bacterium and its environment. Environment: 5M NaCl; 1M glucose Inside cell: 3M NaCl; 10M gl

ucose The bacterium needs more glucose in the cell. True or false. Facilitated diffusion can be utilized to move glucose into the cell.
Chemistry
1 answer:
NARA [144]3 years ago
3 0

Answer:

False

Explanation:

Diffusion uses the concentration gradient that has been set up, this is a naturally occurring phenomena, and using a diffusion gradient some small particles can cross over the cell membrane. Some bigger or polar molecules require facilitated diffusion to move these molecules across the membrane.

Facilitated diffusion still uses the gradient for passive transport. This means that ATP is not used to transport molecules.

The reason that glucose will not move into the cell via passive transport is because there is a higher concentration of glucose inside the cell, meaning it needs active transport to move glucose (against the concentration gradient) into the cell.

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Collisions between gas particles are elastic; there is no net gain or loss of kinetic energy.

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The gas molecules move at very high speed and collide with each other and with the walls of container.

The collision of these particles with each other is perfectly elastic hence the kinetic energy of the colliding gas particles do not change.

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The substance water always has a mass ratio of 11% H to 89% O. If 5.00g of a substance containing H and O was decomposed into .2
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Answer:

                    No the substance is not water.

Explanation:

                   The balance chemical equation for the decomposition of water is as follow;

                                           2 H₂O = 2 H₂ + O₂

Step 1: <u>Calculate moles of H₂O;</u>

               Moles  =  Mass / M.Mass

               Moles  =  5.0 g / 18.01 g/mol

               Moles  =  0.277 moles of H₂O

Step 2: <u>Calculate Moles of O₂ and H₂ produced by 0.277 moles of H₂O:</u>

According to equation,

                        2 moles of H₂O produced  =  1 mole of O₂

So,

                  0.277 moles of H₂O will produce  =  X moles of O₂

Solving for X,

                     X =  0.277 mol × 1 mol / 2 mol

                     X =  0.138 moles of O₂

Also,

According to equation,

                        2 moles of H₂O produced  =  2 mole of H₂

So,

                  0.277 moles of H₂O will produce  =  X moles of H₂

Solving for X,

                     X =  0.277 mol × 2 mol / 2 mol

                     X =  0.227 moles of H₂

Step 3: <u>Calculate Mass of O₂ and H₂ as;</u>

For O₂:

                 Mass  =  Moles × M.Mass

                 Mass  =  0.138 mol × 31.99 g/mol

                 Mass  =  4.44 g of O₂

For H₂:

                 Mass  =  Moles × M.Mass

                 Mass  =  0.227 mol × 2.01 g/mol

                 Mass  =  0.559 g of H₂

Conclusion:

                   From conclusion it is proved that the amount of H₂ produced by decomposition of 5 g of water should be 0.559 g while in statement it is less i.e. 0.290 g.

6 0
3 years ago
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