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algol [13]
2 years ago
13

A scientist performs an experiment in which they create an artificial cell with a selectively permeable membrane through which o

nly water can pass. They inject a 5 M solution of glucose into the cell and then place the cell into a beaker of water. After an hour, what effect do you expect to observe?
Chemistry
1 answer:
lozanna [386]2 years ago
8 0

Answer:

Water moves into the cell

Explanation:

As shown in the question above, the cell is high in glucose and placed in a glass filled with water. This cell has a semi permeable membrane that allows only water to pass through, as the concentration of water within the cell is low, the cell will attempt to strike a balance with the medium it is inserted into. For this reason, what is likely to happen is the passage of water from the most concentrated to the least concentrated medium, that is, the water will pass from the cup to the cell.

water moves into the cell through osmosis.during osmosis water moves from a region of low concentration of solute to a region of high concentration of solute.the glucose introduced into the cell makes it more concentrated.

In this case the cell is hypertonic and water would enter into the cell through the semi permeable membrane.this membrane allows water to pass through but not glucose.this movement of water into the cell causes the cell to become turgid.

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Which of the following statements concerning a metal crystallized in a face-centered cubic cell is/are CORRECT? 1. One metal ato
AleksAgata [21]

Answer:

Only 3 is correct.

Explanation:

The crystal of a metal or an ionic compound is called a cell, and there are 7 types of unit cells: cubic, tetragonal, orthorhombic, monoclinic, hexagonal, rhombohedral, and triclinic.

In a face-centered cubic cell (FCC) all angles are 90º and all lengths are equal. Each cubic cell has 8 atoms in each corner of the cube, and that atom is shared with 8 neighboring cells. So for a metal crystal, the atom is located at each of the eight lattice points, where it is shared equally between eight unit cells.

5 0
2 years ago
In every balanced chemical equation, each side of the equation has the same number of _____.
Ber [7]

The correct answer is option d, that is, atoms of the element.  

As the atoms are neither destroyed nor created in a chemical reaction, the sum of the mass of the products in a reaction must be equivalent to the sum of the mass of the reactants.  

The chemical reactions must be balanced, they must exhibit a similar number of atoms of each element on both the sides of the equation. As a consequence, the mass of the reactants must be equivalent to the mass of the products of the reaction.  


7 0
2 years ago
Read 2 more answers
When the reaction CO2(g) + H2(g) ⇄ H2O(g) + CO(g) is at equilibrium at 1800◦C, the equilibrium concentrations are found to be [C
UNO [17]

Answer:

The new molar concentration of CO at equilibrium will be  :[CO]=1.16 M.

Explanation:

Equilibrium concentration of all reactant and product:

[CO_2] = 0.24 M, [H_2] = 0.24 M, [H_2O] = 0.48 M, [CO] = 0.48 M

Equilibrium constant of the reaction :

K=\frac{[H_2O][CO]}{[CO_2][H_2]}=\frac{0.48 M\times 0.48 M}{0.24 M\times 0.24 M}

K = 4

CO_2(g) + H_2(g) \rightleftharpoons H_2O(g) + CO(g)

Concentration at eq'm:

0.24 M          0.24 M                 0.48 M            0.48 M

After addition of 0.34 moles per liter of CO_2 and H_2 are added.

(0.24+0.34) M    (0.24+0.34) M  (0.48+x)M         (0.48+x)M

Equilibrium constant of the reaction after addition of more carbon dioxide and water:

K=4=\frac{(0.48+x)M\times (0.48+x)M}{(0.24+0.34)\times (0.24+0.34) M}

4=\frac{(0.48+x)^2}{(0.24+0.34)^2}

Solving for x: x = 0.68

The new molar concentration of CO at equilibrium will be:

[CO]= (0.48+x)M = (0.48+0.68 )M = 1.16 M

3 0
2 years ago
What is the kinetic energy acquired by the electron in hydrogen atom, if it absorbs a light radiation of energy 1.08x101 J. (A)
Delvig [45]

Explanation:

The given data is as follows.

            Energy of radiation absorbed by the electron in hydrogen atom = 1.08 \times 10^{-17} J

As energy is absorbed as a photon. Hence, frequency will be calculated will be as follows.

                                    E = h \nu

               1.08 \times 10^{-17} J = 6.626 \times 10^{-34} Js \times \nu

               \nu = 0.163 \times 10^{17} s^{-1}

or,                \nu = 1.63 \times 10^{16} s^{-1}    

It is known that,        \nu = \frac{c}{\lambda}

                1.63 \times 10^{16} s^{-1} = \frac{3 \times 10^{8} m/s}{\lambda}                  

                   \lambda = 1.84 \times 10^{-8} m

And, according to De-Broglie equation \lambda = \frac{h}{p}

as,        p = m \times \nu

So,          \lambda = \frac{h}{m \times \nu}

            m \times \nu = \frac{6.626 \times 10^{-34} Js}{1.84 \times 10^{-8} m}          

                             = 3.6 \times 10^{-26} J/m

Now, on squaring both the sides we get the following.

           (m \times \nu)^{2} = (3.6 \times 10^{-26} J/m)^{2}    

                              = 12.96 \times 10^{-52}  

               m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}

where,   m = mass of electron

So,           m \times \nu^{2} = \frac{12.96 \times 10^{-52}}{m}

                             = \frac{12.96 \times 10^{-52}}{9.1 \times 10^{-31}}

                                   = 1.42 \times 10^{-21} J

Since,  K.E = \frac{1}{2}m \nu^{2}

                 = \frac{1.42 \times 10^{-21} J}{2}

                 = 0.71 \times 10^{-21} J

Thus, we can conclude that kinetic energy acquired by the electron in hydrogen atom is 7.1 \times 10^{-22} J.

4 0
2 years ago
The concentration of Si in an Fe-Si alloy is 0.25 wt%. What is the concentration in kilograms of Si per cubic meter of alloy?
kodGreya [7K]

Answer : The concentration of Si in kilograms is, 19.55kg/m^3

Explanation :

As we are given that, the concentration of Si in an Fe-Si alloy is 0.25 wt% that means:

Weight of Si = 0.25 g = 0.00025 kg

Weight of Fe = 100 - 0.25 = 99.75 g = 0.09975 kg

Density of Si = 2.32g/cm^3=2.32\times 10^6g/m^3

Density of Fe = 7.87g/cm^3=7.87\times 10^6g/m^3

Now we have to calculate the concentration in kilograms of Si per cubic meter of alloy.

Concentration of Si in kilograms =  \frac{\text{Weight of Si in 100 g of alloy}}{\text{Volume of 100 g of alloy}}

Concentration of Si in kilograms =  \frac{\text{Weight of Si in 100 g of alloy}}{\frac{\text{Wight of Fe}}{\text{Density of Fe}}+\frac{\text{Wight of Si}}{\text{Density of Si}}}

Now put all the given values in this expression, we get:

Concentration of Si in kilograms = \frac{0.00025kg}{\frac{99.75g}{7.87\times 10^6g/m^3}+\frac{0.25g}{2.23\times 10^6g/m^3}}

Concentration of Si in kilograms = 19.55kg/m^3

Thus, the concentration of Si in kilograms is, 19.55kg/m^3

5 0
2 years ago
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