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IRINA_888 [86]
2 years ago
14

Please help

Chemistry
1 answer:
Softa [21]2 years ago
6 0

If the colored spheres are solutes, the overall direction of water would be out of the cell based on the principle of osmosis.

<h3>What is osmosis</h3>

It is the movement of water molecules from the region of high water potential (low solutes) to the region of low water potential (high solute) through a selectively permeable membrane.

In the illustration, there are more colored spheres outside of the cell than inside of it. This means that there are more solutes outside the cell (lower water potential) than inside of the cell (higher water potential).

Thus, water will move from the inside of the cell to the outside until an equilibrium is established.

More on osmosis can be found here: brainly.com/question/1799974

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Radioactivity is a random Process, Why?
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Explanation:

Radioactive decay is the set of various processes by which unstable atomic nuclei (nuclides) emit subatomic particles (radiation). Decay is said to occur in the parent nucleus and produces a daughter nucleus. This is a random process, i.e. it is impossible to predict the decay of individual atoms.

HOPE THSI HELPS HAVE A NICE DAY!

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What you can use to represent the valence electrons of an atom
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a calorimeter contained 75.0 g of water at 16.95 C. A 93.3-g sample of iron at 65.58 C was placed in it, giving a final temperat
Nostrana [21]

Answer:- \frac{382.69J}{0C} .

Solution:- Mass of Iron added to water is 93.3 g. Initial temperature of iron metal is 65.58 degree C and final temperature of the system is 19.68 degree C.

temperature change, \Delta T for iron metal = 65.58 - 19.68 = 45.9 degree C

specific heat for the metal is given as 0.444 J per g per degree C.

let's calculate the heat lost by iron metal using the equation:

q=mc\Delta T

where, q is the heat energy, m is mass, c is specific heat and delta T is change in temperature. let's plug in the values and calculate q for iron metal:

q=93.3g(45.9^0C)(\frac{0.444J}{g.^0C})

q = 1901.42 J

Using same equation we will calculate the heat gained by water.

mass of water is 75.0 g.

\Delta T for water = 19.68 - 16.95 = 2.73 degree C

specific heat for water is 4.184 J pr g per degree C. Let's plug in the values:

q=75.0g(\frac{4.184J}{g.^0C})(2.73^0C)

q = 856.674 J

Total heat lost by iron metal is the sum of heat gained by water and calorimeter.

So, heat gained by calorimeter = heat lost by iron metal - geat gained by water

heat gained by calorimeter = 1901.42 J - 856.674 J = 1044.746 J

Change in temperature for calrimeter is same as for water that is 2.73 degree C

For calorimeter, q=C.\Delta T

C=\frac{q}{\Delta T}

C=\frac{1044.746J}{2.73^0C}

C=\frac{382.69J}{0C}

So, the heat capacity of calorimeter is \frac{382.69J}{0C} .


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Answer:

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