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2 years ago

13

A 0.1 M NaCl solution and a 0.2 M glucose solution are separated by a membrane that is impermeable to these solutes but is perme

able to water. What will happen

1 answer:

katovenus [111]2 years ago

7 0

nothing will happen

if you calculate the water potiential for both solutions

you get 2(0.1)(R)(T) and 1(0.2)(R)(T)

which is the same thing, so its isotonic.

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A gas occupies a volume of 30.0L, a temperature of 25°C and a pressure of 0.600atm. What will be the volume of the gas at STP?

Shalnov [3]

Answer:

=16.49 L

Explanation:

Using the equation

P1= 0.6atm V1= 30L, T1= 25+273= 298K, P2= 1atm, V2=? T2= 273

P1V1/T1= P2V2/T2

0.6×30/298= 1×V2/273

V2=16.49L

5 0

3 years ago

Susan made observations about outside events she noticed throughout the day. The day began with rain. Susan saw puddles on the r

S_A_V [24]

Answer:

Rainfall - precipitation

disappeared puddles - evaporation

cloud formation - condensation

Explanation:

Rainfall that is observed by the Susan is the precipitation of the water cycle in which the water vapor that was condensed become heavy and form droplets of water and fall from sky to the earth surface.

Puddles that formed due to rainfall are the collection of water in the water cycle which is evaporated (process: evaporation) into the atmosphere n the form of water vapor which condensed to form clouds.

4 0

3 years ago

If you have 100 ml of a 0.10 m tris buffer (pka 8.3) at ph 8.3 and you add 3.0 ml of 1.0 m hcl, what will be the new ph?

sukhopar [10]

The new pH is 7.69.

According to Hendersen Hasselbach equation;

The Henderson Hasselbalch equation is an approximate equation that shows the relationship between the pH or pOH of a solution and the pKa or pKb and the ratio of the concentrations of the dissociated chemical species. To calculate the pH of the buffer solution made by mixing salt and weak acid/base. It is used to calculate the pKa value. Prepare buffer solution of needed pH.

pH = pKa + log10 ([A–]/[HA])

Here, 100 mL of 0.10 m TRIS buffer pH 8.3

pka = 8.3

0.005 mol of TRIS.

∴ $8.3 = 8.3 + log \frac{[0.005]}{[0.005]}$

<em> </em>inverse log 0 = $\frac{[B]}{[A]}$

$\frac{[B]}{[A]} = 1$

Given; 3.0 ml of 1.0 m hcl.

pka = 8.3

0.003 mol of HCL.

$pH = 8.3 + log \frac{[0.005-0.003]}{[0.005+0.003]}\\pH = 8.3 + log \frac{[0.002]}{[0.008]}\\\\pH = 8.3 + log {0.25}\\\\pH = 8.3 + (-0.62)\\pH = 7.69$

Therefore, the new pH is 7.69.

Learn more about pH here:

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8 0

1 year ago

The specific heat of copper metal is 0. 385 J/(g °C). How much energy must be added to a 35. 0-gram sample of copper to change t

Rus_ich [418]

The amount of heat required for changing the temperature of copper has been 606 J. Thus, option B is correct.

Specific heat has been defined as the amount of heat required to raise the temperature of 1 gram of substance by 1 degree Celsius.

The heat required to raise the temperature has been expressed as:

$\rm Heat=mass\;\times\;specific\;heat\;\times\;Change\;in\;temperature$

<h3>Computation for the heat energy required</h3>

The given specific heat of copper has been $\rm 0.385\;J/g^\circ C$

The mass of copper has been, $\rm 35\;g$

The initial temperature of copper has been, $\rm 20^\circ C$

The final temperature of copper has been, $\rm 65^\circ C$

The change in temperature has been, $\Delta T$

$\Delta T=\text{Final\;temperature-Initial\;temperature}\\\Delta T =65^\circ \text C-20^\circ \text C\\\Delta T=45^\circ \text C$

Substituting the values for the heat required as:

$\rm Heat=35\;g\;\times\;0.385\;J/g^\circ C\;\times\;45^\circ C\\Heat=606\;J$

The amount of heat required for changing the temperature of copper has been 606 J. Thus, option B is correct.

Learn more about specific heat, here:

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7 0

2 years ago

The partial negative charge at one end of a water molecule is attracted to the partial positive

Neporo4naja [7]

Answer:

Hydrogen Bond

Explanation:

Hydrogen bond interactions are formed between the hydrogen atom bonded to most electronegative atoms (i.e. F, O and N) of one molecule and most electronegative atom (i.e. F, O and N) of another molecule.

In this interaction the hydrogen atom has partial positive charge and electronegative atom has partial negative charge.

6 0

2 years ago