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a_sh-v [17]
2 years ago
12

A word which describes a solution which is more dilute than cytoplasm and would cause a cell to swell with water

Chemistry
1 answer:
Snowcat [4.5K]2 years ago
6 0

Answer:

A hypertonic solution has increased solute, and a net movement of water outside causing the cell to shrink. A hypotonic solution has decreased solute concentration, and a net movement of water inside the cell, causing swelling or breakage.

Explanation: hope this helps :) sorry if it's wrong :(

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Aqua regia, a mixture of HCl and HNO₃, has been used since alchemical times to dissolve many metals, including gold. Its orange
marissa [1.9K]

Aqua regia is an oxidative mixture that is highly corrosive and is composed of hydrochloric acid and nitric acid. The Ea (rev) for the reaction is 3 kJ.

<h3>What is activation energy?</h3>

The activation energy is the minimum required energy by the reactant to undergo changes to form the product. The activation energy of the reverse reaction is given by the difference in the production state and transition state.

It is given as,

Ea (rev) = Ea (fwd) − ΔHrxn

Given,

ΔH° = 83KJ

Ea (fwd) = 86 kJ/mol

Substituting the values above as:

Ea (rev) = 86 - 83

= 3 kJ

Therefore, the activation energy of the reverse reaction is 3 kJ.

Learn more about activation energy, here:

brainly.com/question/14287952

#SPJ4

4 0
1 year ago
Please help I’ll mark brainliest!
Anni [7]

Answer:

D because excited electrons fall back.....

6 0
3 years ago
Read 2 more answers
What is true of a scientific law?
Minchanka [31]

Answer: d

Explanation:

5 0
3 years ago
1 mol super cooled liquid water transformed to solid ice at -10 oC under 1 atm pressure.
Arada [10]

Answer:

Explanation:

Given that:

number of moles of super cooled liquid water = 1

Melting enthalpy of ice = 6020 J/mol

Freezing point =0 °C = (0 + 273 K)= 273 K

The decrease in entropy of the system during freezing for 1 mol (i.e during transformation from liquid water to solid ice )  = - 6020 J/mol × 1 mol /273 K = -22.051 J/K

Entropy change during further cooling from 0 °C (273 K) to -10 °C (263 K)

\Delta \ S = \int\limits^{T_2}_{T_1}\dfrac{nC_p(s)dT}{T}

\Delta \ S = {nC_p(s)In \dfrac{T_2}{T_1}

\Delta \ S = {(1*37.7)In \dfrac{263}{273}

Δ S = -1.4 J/K

Total entropy change of the system = - 22.05 J/K - 1.4 J/K = - 23.45 J/K

Entropy change of universe = entropy change of the system+ entropy change of the surrounding

According to the second law of thermodynamics

Entropy change of universe  >0

SO,

Entropy change of the system + entropy change in the surrounding > 0

Entropy change in the surrounding > - entropy change of the system

Entropy change in the surrounding > - (- 23.53 J/K)

Entropy change in the surrounding > 23.53 J/K

b) Make some comments on entropy changes from the obtained data.

From the data obtained; we will realize that the entropy of the system decreases as cooling takes place when water is be convert to ice , randomness of these molecules reduces and as cooling proceeds , hence, entropy reduces more as well and the liberated heat will go into the surrounding due to this entropy of the surrounding increasing.

4 0
3 years ago
30 POINTS PLEASE HELP ASAP
lina2011 [118]

Hello!

We have the following data:

f (radiation frequency) = 3.0*10^{19}\:Hz

v (speed of light) = 3.0*10^8\:m/s

λ (wavelength) = ? (in m)

Let's find the wavelength, let's see:

f = \dfrac{v}{\lambda}

3.0*10^{19} = \dfrac{3.0*10^8}{\lambda}

\lambda = \dfrac{3.0*10^8}{3.0*10^{19}}

\boxed{\boxed{\lambda = 1*10^{-11}\:m}}\Longleftarrow(wavelenght)\end{array}}\qquad\checkmark

I Hope this helps, greetings ... DexteR! =)

4 0
3 years ago
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