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lys-0071 [83]
2 years ago
11

Name two properties that make solutions different from colloids.

Chemistry
2 answers:
Airida [17]2 years ago
3 0
Tyndall effect or Brownian movement and Particles of colloidal solution are the two properties that make solutions different from colloids.
Andrews [41]2 years ago
3 0

Answer:

1. the apperance of a solution is clear while that of a colloid is dull

2. colloids are heterogenou while solutions are homogenous

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What is the molarity of a strong acid solution with a pH of 3?
Anna35 [415]

Answer:

Since HCl is a strong acid, it completely ionizes, and the pH of HCl in solution can be found from the concentration (molarity) of the H+ ions, by definition equal to 0.100 M. (The conjugate base of the acid, which is the chloride ion Cl–, would also have a concentration of 0.100 M.) The pH is thus –log(0.100) = 1.000.

Explanation:

6 0
3 years ago
What would happen to the block if it had a density of 0.500 kg/L and was placed in the same 100.0 L tank of water?
raketka [301]
I’m not sure if there was important information in the question before this one, but the answer based on the info I have is B.

The density of water is 1kg/L. Since the density of the block is less, it will float.
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What is the only planet in our solar system that rotates in a different direction from the other planets??​
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Venus so that the sun rises in the west
4 0
3 years ago
It takes 495.0 kJ of energy to remove 1 mole of electron from an atom on the surface of sodium metal. How much energy does it ta
Zigmanuir [339]

Answer:

\lambda=241.9\ nm

Explanation:

The work function of the sodium= 495.0 kJ/mol

It means that  

1 mole of electrons can be removed by applying of 495.0 kJ of energy.

Also,  

1 mole = 6.023\times 10^{23}\ electrons

So,  

6.023\times 10^{23} electrons can be removed by applying of 495.0 kJ of energy.

1 electron can be removed by applying of \frac {495.0}{6.023\times 10^{23}}\ kJ of energy.

Energy required = 82.18\times 10^{-23}\ kJ

Also,  

1 kJ = 1000 J

So,  

Energy required = 82.18\times 10^{-20}\ J

Also, E=\frac {h\times c}{\lambda}

Where,  

h is Plank's constant having value 6.626\times 10^{-34}\ Js

c is the speed of light having value 3\times 10^8\ m/s

So,  

79.78\times 10^{-20}=\frac {6.626\times 10^{-34}\times 3\times 10^8}{\lambda}

\lambda=\frac{6.626\times 10^{-34}\times 3\times 10^8}{82.18\times 10^{-20}}

\lambda=\frac{10^{-26}\times \:19.878}{10^{-20}\times \:82.18}

\lambda=\frac{19.878}{10^6\times \:82.18}

\lambda=2.4188\times 10^{-7}\ m

Also,  

1 m = 10⁻⁹ nm

So,  

\lambda=241.9\ nm

6 0
3 years ago
How does the amount of salt dissolve in water affect the density of water
ad-work [718]

Answer:

Adding salt to the water increases the density of the solution because the salt increases the mass without changing the volume very much. When you add table salt (sodium chloride, NaCl) to water, the salt dissolves into ions, Na+ and Cl-. The volume increases by a small factor, but the mass increases by a bigger factor.

Explanation:

3 0
4 years ago
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