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3 years ago

How is a suspension different from a colloid?

Chemistry

1 answer:

Mashcka [7]3 years ago

3 0

Answer:

Explanation:

Colloidal:

Colloid consist of the particles having size between 1 - 1000 nm i.e, 0.001- 1μm.

The particles in colloid can not be seen through naked eye.

It is homogeneous.

We can not separate the colloidal through the filtration. The pore size of filter paper is 2μm. However it can be separated through the ultra filtration. In ultra filtration the pore size is reduced by soaking the filter paper in gelatin and then in formaldehyde. This is only in case of when solid colloidal is present, if colloid is liquid , there is no solid particles present and ultra filtration can not be used in this case.

Suspension:

The particle size in suspension is greater than 1000 nm.
These particles can seen through naked eye.

It is heterogeneous.
The particles in suspension can be separated through the filtration.

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If you dispense 40 ml of hexane, but it turns out you only need 5 ml, what should you do with the remainder?

Natasha2012 [34]

After subtracting the volume needed from the volume dispensed, we got a remainder of 35ml

<h3>Subtraction of Numbers</h3>

Given Data

Volume of Hexane dispensed = 40ml

Volume needed = 5 ml

Let us compute the amount of excess hexane/ the volume that will remain

Remainder = The difference in volume dispensed and the volume needed

Remainder = 40-5

Remainder = 35 ml

The remainder is 35ml

Learn more about subtraction of numbers here:

brainly.com/question/4721701

7 0

2 years ago

It is found that a gas undergoes a first-order decomposition reaction. If the rate constant for this reaction is 8.1 x 10-2 /min

kap26 [50]

Answer:

28.43 min

Explanation:

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given that:

The rate constant, k = $8.1\times 10^{-2}$ min⁻¹

Initial concentration $[A_0]$ = 0.1 M

Final concentration $[A_t]$ = $1.0\times 10^{-2}$ M

Time = ?

Applying in the above equation, we get that:-

$1.0\times 10^{-2}=0.1e^{-8.1\times 10^{-2}\times t}$

$0.1e^{-8.1\times \:10^{-2}t}=10^{-2}$

$e^{-8.1\times \:10^{-2}t}=\frac{1}{10}$

$\ln \left(e^{-8.1\times \:10^{-2}t}\right)=\ln \left(\frac{1}{10}\right)$

$t=28.43\ min$

3 0

3 years ago

Calculate the volume in liters of a 0.00231M copper(II) fluoride solution that contains 175.g of copper(II) fluoride CuF2. Be su

Free_Kalibri [48]

Answer:

Volume = 746 L

Explanation:

Given that:- Mass of copper(II) fluoride = 175 g

Molar mass of copper(II) fluoride = 101.543 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{175\ g}{101.543\ g/mol}$

$Moles_{copper(II)\ fluoride}= 1.7234\ mol$

Also,

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

So,,

$Volume =\frac{Moles\ of\ solute}{Molarity}$

Given, Molarity = 0.00231 M

So,

$Volume =\frac{1.7234}{0.00231}\ L$

<u>Volume = 746 L</u>

7 0

3 years ago

Find the density of an object that has a mass of 28 grams and a volume of 2 points<br> 38.6 cm^3. *

Shtirlitz [24]

Answer:

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question we have

$density = \frac{28}{38.6} \\ =0.72538...$

We have the final answer as

Hope this helps you

4 0

3 years ago

If an element is more reactive, is it more likely to be found as an element or a compound?

laila [671]

It is more likely to be found as a COMPOUND, as it is more reactive, by the time we found them, they're already reacted with other elements or compounds to form new compounds.
Example is oxygen, it is very reactive, therefore we often found oxygen in water, which is H2O, in earth, instead of just pure oxygen.

5 0

3 years ago