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3 years ago

Solutions and colloids which contain similarly sized particles can be effectively separated using paper filtration.

Chemistry

2 answers:

mojhsa [17]3 years ago

6 0

False cause there particles arent large enough to be filtered

oee [108]3 years ago

4 0

Answer:

The given statement is false.

Explanation:

A solution is defined as the solution in which the components gets completely dissolved in it. Particles are evenly spread in these solutions. These solution does not scatter light falling on it.

A colloid is defined as the mixtures when physical boundary is seen between the dispersed phase and dispersed medium. They scatter the light falling on it.

Particle size of solutions are smaller than that of the colloids.But size of colloids particles is smaller than the size of pores filter paper used in filtration process due to which they easily escape through filter and results in no separation.

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Answer:

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3 years ago

Separate this redox reaction into its balanced component half-reactions. What is the oxidation and reduction half reactions

erma4kov [3.2K]

Separate this redox reaction into its component half-reactions.
Cl2 + 2Na ----> 2NaCl

reduction: Cl2 + 2 e- ----> 2Cl-1
oxidation: 2Na ----> 2Na+ & 2 e-

2) Write a balanced overall reaction from these unbalanced half-reactions:

oxidation: Sn ----> Sn^2+ & 2 e-
reduction: 2Ag^+ & 2e- ----> 2Ag

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3 0

3 years ago

Read 2 more answers

One kilogram of water at 100 0C is cooled reversibly to 15 0C. Compute the change in entropy. Specific heat of water is 4190 J/K

mina [271]

Answer:

The change in entropy is -1083.112 joules per kilogram-Kelvin.

Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy ( $s_{2} - s_{1}$ ), in joules per gram-Kelvin, by the following model:

$s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}}$ (1)

Where:

$m$ - Mass, in kilograms.

$c_{w}$ - Specific heat of water, in joules per kilogram-Kelvin.

$T_{1}$ , $T_{2}$ - Initial and final temperatures of water, in Kelvin.

If we know that $m = 1\,kg$ , $c_{w} = 4190\,\frac{J}{kg\cdot K}$ , $T_{1} = 373.15\,K$ and $T_{2} = 288.15\,K$ , then the change in entropy for the entire process is:

$s_{2} - s_{1} = (1\,kg) \cdot \left(4190\,\frac{J}{kg\cdot K} \right)\cdot \ln \frac{288.15\,K}{373.15\,K}$

$s_{2} - s_{1} = -1083.112\,\frac{J}{kg\cdot K}$

The change in entropy is -1083.112 joules per kilogram-Kelvin.

7 0

3 years ago

Consider the following reaction: 3Fe(s) + 4H₂O(g) ➞ Fe₃O₄(s) + 4H₂(g). To answer the following question: "How many moles of hot

UkoKoshka [18]

X moles of hot water vapor(H2O) * 1 mol Fe3O4 / 4 mol H2O * 233.55g Fe3O4 / 1 mol = 275 g

Solve the equation

x = 4.71 mol of hot water vapor (steam)

Can you mark me as brainliest?

7 0

2 years ago

The powder mixture (Cu, Al and Fe) = 10g was oxidized from sufficient chloride acid. 1) What are the possible reactions to the p

Mandarinka [93]

Answer:

1) 2Al + 6HCl ⟶ 2AlCl₃ + 3H₂

Fe + 2HCl ⟶ FeCl₂ + H₂

2) Cu = 2.5 g; Al = 3.5 g; Fe = 4.0 g

Explanation:

1) Possible reactions

2Al + 6HCl ⟶ 2AlCl₃ + 3H₂

Fe + 2HCl ⟶ FeCl₂ + H₂

2) Mass of each metal

a) Mass of Cu

The waste was the unreacted copper.

Mass of Cu = 2.5 g

b) Masses of Al and Fe

We have two relations :

Mass of Al + mass of Fe = 10 g - 2.5 g = 7.5 g

H₂ from Al + H₂ from Fe = 6.38 L at NTP

i) Calculate the moles of H₂

NTP is 20 °C and 1 atm.

$\begin{array}{rcl}pV & = & n RT\\\text{1 atm} \times \text{6.38 L} & = & n \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{293.15 K}\\6.38 & = & 24.06n \text{ mol}^{-1} \\n & = & \dfrac{6.38}{24.06 \text{ mol}^{-1} }\\\\ & = & \text{0.2652 mol}\\\end{array}$

(ii) Solve the relationship

Let x = mass of Al. Then

7.5 - x = mass of Fe

Moles of Al = x/27

Moles of Fe = (7.5 - x)/56

Moles of H₂ from Al = (3/2) × Moles of Al = (3/2) × (x/27) = x /18

Moles of H₂ from Fe = (1/1) × Moles of Fe = (7.5 - x)/56

∴ x/18 + (7.5 - x)/56 = 0.2652

56x + 18(7.5 - x) = 267.3

56x + 135 - 18x = 267.3

38x = 132.3

x = 3.5 g

Mass of Al = 3.5 g

Mass of Fe = 7.5 g - 3.5 g = 4.0 g

The masses of the metals are Cu = 2.5 g; Al = 3.5 g; Fe = 4.0 g

4 0

3 years ago