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Pie
3 years ago
10

explain why particles of colloidal solution donot settle down when left undisturbed , while in the case of suspensions they do .

?​
Chemistry
1 answer:
Leni [432]3 years ago
7 0

Answer:

the partial of colloidal are smaller and they are not heavy whereas the partial of suspension are larger but they are heavy and less movement thus they settle down due to gravity

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Nimfa-mama [501]

Answer:

The rotational period of the moon is equal to one revolution around Earth.

I think  this is the answer am not that positive.

Explanation

Only one side of the Moon is visible from Earth because the Moon rotates on its axis at the same rate that the Moon orbits the Earth – a situation known as synchronous rotation, or tidal locking. The Moon is directly illuminated by the Sun, and the cyclically varying viewing conditions cause the lunar phases.

7 0
3 years ago
Please help with both of these lol
ZanzabumX [31]

Answer: 1. Unbalanced

2. Atoms cannot be created or destroyed only rearranged

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7 0
3 years ago
Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho
Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3

c) The leftover is computed as follows:

m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0
3 years ago
If liquid carbon disulfide reacts with 450 mL of oxygen to produce the gases carbon dioxide and sulfur dioxide, what volume of e
ivolga24 [154]

If excess carbon disulfide reacts with 450 mL of oxygen, 150 mL of carbon dioxide and 300 mL of sulfur dioxide gases would be produced respectively.

<h3>Stoichiometric calculation</h3>

The reaction between liquid carbon disulfide and oxygen is represented by the equations below:

CS_2 + 3O_2 -- > CO_2 + 2SO_2

The mole ratio of oxygen to carbon dioxide and sulfur dioxide produced is 3:1:2.

Thus, for 450 mL oxygen, 1/3 x 450 = 150 mL of carbon dioxide will be required.

Also for 450 mL of oxygen, 2/3 x 450 = 300 mL of sulfur dioxide will be required.

More on stoichiometric calculations can be found here: brainly.com/question/27287858

#SPJ1

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2 years ago
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