Explanation:
#Medicine
Compounds used as medicines are most often organic compounds, which are often divided into the broad classes of small organic molecules (e.g., atorvastatin, fluticasone, clopidogrel) and "biologics" (infliximab, erythropoietin, insulin glargine),
#Industry
Polymers and plastics such as polyethylene, polypropylene, polyvinyl chloride, polyethylene terephthalate, polystyrene and polycarbonate
Shale actually forms in the part of the rock cycle called compaction.
Given what we know, we can confirm that in a voltaic cell, the anode loses electrons and is oxidized, meanwhile, the cathode is reduced by gaining electrons.
<h3 /><h3>What is a voltaic cell?</h3>
- It is described as an electrochemical cell.
- These cells use chemical reactions to produce electrical energy.
- During this reaction, an anode loses electrons, thus oxidizing.
- Meanwhile, the cathode gains electrons and is reduced.
Therefore, given the nature of the voltaic cell, we can confirm that during its reaction, the anode is oxidized by losing electrons while the cathode becomes reduced by gaining them.
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Answer:
163.2g
Explanation:
First let us generate a balanced equation for the reaction. This is shown below:
4Al + 3O2 —> 2Al2O3
From the question given, were were told that 3.2moles of aluminium was exposed to 2.7moles of oxygen. Judging by this, oxygen is excess.
From the equation,
4moles of Al produced 2moles of Al2O3.
Therefore, 3.2moles of Al will produce = (3.2x2)/4 = 1.6mol of Al2O3.
Now, let us covert 1.6mol of Al2O3 to obtain the theoretical yield. This is illustrated below:
Mole of Al2O3 = 1.6mole
Molar Mass of Al2O3 = (27x2) + (16x3) = 54 + 48 =102g/mol
Mass of Al2O3 =?
Number of mole = Mass /Molar Mass
Mass = number of mole x molar Mass
Mass of Al2O3 = 1.6 x 102 = 163.2g
Therefore the theoretical of Al2O3 is 163.2g
Answer:
\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{2}
Explanation:
Hello,
The suitable differential equation for this case is:
As we're looking for the change in height with respect to the time, we need a relationship to achieve such as:
Of course, 
.
Now, since the volume of a cone is 
and the ratio 
or 
, the volume becomes:
We proceed to its differentiation:
Then, we compute 
Finally, at h=2:
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