Write out the eqn of magnesium and oxygen. this should be under “metals” chapter. do revise.
next, find the mols of both oxygen and magnesium. compare the ratios and find the LIMITING REAGENT.
use the mols of the limiting reagent to compare with the mols of the product.
take the mols of the product/mr of the product.
this will give u the mass.
Answer:
No they are not the same the are both di
1. A mixture of ammonium chloride, sand, and zinc chloride should be separated by sublimation.
2. A mixture of zinc chloride and silver chloride should be separated through crystallization.
<h3>What is a separation technique?</h3>
A separation technique can be defined as a technique that is typically used to separate or convert two (2) or more mixture and solution of chemical substances into distinct product such as chemical compounds or elements.
<h3>The types of
separation technique.</h3>
In Chemistry, there are various types of separation technique used for the separation of mixtures or solutions and these include:
In this scenario, the most effective and efficient means to separate a mixture of ammonium chloride, sand, and zinc chloride is by sublimation from solid to gas state.
On the other hand, the best means to separate a mixture of zinc chloride and silver chloride is through crystallization.
Read more on crystallization here: brainly.com/question/4980962
The formula we use would be the graham's law. We do as follows:
<span>E_Kr / E_Ne = sqrt ( M_Ne / M_Kr)
</span>
<span>= sqrt ( 20.1797 g/mol / 83.798 g/mol ) </span>
<span>= sqrt (0.24081) </span>
<span>= 0.4907
</span>
Hope this answers the question. Have a nice day.
The relationship of radiation with distance obeys the inverse square law. Therefore, doubling the distance decrease the radiation by a factor of 4. The new count is 250.
1) Applying the same principle, the count decreases by a factor of 100. The new count is 10
2) An alpha particle is 4He2 and the Hydrogen can be represented as 1H1
14N7 + 4He2 - 1H1
= 17X8
Proton number 8 belongs to Oxygen. Therefore, the resultant nucleus is:
17O8
3) 185Au79 - 4He2
= 181Ir77
4) X - 4He2 = 234Th90
X = 238U92
5) Beta emission results in the same nucleon number but an increase in the proton number; therefore, the result is:
234Pa91
