1 — Element D and A ( Which are Sodium and aluminium )
2 — ( 2 + 8 + 3 = 13 electrons total ) Element A Because it's atomic number is 13.
3 — Element E Is stable. ( Which is Argon )
( Note, Elements which has 8 election on its outermost cell is stable. ( Helium is exception which is a noble gas but have 2 electrons in outermost cell )
4 — Element F , which is hydrogen. ( Hydrogen is the only element to not have any neutron )
5 — Element D ( Which is sodium )
6 — The element F ( Which is hydrogen ) Don't contain any neutron.
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The balanced equation is:
2NaCl + 2 H2o → 2 NaOH + Cl2 + H2
1- we put 2 NaCl to balance Cl2 on the other side.
2- we put 2 NaOH to balance Na
3- We put 2 H2O to balance H & O on the to sides
5.20 mol C6H12* (84.2 g/mol C6H12)= 4.38*10^(2) g C6H12.
The answer should only have three significant figures, according to the numbers in the problem.
Hope this is helpful~
In an average mass, each entry has equal weight. In a weighted average, we multiply each entry by a number representing its relative importance.
Assume that your class consists of 15 girls and 5 boys. Each girl has a mass of 54 kg, and each boy has a mass of 62 kg.
<em>Average mass</em> = (girl + boy)/2 = (54 kg + 62 kg)/2 = <em>58 kg</em>
<em>Weighted average (Method 1)
</em>
Use the <em>numbers of each</em> gender (15 girls + 5 boys)
,
Weighted average = (15×54 kg + 5×62 kg)/20 = (810 kg + 310 kg)/20
= 1120 kg/20 = <em>56 kg</em>.
If you put all the students on one giant balance, their total mass would be
1120 kg and the average mass of a student would be <em>56 kg.
</em>
<em>Weighted average (Method 2)
</em>
Use the <em>relative percentages</em> of each gender (75 % girls and 25 % boys).
Weighted average = 0.75×54 kg + 0.25×62 kg = 40.5 kg + 15.5 kg = <em>56 kg</em>
Each girl contributes 40.5 kg and each boy contributes 15.5 kg to the <em>weighted average</em> mass of a student.
