Answer:
Explanation:
Group one:
The elements of group one shows +1 charge because these all are metals and lose their one valance electron.
Hydrogen lithium sodium potassium rubidium cesium francium
Group 2:
The elements of group two shows +2 charge because these all alkali metals and lose their two valance electrons.
beryllium magnesium calcium strontium barium radium
Group 3:
The elements of group three-B shoes +3 charge by losing three valance electrons.
Scandium yttrium lanthanum actinium
Group 4:
The elements of group 4th A and 4th B lose four electrons or gain four electrons to complete the octet and shows +4 or -4 charge.
Group 5:
Group 5th elements gain three electrons and shows -3 charge to complete the 8 electrons. (octet).
It involve the elements of group 5th A.
Group 6:
The elements of group 6A gain two electrons to complete the octet and shows -2 charge.
Group 7:
The elements of group 7A gain one electron to complete the octet and shows -1 charge.
Group 8:
The elements of group 8A are noble gases and have complete octet. That's why shows 0 charge.
Answer : The correct option is, 13.7 mole
Solution : Given,
Moles of 
= 27.4 moles
The given balanced chemical reaction is,
From the balanced chemical reaction, we conclude that
As, 2 moles of react with 1 moles of 
So, 27.4 moles of react with 
moles of
Therefore, the number of moles of oxygen required are, 13.7 moles
Answer:
29,200 cal = 1.22 E 5 joules
Explanation:
hope this helps
The answer is 33.2 moles of a solute is present in 4.00 L of an 8.30 M solution , Option A is correct .
<h3>
What is Molarity ?</h3>
Molarity is defined as the amount of solute (in moles)in per litre of solution.
It is also known as molar concentration of a solution , It is expressed in mol/l
We can rearrange this equation to get the number of moles:
n= M * V
The molarity of solution is 8.3 M and the volume given is 4 litres
the moles will be n = 8.30 * 4 = 33.2 moles
Therefore 33.2 moles of a solute is present in 4.00 L of an 8.30 M solution , Option A is correct .
To know more about molarity
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