Answer: Theoretical yield is 313.6 g and the percent yield is, 91.8%
Explanation:
To calculate the moles :


According to stoichiometry :
1 mole of
require 3 moles of 
Thus 2.8 moles of
will require=
of 
Thus
is the limiting reagent as it limits the formation of product and
is the excess reagent.
As 1 mole of
give = 2 moles of 
Thus 2.8 moles of
give =
of 
Mass of 
Theoretical yield of liquid iron = 313.6 g
Experimental yield = 288 g
Now we have to calculate the percent yield

Therefore, the percent yield is, 91.8%
1s^2 2s^2 2p^6 for the Mg2+ ion.
Stationary Front: a front that is not moving. When a warm or cold front stops moving, it becomes a stationary front.
When 6.85×10⁵ cal is converted to kilojoules, the result obtained is 2866.04 KJ
<h3>Data obtained from the question </h3>
- Energy (cal) = 6.85×10⁵ cal
- Energy (KJ) =?
<h3>Conversion scale </h3>
1 cal = 0.004184 KJ
<h3>How to convert 6.85×10⁵ cal to kilojoules</h3>
1 cal = 0.004184 KJ
Therefore,
6.85×10⁵ cal = 6.85×10⁵ × 0.004184
6.85×10⁵ cal = 2866.04 KJ
Thus, 6.85×10⁵ cal is equivalent to 2866.04 KJ
Learn more about conversion:
brainly.com/question/2139943
Answer:
Chemical properties, such as combustibility, are generally observed as the identity of a substance changes and one or more new substances form.
Explanation:
Chemical change involves formation of new substances. Therefore, the correct answer is "Chemical properties, such as combustibility, are generally observed as the identity of a substance changes and one or more new substances form."