Additional information
Relative atomic mass(Ar) : A=7, G=16
The empirical formula : A₂G
<h3>Further explanation</h3>Given
3.5g of element A
4.0g of element G
Required
the empirical formula for this compound
Solution
The empirical formula is the smallest comparison of atoms of compound forming elements.
The empirical formula also shows the simplest mole ratio of the constituent elements of the compound
mol of element A :
mol of element G :
mol ratio A : G = 0.5 : 0.25 = 2 : 1
Answer:
Explanation:
Hello,
In this case, we write the reaction again:
In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:
Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:
But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:
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Answer:
The reaction is exothermic.
Yes, released.
The heat released is 4,08x10³ kJ.
Explanation:
For the reaction:
C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l)
The ΔH is -2220 kJ, As ΔH is <0, <em>The reaction is exothermic.</em>
As the reaction is exothermic, the heat of the reaction will be <em>released.</em>
The heat released in 81,0g is:
81,0g C₃H₈××
= <em>4,08x10³ kJ</em>
<em>-Using molar mass of C₃H₈ to convert mass to moles and knowing that there are released 2220 kJ per mole of C₃H₈-</em>
I hope it helps!