☃️ Chemical formulae ➝ 
How to find?
For solving this question, We need to know how to find moles of solution or any substance if a certain weight is given.
Solution:
❍ Molecular weight of
= 2 × 126.90
= 253.80
= 254 (approx.)
❍ Given weight: 12.7
Then, no. of moles,
⇛ No. of moles = 12.7 / 254
⇛ No. of moles = 0.05 moles
⚘ No. of moles of Iodine molecule in the given weight = <u>0.05</u><u> </u><u>moles </u>
<u>━━━━━━━━━━━━━━━━━━━━</u>
Answer:
0.641 moles of ethane
Explanation:
Based on the equation:
C2H6(g) + 7/2O2(g) → 2CO2(g) + 3H2O(l)
We can determine ΔH of reaction using Hess's law. For this equation:
<em>Hess's law: ΔH products - ΔH reactants</em>
ΔH = {2ΔHCO2 + 3ΔHH2O} - {ΔHC2H6}
<em>Pure monoatomic substances have a ΔH = 0kJ/mol; ΔHO2 = 0kJ/mol</em>
<em />
ΔH = {2*-393.5kJ/mol + 3*-285.8kJ/mol} - {-84.7kJ/mol}
ΔH = -1559.7kJ/mol
That means when 1 mole of ethane is in combustion there are released 1559.7kJ of heat. To produce 1.00x10³kJ there are needed:
1.00x10³kJ * (1mole ethane / 1559.7kJ) =
<h3>0.641 moles of ethane</h3>
Answer: good luck with that F
Explanation:
Answer:
Explanation:
Hello,
In this case, for first order reactions, we can use the following integrated rate law:
![ln(\frac{[A]}{[A]_0} )=kt](https://tex.z-dn.net/?f=ln%28%5Cfrac%7B%5BA%5D%7D%7B%5BA%5D_0%7D%20%29%3Dkt)
Thus, we compute the time as shown below:
![t=-\frac{ln(\frac{[A]}{[A]_0} )}{k}=- \frac{ln(\frac{0.220M}{0.690M} )}{0.55s^{-1}} \\\\t=-\frac{-1.14}{0.550s^{-1}}\\ \\t=2.08s](https://tex.z-dn.net/?f=t%3D-%5Cfrac%7Bln%28%5Cfrac%7B%5BA%5D%7D%7B%5BA%5D_0%7D%20%29%7D%7Bk%7D%3D-%20%5Cfrac%7Bln%28%5Cfrac%7B0.220M%7D%7B0.690M%7D%20%29%7D%7B0.55s%5E%7B-1%7D%7D%20%5C%5C%5C%5Ct%3D-%5Cfrac%7B-1.14%7D%7B0.550s%5E%7B-1%7D%7D%5C%5C%20%5C%5Ct%3D2.08s)
Best regards.
The atoms which make up the ion are covalently bonded to one another. 19) It is possible for a compound<span> to possess </span>both ionic and covalent bonding<span>. a. If one of the </span>ions<span> is polyatomic then there will be </span>covalent bonding<span> within it.</span>
