Answer: Now it is proved that the atom consists of particles like electrons, protons, and neutrons.
Explanation: because he did not have the tech we do to understand atoms
Combustion can be defined as the reaction of a compound with oxygen. The enthalpy of combustion of octane is 
for 
.
<h3>What is the enthalpy of reaction?</h3>
The enthalpy of reaction is the amount of heat energy absorbed or lost by the molecules in the chemical reaction.
The enthalpy of combustion is the amount of heat energy released by the compound in the reaction with oxygen.
The reaction in which heat is liberated with the reaction of a compound with oxygen has an enthalpy of combustion, equivalent to the enthalpy of reaction.
The combustion of octane can be given as:
Thus, the reaction has combustion energy equivalent to the enthalpy of the reaction is . Thus, option B is correct.
Learn more about enthalpy of reaction, here:
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Aueyeowosy yo this is my answer
Hi!
<u>HCl is the limiting reactant.</u>
From the equation, we understand that 1 mole of Mg(OH)2 will react with 2 moles of HCl for the reaction to be complete. To find the limiting reactant we need to see which molecule is in lower quantity (in terms of moles and not mass)
Calculating the number of moles of Mg(OH)2 given
Using the formula, number of moles= mass / molecular mass
number of moles = 50.6 / 58 = ~ 0.872
Calculating the number of moles of HCl given
Number of moles = 45 / 36.5 = ~ 1.232
There are several ways to use this information to calculate the limiting reactant. In our case, we can subtract the number of moles of Mg(OH)2 from the number of moles of HCl and check for the remainder.
If it is more than 0.872, then HCl would be in excess, with Mg(OH)2 being the limiting reactant; however, if the remainder is less than 0.872, it would mean <em>HCl is the limiting reactant.</em>
1.232- 0.872 = 0.36
As 0.36 < 0.872 , we see that HCl is the limiting reagent.
NOTE: this is in the case that the moles of HCl needed to complete the reaction is twice that of Mg(OH)2.
Hope this helps!
Answer:
They don't have values there because they aren't on the pauling scale of electronegativity, as they don't form any compounds with other elements.
Explanation:
