Answer:
4 ClO-
Explanation:
Bond length ∝ 1/ Bond order
Bond order of ClO − =1,
Bond order of ClO 2− =1.5,
Bond order of ClO 3− =1.66 ,
Bond order of ClO 4− =1.75.
Thus the order of bond length is
ClO 4− <ClO3− <ClO2−
<ClO −
Seeing as there are 12 Carbon atoms and 12 Oxygen atoms, the ratio is 12:11
12/11 etc.
Answer:
77 L of water can be made.
Explanation:
Molar mass of 
= 32 g/mol
So, 55 g of = 
mol of = 1.72 mol of
As hydrogen is present in excess amount therefore is the limiting reagent.
According to balanced equation, 1 mol of produces 2 mol of 
.
So, 1.72 mol of produce 
mol of or 3.44 mol of .
Let's assume gas behaves ideally at STP.
Then, 
, where P, V, n, R and T represents pressure, volume, no. of moles, gas constant and temperature in kelvin scale respectively.
At STP, pressure is 1 atm and T is 273 K.
Here, 
= 3.44 mol and R = 0.0821 L.atm/(mol.K)
So, 
Option (b) is correct.
The enthalpy of the reaction can be obtained from the enthalpies of formation as -16.2 kJ/mol.
<h3>What is the enthalpy of reaction?</h3>
We know that any time that there is a chemical reaction, there is an interaction that place between the reactants and the products and as such we are going to get new substances and these are the substances that I have referred to here as the products of the reaction.
In this case, we are asked to obtain the enthalpy change of the reaction. This tells us the heat that could have been absorbed or evolved in the reaction. We have to at this point know that the enthalpy change of the reaction gotten from;
Sum of enthalpy of the products - Sum of enthalpy of the reactants
ΔH = [(-484.5)] - [(-393.5) + (-74.8)]
ΔH = (-484.5) + 468.3
ΔH = -16.2 kJ/mol
Learn more about enthalpy:brainly.com/question/3393755
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To balance a reaction, we must say to it that the number of elements in one side is equal to the other side. For a combustion reaction such as the one given, we first need to balance the number of carbon atoms, then the hydrogen atoms and lastly the oxygen atoms.
<span>C4H10 + 13/2O2 → 4CO2 + 5H2O.</span>
