Answer:
C5H12 + 8O2 --> 5CO2 + 6H2O
Explanation:
Complete question
Write a balanced chemical equation to represent the combustion of CH3CH2CH2CH2CH3
Solution
The given compound is pentane
C5H12
The empirical equation representing combustion of pentane is
C5H12 + O2 --> CO2 + H2O
We will first balance the carbon atoms
C5H12 + O2 --> 5CO2 + H2O
Now we will balance the Hydrogen molecule
C5H12 + O2 --> 5CO2 + 6H2O
Now we will balance the oxygen molecule
C5H12 + 8O2 --> 5CO2 + 6H2O
Answer:
( B) They all have their valence electrons in the same type of subshell.
Explanation:
With each period, a new shell is added to the atom.
Further, the groups are classified based on the type of subshell the last electron enters and number of valence electrons.
For all elements of same group, the last electron enters the same type of subshell.
Say, for group 1, last electron enters s orbital and they have 1 valence elctron.
for group 17, last electron enters p orbital and they have 7 valence electrons.
(A) and (D) are wrong because, energy level of the valence electrons is determined by the principle quantum number n and l and not by the type of subshell(only l) they enter.
(C) if the valence electron enters p orbital, then the elements will be placed in the p- block.
Answer:
The molar heat of combustion of hydrazine is -663.82 kJ/mole.
Explanation:
First we have to calculate the heat gained by the calorimeter.
where,
q = heat gained = ?
c = specific heat =
= final temperature =
= initial temperature =
Now put all the given values in the above formula, we get:
Now we have to calculate the enthalpy change during the reaction.
where,
= enthalpy change = ?
q = heat gained = 20.7444 kJ
n = number of moles fructose =
The molar heat of combustion of hydrazine is -663.82 kJ/mole.
Answer:
The gravitational pull of the moon and the rotational force of the Earth cause tides to rise and fall across the planet. The species living in coastal areas most affected by changing tides have unique ways of surviving.
So, the answer is B.
Answer:
H₂ is the limiting reactant
Explanation:
Balanced equations demonstrate the ratios of all reactants and products in terms of the number of atoms/molecules or moles;
In the given reaction:
1 mol of N₂ reacts with 3 mol of H₂ to produce 2 mol of NH₃
So the ratio of N₂ to H₂ to NH₃ will be 1:3:2
If we have 3 moles of N₂, we can apply the ratio to find that it will react with 9 moles of H₂ (3× the moles of N₂) to produce 6 moles of NH₃ (2× the moles of N₂);
Similarly, if we have 5 moles of H₂, then applying the ratio, we find that ⁵/₃ or 1.66... moles of N₂ (¹/₃× the moles of H₂) react with 5 moles of H₂ to produce ¹⁰/₃ or 3.33... moles of NH₃ (²/₃× the moles of H₂);
In order for all of the 3 moles of N₂ to react, it would require 6 moles of H₂;
There is only 5 moles of H₂ available, meaning there will be an excess of N₂;
5 moles of H₂ will react with 1.66... moles of the N₂, leaving 1.33... or ⁴/₃ moles of N₂ unreacted;
If the N₂ is in excess, then the H₂ is limited (i.e. the limiting reactant)