Answer:
The volume of NaOH required is - 0.01 L
Explanation:
At equivalence point
,
Moles of 
= Moles of NaOH
Considering
:-
Given that:
So,
<u>The volume of NaOH required is - 0.01 L</u>
Answer:
the size of Ca is the greatest ,then Mg is the greater on size than Be
Explanation:
if you make the electron configuration for each of the elements, what is the main difference u gonna see ?
Be 4 1s2/2s2
Mg 12 1s2 /2s2 2p6/3s2
Ca 20 1s2 /2s2 2p6/3s2 3p6 3d/4s2
see that all the elements are in the same group but are in different period
u gonna see the last electron valance shell in Ca are too far from its nucleus but in Be the last electrons are too close and more attracted to the atom's nucleus , so the size of Ca is the biggest then Mg then Be
Answer:
the entropy change for the surroundings when 1.68 moles of Fe2O3(s) react at standard conditions = 49.73 J/K.
Explanation:
3Fe2O3(s) + H2(g)-----------2Fe3O4(s) + H2O(g)
∆S°rxn = n x sum of ∆S° products - n x sum of ∆S° reactants
∆S°rxn = [2x∆S°Fe3O4(s) + ∆S°H2O(g)] - [3x∆S°Fe2O3(s) + ∆S°H2(g)]
∆S°rxn = [(2x146.44)+(188.72)] - [(3x87.40)+(130.59)] J/K
∆S°rxn = (481.6 - 392.79) J/K =88.81J/K.
For 3 moles of Fe2O3 react, ∆S° =88.81 J/K,
then for 1.68 moles Fe2O3 react, ∆S° = (1.68 mol x 88.81 J/K)/(3 mol) = 49.73 J/K the entropy change for the surroundings when 1.68 moles of Fe2O3(s) react at standard conditions.
Gravitational Pull, is the answer. just like we orbit the sun, the moon orbits us. we act almost like a magnet, we follow the biggest guy as leader. if the sun dissapeared, we would die, but probably, we would start to revolve around jupiter as it is the biggest
The balanced chemical equation that illustrates this reaction is:
<span>C2H4 + 3O2 --> 2CO2 + 2H2O
</span>
From the periodic table:
mass of carbon = 12 grams
mass of hydrogen = 1 gram
Therefore:
molar mass of C2H4 = 12(2) + 4(1) = 24 + 4 = 28 grams
number of moles = mass / molar mass
number of moles of C2H4 = 54.7 / 28 = 1.95 moles
From the balanced equation above:
3 moles of oxygen are required to react with one mole of C2H4, therefore, to know the number of moles required to react with 1.95 moles of C2H4, all you have to do is cross multiplication as follows:
number of oxygen moles = (1.95*3) / 1 = 5.85 moles
