The balanced chemical equation for the above reaction is as follows ;
Mg + 2HCl —> MgCl2 + H2
The stoichiometry of Mg to HCl is 1:2
This means that 1 mol of Mg reacts with 2 mol of HCl
Equal amounts of both Mg and HCl have been added. One reagent is the limiting reactant and other reactant is in excess.
Limiting reactant is the reagent that is fully used up in the reaction and the amount of Product formed depends on the amount of limiting reactant present.
In this reaction if Mg is the limiting reactant, 4.40 moles of Mg should react with 4.40x2 -8.80 moles of HCl.
But only 4.40 moles of HCl present therefore HCl is the limiting reactant that reacts with 4.40/2 = 2.20 moles of Mg
Stoichiometry of HCl to MgCl2 is 2:1
Since HCl moles reacted -4.40 mol
Then MgCl2 moles formed are 4.40/2 = 2.20 mol of MgCl2
The atom has only one isotope which means 100 % of same atom is present in nature. The atomic mass of an element is the number of times an atom of that element is heavier than an atom of carbon taken as 12. Mass of one atom of that isotope is 9.123 ✕ 10⁻²³ g, so mass of one mole of atom that is Avogadro's number of atom is 6.023 X 10²³ X 9.123 X 10⁻²³ g=54.94 g = 55 g (approximate).
So, the atom having atomic mass 55 will be Cesium (Cs). Only one isotope of Cesium is stable in nature.
Answer:
334J/g
Explanation:
Data obtained from the question include:
Mass (m) = 1g
Specific heat of Fusion (Hf) = 334 J/g
Heat (Q) =?
Using the equation Q = m·Hf, we can obtain the heat released as follow:
Q = m·Hf
Q = 1 x 334
Q = 334J
Therefore, the amount of heat released is 334J
Question:
The question is incomplete. What is required to calculate was not added.The equilibrium data was not also added. Below is the additional questions and the answers.
1. Calculate the minimum solvent that can be used.
2.Using a solvent rate of 1.5 times the minimum, calculate the number of
theoretical stages.
Answer:
1. Minimum solvent = 411.047
2. N = 5
Explanation:
See the attached files for explanations.
Answer:
See the explanation
Explanation:
1) The Lewis structure for
has a central Carbon<em> </em>atom attached to Oxygen atoms.
In the
we will have a structure: O=C=O the <u>central atom</u> "carbon" we will have <u>2 sigma bonds and 2 pi bonds</u>, therefore, we have an <u>Sp hybridization</u>. For O we have <u>1 pi and 1 sigma bond</u>, therefore, we have an <u>Sp2 hybridization</u>.
2) These atoms are held together by <u>double bonds.</u>
<u></u>
Again in the structure of
: O=C=O we only have double bonds.
3. Carbon dioxide has a Carbon dioxide has a <u>Linear</u> electron geometry.
Due to the double bonds we have to have a linear structure because in this geometry the atoms will be further apart from each other.
4. The carbon atom is <u>Sp</u> hybridized.
We will have for carbon 2 pi bonds, so we will have an <u>Sp</u> hybridization.
5. Carbon dioxide has two Carbon dioxide has two C(p) - O(p) π bonds and two C(sp) - O(Sp2) σ bonds.
(See figures)
Figure 1: Carbon hybridization
Figure 2: Oxygen hybridization