Answer:
0.10M HCN < 0.10 M HClO < 0.10 M HNO₂ < 0.10 M HNO₃
Explanation:
We are comparing acids with the same concentration. So what we have to do first is to determine if we have any strong acid and for the rest ( weak acids ) compare them by their Ka´s ( look for them in reference tables ) since we know the larger the Ka, the more Hydronium concentration will be in these solutions at the same concentration.
HNO₃ is a strong acid and will have the largest hydronium concentration.
HCN Ka = 6.2 x 10⁻¹⁰
HNO₂ Ka = 4.0 x 10⁻⁴
HClO Ka = 3.0 x 10⁻⁸
The ranking from smallest to largest hydronium concentration will then be:
0.10M HCN < 0.10 M HClO < 0.10 M HNO₂ < 0.10 M HNO₃
Answer:
1. Ionic bonding
2. Covalent bonding
3. Metallic bonding
Explanation:
Ionic bonding also referred to as electrovalent bonding is a kind of chemical bonding that involves the transfer of electrons between the valence shells of two elements with a large electronegativity difference usually a metal and a nonmetal.
For example an ionic bonding scenario might play out between a group one metal and a group seven halogen. While group one metals have one electron hindering their stability, group seven halogens need that one electron that could make them achieve this stability. It is this that causes them to come together in a way where the electron is transferred completely from the valence shell of the group 1 atom and accepted into the valence shell of the group 7 halogen.
Covalent bonding involves the sharing of electrons between atoms of comparable electronegativities. The electro negativity difference is not large enough to permit the total movement of the electrons and hence the electrons are then controlled by the nuclei of the two atoms
Between two metals, what we have is called the metallic bonding
Answer:
= 9.28 g CO₂
Explanation:
First write a balanced equation:
CH₄ + 2O₂ -> 2H₂O + CO₂
Convert the information to moles
7.50g CH₄ = 0.46875 mol CH₄
13.5g O₂ = 0.421875 mol O₂
Theoretical molar ratio CH₄:O₂ -> 1:2
Actual ratio is 0.46875 : 0.421875 ≈ 1:1
If all CH₄ is used up, there would need to be more O₂
So O₂ is the limiting reactant and we use this in our equation
Use molar ratio to find moles of CO₂
0.421875 mol O₂ * 1 mol CO₂/2 mol O₂=0.2109375 mol CO₂
Then convert to grams
0.2109375 mol CO₂ = 9.28114 g CO₂
round to 3 sig figs
= 9.28 g CO₂
Answer:
D
Explanation:
This creates a gap that we call an oceanic trench
:)
= 5.54