A balloon of helium gas in Alabama has a volume of 22.4 L at 26°C and a 1.03 ATM how big with a balloon get if is transported to
1 answer:
Answer:
The final volume of the balloon is = 28.11 L
Explanation:
Initial pressure = 1.03 atm = 104.325 K pa
Initial temperature = 26 °c = 299 K
Initial volume = 22.4 L
Final temperature = 22 °c = 295 K
Final pressure = 0.81 atm = 82 K pa
We know that
Put all the values in above formula we get
= 28.11 L
This is the final volume of the balloon.
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Answer:
is the symbol for an oxide ion
Answer:
0.0714 M for the given variables
Explanation:
The question is missing some data, but one of the original questions regarding this problem provides the following data:
Mass of copper(II) acetate:
Volume of the sodium chromate solution:
Molarity of the sodium chromate solution:
Now, when copper(II) acetate reacts with sodium chromate, an insoluble copper(II) chromate is formed:
Find moles of each reactant. or copper(II) acetate, divide its mass by the molar mass:
Moles of the sodium chromate solution would be found by multiplying its volume by molarity:
Find the limiting reactant. Notice that stoichiometry of this reaction is 1 : 1, so we can compare moles directly. Moles of copper(II) acetate are lower than moles of sodium chromate, so copper(II) acetate is our limiting reactant.
Write the net ionic equation for this reaction:
Notice that acetate is the ion spectator. This means it doesn't react, its moles throughout reaction stay the same. We started with:
According to stoichiometry, 1 unit of copper(II) acetate has 2 units of acetate, so moles of acetate are equal to:
The total volume of this solution doesn't change, so dividing moles of acetate by this volume will yield the molarity of acetate:
Answer:
a. the maximum number of σ bonds that the atom can form is 4
b. the maximum number of p-p bonds that the atom can form is 2
Explanation:
Hybridization is the mixing of at least two nonequivalent orbitals, in this case, we have the mixing of one <em>s, 3 p </em> and <em> 2 d </em> orbitals. In hybridization the number of hybrid orbitals generated is equal to the number of pure atomic orbital, so we have 6 hybrid orbital.
The shape of this hybrid orbital is octahedral (look the attached image) , it has 4 orbital located in the plane and 2 orbital perpendicular to it.
This shape allows the formation of maximum 4 σ bond, because σ bonds are formed by orbitals overlapping end to end.
And maximum 2 p-p bonds, because p-p bonds are formed by sideways overlapping orbitals. The atom can form one with each one of the orbitals located perpendicular to the plane.
