There is a very simple relationship between the three. First off, power is the amount of energy used over a certain amount of time. Energy is the capacity of carrying out that power. Lastly, time depends on how much energy you have to exert the work.
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Answer:
See Explanation
Explanation:
An ionic bond occurs due to electrostatic attraction between a positively charged ion and a negatively charged ion.
A metal and a ligand are bound by a coordinate covalent bond or a dative bond. This bond occurs due to donation of electron pairs from ligands to available orbitals on metals.
The formation of coordinate bonds is evident when neutral molecules or negative ions with non bonding electrons donate same to empty metal orbitals. This is sometimes shown by an arrow pointing from the ligands to the metal center.
For instance; tetraammine copper II ion is formed when four ammonia molecules donate a lone pair each to available vacant orbitals of the copper metal center to form [Cu(NH3)4]^2+.
Answer:
The concentration of the solution is 5.8168 × 
mol.
Explanation:
Here, we want to calculate the concentration of the solution.
The unit of this is mol/dm^3
So the first thing to do here is to calculate the number of moles of the solute present, which is the number of moles of AlCO3
The number of moles = mass/molar mass
molar mass of AlCO3 = 27 + 12 + 3(16) = 27 + 12 + 48 = 87g/mol
Number of moles = 33.4/87 = 0.384 moles
This 0.384 moles is present in 660 L
x moles will be present in 1 dm^3
Recall 1 dm^3 = 1L
x * 660 = 0.384 * 1
x = 0.384/660 = 0.00058168 = 5.8168 * 10^-4 mol/dm^3
Answer:
A link is a fastening unit that attaches two parts of an object together
Different types of links have different characteristics
The balanced chemical equation for the above reaction is as follows;
2Ca + O₂ --> 2CaO
stoichiometry of Ca to O₂ is 2:1
this means that 2 mol of Ca reacts with 1 mol of O₂.
If O₂ is the limiting reactant,
4 mol of O₂ should react with (4x2) - 8 mol of Ca
however only 7.43 mol of Ca is present. Therefore Ca is the limiting reactant.
7.43 mol of Ca reacts with - 7.43/2 = 3.715 mol of O₂
therefore there's excess O₂₂ remaining after the reaction
Since Ca is the limiting reactant, it is fully used up in the reaction and there is no Ca remaining after the reaction is completed.
