Explanation:
An electron current, the flow of electrons, contributes to an electric current since the electron 'carries' negative electric charge. ... The flow of ions (either positively or negatively charged) also contributes to an electric current in, for example, the electrolyte of an electrochemical cell.
hope it will work well?!
When calcium joins with calcium its follows the following equation:
2Ca + O2 = 2CaO
hope that helps
The increase in the boiling point of a solvent is a colligative property.
That means that the increase in the boling point will be related to the number of particles (molecules or ions) present in the solution.
The higher the number of particles (molecules or ions) the higher the increase in the boiling point.
All the aqueous solutions presented are electrolytes, i.e. the solutes are ionic compounds.
Then, you have to compare the number of ions that you have in each solution.
A) 1.0 M KCl ---> 1.0 M K+ + 1.0 MCl- = 2 moles of particles / liter
B) 1.0 M CaCl2 --> 1.0M Ca(2+) + 1.0M * 2 Cl (-) = 3 moles of particle / liter
C) 2.0M KCl ---> 2.0 M K+ + 2.0 M Cl- = 4 moles of particle / liter
D) 2.0 M CaCl2 ----> 2.0 M Ca (2+) + 2.0M * 2 Cl (-) = 6 moles of particle / liter.
Then, the solution 2.0M CaCl2(aq) has the highest increase in the boiling point.
Answer: option D) 2.0 M Ca Cl2(aq)
Answer:
Explanation:
1) Find number of each of the type of atom that is present in the compound, using the chemical formula .
2) Then multiply number of atoms of each element that is present in the compound with the atomic weight of each of the element
3) Add everything together and add the units (grams/mole ) after the number
Let finds that of water
Chemical formula of water is (H20 )
hydogens atoms= 2
oxygen atom= 1
Atomic weight for Hydrogen= 1
Atomic weight for Oxygen= 16
Total number of atoms of Hydrogen from the formula (H2O)= 2
Total number of atoms of Oxygen from the formula (H2O)= 1
the molar mass=
Hydrogen: ( 2 x 1)= 2
Oxygen: ( 1 x 16)= 16
Add together= (16+2)
= 18
Then add the unit, we have(18 g/mol.)
