To determine the amount of 6.0 M H2SO4 needed for the preparation, equate the number of moles of the 6.0 M and 2.5 M H2SO4 solution. This is done as follows
M1 x V1 = M2 x V2
Substituting the known variables,
(6.0 M) x V1 = (2.5 M) x (4.8 L)
Solving for V1 gives an answer of V1 = 2 L. Thus, to prepare the needed solution, dilute 2 L of 6.0 M H2SO4 solution with water until the volume reach 4.8 L.
Answer: Ions are an atom or molecule with a net electric charge due to the loss or gain of one or more electrons. They are formed when atoms lose or gain electrons in order to fulfill the octet rule and have full outer valence electron shells.
Explanation:
When they lose electrons, they become positively charged and are named cations. When they gain electrons, they are negatively charged and are named anions.
The molar concentration of a sucrose solution prepared by dissolving 350.25 g of sucrose is enough deionized water to yield a final solution volume of 500.00 mL is 2.048 M.
We must first obtain the number of moles of sucrose in the solution as follows;
Number of moles = mass/ molar
Mass of sucrose = 350.25 g
Molar mass of sucrose = 342 g/mol
Number of moles of sucrose= 350.25 g / 342 g/mol
= 1.024 moles
Recall that;
Number of moles = concentration × volume
concentration = Number of moles/volume
volume of solution = 500.00 mL or 0.5 L
concentration = 1.024 moles/0.5 L
concentration = 2.048 M
Learn more: brainly.com/question/13385951
Answer:
2HCl(aq)+K2CO3(aq)→H2O(l)+CO2(g)+2KCl(aq)
Explanation:
HCl(aq)+K2CO3(aq)→H2O(l)+CO2(g)+KCl(aq)
2HCl(aq)+K2CO3(aq)→H2O(l)+CO2(g)+2KCl(aq)
H-1*2=2 H-2
Cl-1*2=2 Cl- 1*2=2
K -2 K-1*2=2
C- 1 C-1
O - 3 O-3
