Answer:
H^+(aq) + OH^-(aq) —> H2O(l)
Explanation:
We'll begin by writing the balanced equation for the reaction.
2HCl(aq) + Ca(OH)2(aq) —> CaCl2(aq) + 2H2O(l)
Ca(OH)2 is a strong base and will dissociates as follow:
Ca(OH)2(aq) —> Ca^2+(aq) + 2OH^-(aq)
HCl is a strong acid and will dissociates as follow:
HCl(aq) —> H^+(aq) + Cl^-(aq)
Thus, In solution a double displacement reaction occurs as shown below:
2H^+(aq) + 2Cl^-(aq) + Ca^2+(aq) + 2OH^-(aq) —> Ca^2+(aq) + 2Cl^-(aq) + 2H2O(l)
To get the net ionic equation, cancel out Ca^2+ and 2Cl^-
2H^+(aq) + 2OH^-(aq) —> 2H2O(l)
H^+(aq) + OH^-(aq) —> H2O(l)
Answer:
See the answers below
Explanation:
1) 100. mL of solution containing 19.5 g of NaCl (3.3M)
2) 100. mL of 3.00 M NaCl solution (3 M)
3) 150. mL of solution containing 19.5 g of NaCl (2.2 M)
4) Number 1 and 5 have the same concentration (1.5M)
MW of NaCl = 23 + 36 = 59 g
For number 3
59 g ------------------- 1 mol
19,5 g ----------------- x
x = 19.5 x 1/59 = 0.33 mol
Molarity (M) = 0.33 mol/0.150 l = 2.2 M
For number 4,
Molarity (M) = 0.33mol/0.10 l = 3.3 M
For number 5
Molarity (M) = 0.450/0.3 = 1.5 M
Molarity = moles of solute/volume of solution in liters.
The solute here is NaCl, of which we have 46.5 g. To calculate the molarity of an NaCl solution, we need to know the number of moles of NaCl. To convert from grams to moles, we divide the mass by the molar mass of NaCl. The molar mass of NaCl is the sum of the atomic masses of Na and Cl: 23 amu + 35 amu = 58 amu. For our purposes, we can regard amu as equivalent to grams/mole.
(46.5 g)/(58 g/mol) = 0.8017 moles NaCl.
Now that we know both the number of moles of our NaCl solute and the volume of the solution, we can calculate the molarity:
(0.8017 moles NaCl)/(2.2 L) = 0.364 M.
Every 2 million years the amount is halved
0 million = 200g
2 million = 200g/2 = 100g
4 million = 100g/2 = 50g
H2SO4 + 2RbOH -> Rb2SO4 + 2H2O
If you want an explanation, keep reading.
In the first portion, there are two hydrogen ions and four sulfate ions.
The second portion has one rubidium ions and one hydroxide ion.
On the other side of the equation, in order to keep those two rubidiums balanced, you'll need to add a two at the beginning of the second portion, but in that process you are giving a second hydroxide value.
Back to the right side, there is there is water (H2O).
On the first portion, there were two hydrogen ions. The second portion also has two hydroxides because of the value change (adding the two to the front).
So on the fourth portion, you'd have to add another two so you could balance the four hydrogen ions (H2 and 2OH) and the two oxygen ions (2OH).
I hope this was easy to understand.
