Answer:
CaCl2 + H2O Ca2+(aq) + 2Cl-(aq)
Explanation:
CaCl2 + H2O Ca2+(aq) + 2Cl-(aq)
When CaCl2 is dissolved in H2O (water) it will dissociate (dissolve) into Ca+2 and Cl- ions.
The dissolution of calcium chloride is an exothermic process.
Answer:
Ag⁺(aq) + Cl⁻(aq) ==> AgCl(s)
Explanation:
The net ionic equation can be described as the equation that contains only those species which would be participated in the chemical reaction. The spectator ions are the type of the ions that are present in both sides of the chemical equation these ions could not be present in the net ionic equation
First, it is easiest if you write the compete molecular equation:
AgNO₃(aq) + KCl(aq) ⇔ AgCl(s) + KNO₃(aq)
we look up which compounds are soluble (aq) and which are not (s). In this case, silver chloride (AgCl) is not soluble. Thus, the net ionic equation is...
Ag⁺(aq) + Cl⁻(aq) ==> AgCl(s)
To solve this problem, we begin by first calculating the area of the front lawn. The length and width of the lawn was given and the area of a rectangle is given by the formula: Area = length x width. Thus, the area of the front lawn can be obtained by multiplying 18 ft by 20 ft, wherein we get 360 ft^2 as the area.
Second, the problem indicated that each square foot of lawn accumulates 1450 new snow flakes per minute. This can be translated into the expression 1450 snow flakes/ (minute·ft^2). In this way, we can convert it to units of mass (kg). Afterwards, we simply need to multiply it to the area of the lawn and convert minute to hour. The following expression is then used:
1450 snow flakes/ (minute·ft^2) x 1.90 mg/snow flake x 1 g/1000 mg x 1kg/1000 g x 360 ft^2 x 60 minutes/hour = 59.508 kg snow flake/hour
It is then calculated that 59.508 kg of snow flake accumulates in the lawn every hour.
1 mole = 18g water has 6.023*10^23 molecules
And therefore 12 g has 6.023*10^23/18*12 = 4*10^23
C ans
Explanation:
Let us assume that the ratio for the given reaction is 1:1.
Therefore, we will calculate the moles of 
as follows.
Moles of solution = molarity × volume (L)
= 0.0440 M × 0.014 L
= 0.000616 moles
Moles of excess EDTA = 0.000616 moles
Also, the initial moles of EDTA will be calculated as follows.
Total initial moles of EDTA = 0.0600 M × 0.025 L
= 0.0015
Therefore, moles of EDTA reacted with 
will be as follows.
= 0.0015 - 0.000616
= 0.00088 moles
Since, we have supposed a 1 : 1 ratio between and EDTA
.
So, moles of = 0.00088 moles
Now, we will calculate the molarity of as follows.
Molarity of solution = 
= 
= 0.015 M
Thus, we can conclude that the original concentration of the solution is 0.015 M.
