Answer:
Answer to you need to make a 6.00 x 10-4 M KSCN solution starting with a 2.00 x 10-3 M KSCN solution. You will be making up the so. ... X 10-3 M KSCN Solution. You Will Be Making Up The Solution In A 25 ML Volumetric Flask And Using 0.5 M HNO3 As The Diluent. What Volume Of 2.00 X 10-3 M KSCN Will You Need?
Explanation:
Stoichiometry is not really hard if you are good with your numbers. All you have to do is pay close attention and follow you teacher because if you don't you will end up like me not knowing what you are doing. Listen and pay real close attention and I can guarantee you it will be just fine!
Answer:
Complete ionic: 
.
Net ionic: 
.
Explanation:
Start by identifying species that exist as ions. In general, such species include:
- Soluble salts.
- Strong acids and strong bases.
All four species in this particular question are salts. However, only three of them are generally soluble in water: 
, 
, and 
. These three salts will exist as ions:
- Each
formula unit will exist as one 
ion and one 
ion. - Each formula unit will exist as one
ion and two 
ions (note the subscript in the formula 
.) - Each formula unit will exist as one and two ions.
On the other hand, 
is generally insoluble in water. This salt will not form ions.
Rewrite the original chemical equation to get the corresponding ionic equation. In this question, rewrite , , and (three soluble salts) as the corresponding ions.
Pay attention to the coefficient of each species. For example, indeed each formula unit will exist as only one ion and one ion. However, because the coefficient of 
in the original equation is two, 
alone should correspond to two 
ions and two 
ions.
Do not rewrite the salt because it is insoluble.
.
Eliminate ions that are present on both sides of this ionic equation. In this question, such ions include one unit of and two units of . Doing so will give:
.
Simplify the coefficients:
.
Aqueous nitric acid and aqueous ammonia reacts to form ammonium nitrate
HNO₃(aq)+NH₃(aq) = NH₄NO₃(aq)
HNO₃ +NH3 = NH₄ (+) + NO₃ (-)
Therefore the net ionic equation will be;
H⁺(aq) + NH₃ = NH₄⁺ (aq)
<span>294400 cal
The heating of the water will have 3 phases
1. Melting of the ice, the temperature will remain constant at 0 degrees C
2. Heating of water to boiling, the temperature will rise
3. Boiling of water, temperature will remain constant at 100 degrees C
So, let's see how many cal are needed for each phase.
We start with 320 g of ice and 100 g of liquid, both at 0 degrees C. We can ignore the liquid and focus on the ice only. To convert from the solid to the liquid, we need to add the heat of fusion for each gram. So multiply the amount of ice we have by the heat of fusion.
80 cal/g * 320 g = 25600 cal
Now we have 320 g of ice that's been melted into water and the 100 g of water we started with, resulting in 320 + 100 = 420 g of water at 0 degrees C. We need to heat that water to 100 degrees C
420 * 100 = 42000 cal
Finally, we have 420 g of water at the boiling point. We now need to pump in an additional 540 cal/g to boil it all away.
420 g * 540 cal/g = 226800 cal
So the total number of cal used is
25600 cal + 42000 cal + 226800 cal = 294400 cal</span>
