When 50.0 cm3 of 0.100 mol dm-3 AgNO3 is mixed with 50.0 cm3 of 0.0500 dm-3 HCl in a polystyrene cup, the temperature increases from 20.00°C to 20.35°C. Assume that the specific heat capacity of the solution is the same as that of water, and that the density of the solution is 1.0 g cm-3. What is the value of ΔH for the reaction shown below?
AgNO3(aq) + HCl(aq) AgCl(s) + HNO3(aq)
The answer says...
To find heat.. q=100.0x4.18x0.35
And so on.....
BUT WHAT I DON'T UNDERSTAND IS WHERE DOES 100.0 COME FROM.... THE PROBLEM STATES 1.0 g cm^-3....SO IS IT 100.0 GRAMS??? HOW MANY GRAMS ARE IN 1 g cm ^-3??? DO I HAVE TO CONVERT... IF SO... HOW AND TO WHAT UNIT??????
Yes it does.
From the problem, it says that 50 cm^3 of AgNO3 and 50 cm^3 of HCl are mixed. Therefore, the total volume of the solution is 50 + 50 = 100 cm^3.
Then an assumption is that the density of the solution is 1 g/ cm^3. We have the volume, we have the density, we can get the mass:
d = m/V
m = d * V
m = 1 g/cm^3 * 100 cm^3
m = 100 g (mass of the solution)
To find the heat, the formula for heat absorbed/released is
q = mc(T2 - T1)
where
m = mass (in g)
c = specific heat capacity (J/g-K) = 4.184 J/g-K (for water)
T = temperature
Substituting, you'll get the expression you typed for the answer:
q = 100 * 4.184 * (20.35 - 20)
Hope this helps :3
I apologize for the confusion. The value of 100.0 is derived from the volume of the solution used, which is the sum of the volumes of AgNO3 and HCl, both at 50.0 cm3. To calculate the mass of the solution, we can use the density of the solution, which is given as 1.0 g cm-3.
The formula for density is mass divided by volume (d = m/v), where mass is in grams and volume is in cm3. In this case, the volume is 100.0 cm3.
To find the mass, we can rearrange the formula: m = d × v. Plugging in the values, we get: m = 1.0 g cm-3 × 100.0 cm3 = 100.0 grams.
Therefore, the mass of the solution is indeed 100.0 grams. We then proceed with the heat calculation using this mass in the equation q = m × c × ΔT, where m is the mass, c is the specific heat capacity (given as 4.18 J g-1 °C-1), and ΔT is the change in temperature (0.35°C in this case).
Substituting the values, we get: q = 100.0 g × 4.18 J g-1 °C-1 × 0.35°C = 146.3 J.
Hence, the value of q is 146.3 J, which represents the amount of heat transferred in this reaction.