This question is asking for a method for the determination of the freezing point in a solution that does not have a noticeable transition in the cooling curve, which is basically based on a linear fit method.
The first step, would be to understand that when the transition is well-defined as the one on the attached file, we can just identify the temperature by just reading the value on the graph, at the time the slope has a pronounced change. For instance, on the attached, the transition occurs after about 43 seconds and the freezing point will be about 4 °C.
However, when we cannot identify a pronounced change in the slope, it will be necessary to use a linear fit method (such as minimum squares) to figure out the equation for each segmented line having a significantly different slope and then equal them so that we can numerically solve for the intercept.
As an example, imagine two of the segmented lines have the following equations after applying the linear fit method:
First of all, we equal them to find the x-value, in this case the time at which the freezing point takes place:
Next, we plug it in in any of the trendlines to obtain the freezing point as the y-value:
This means the freezing point takes place after 7.72 second of cooling and is about 1.84 °C. Now you can replicate it for any not well-defined cooling curve.
Learn more:
Answer:
5.50 moles of magnesium oxide is 221.6742 grams
Explanation:
to do this you multiply the number of moles by the molar mass
Answer:
1.7 bar
Explanation:
We can use the <em>Ideal Gas Law</em> to calculate the individual gas pressure.
pV = nRT Divide both sides by V
p = (nRT)/V
Data: n = 1.7 × 10⁶ mol
R = 0.083 14 bar·L·K⁻¹mol⁻¹
T = 22 °C
V = 2.5 × 10⁷ L
Calculations:
(a) <em>Change the temperature to kelvins
</em>
T = (22 + 273.15) K
= 295.15 K
(b) Calculate the pressure
p = (1.7 × 10⁶ × 0.083 14 × 295.15)/(2.5× 10⁷)
= 1.7 bar
Answer: Heat of the solution = mass water × specific heat water × change in temperature
mass water = 260ml (1.00g/ml ) = 260g
specific heat of water = c(water) = 4.184J/ g°C
Heat change of water = final temperature - initial temperature
= 26.5 - 21.2
= 5.3 °C
H = 260 g ( 4.184J/g°C ) (5.3°C) = 5765J
Molar heat = 
= 16473J/mol
Explanation: finding molar heat requires first to look at specific heat of water and the change of water temperature
To find the net ionic equation we must first write the balanced equation for the reaction. We must bear in mind that the reagents Ca(NO3)2 and Na2S are in the aqueous state and as product we will have CaS in the solid state, since it is not soluble in water and NaNO3 in the aqueous state.
The balanced equation of the reaction will be:
Ca(NO3)2(aq) + → Ca(aq) + 2Na(s)NO3Now, c(aq)ompounds in the aqueous state can be written in their ionic form, so the reaction will transform into:Na2S +
So, the answer will be option A
