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.
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NaCI is the answer, in what my teacher taught me.
Hope I helped :)
Answer:
The granite block transferred <u>4080 joules</u> of energy, and the mass of the water is <u>35.84 grams</u>.
Explanation:
The equation needed to answer both parts of the question is:
Q = mcΔT
In this equation,
-----> Q = energy/heat (J)
-----> m = mass (g)
-----> c = specific heat (J/g°C)
-----> ΔT = change in temperature (°C)
<u>Part #1:</u>
First, you need to find the energy transferred from granite block using the previous equation. You have been given the mass, specific heat, and change in temperature.
Q = ? J c = 0.795 J/g°C
m = 126.1 g ΔT = 92.6 °C - 51.9 °C = 40.7 °C
Q = mcΔT
Q = (126.1 g)(0.795 J/g°C)(40.7 )
Q = 4080
<u>Part #2:</u>
Secondly, using the energy calculated in Part #1, you need to calculate the mass of the water. You have calculated the energy transferred, and have been given the specific heat and change in temperature.
Q = 4080 J c = 4.186 J/g°C
m = ? g ΔT = 51.9 °C - 24.7 °C = 27.2 °C
Q = mcΔT
4080 J = m(4.186 J/g°C)(27.2 °C)
4080 J = m(113.8592)
35.84 = m
Answer:
Explanation:
C = 49.48
H = 5.19
O = 16.48
N = 28.85
ratio of moles
= 49.48 / 12 : 5.19 / 1 : 16.48 / 16 : 28.85 / 14
= 4.123 : 5.19 : 1.03 : 2.06
= 4 : 5 : 1 : 2
so the empirical formula = C₄ H₅O N₂
Let molecular formula = ( C₄ H₅ON₂ )ₙ ,
n ( 48 + 5 + 16 + 28 ) = 119.19
97 n = 194.19
n = 2 ( approx )
molecular formula = C₈ H₁₀O₂ N₄
