Hello, This is a practice test. I needed help with questions that are wrong for
ID: 304519 • Letter: H
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Hello, This is a practice test. I needed help with questions that are wrong for example; linear “function of this data and exponential” Also, I don’t know how this person got the data for x axis and y axis using the temp and solubility. Lastly, how would I go on to graphing this? With an appropriate equation. Thank you L. Mr. "P-Funk" obtained the following data and believes that the relationship between solubility and temperature of KI is EXPONENTIAL. Please help him determine whether this relationship truly is EXPONENTIAL by graphing it LINEARI.Y first then express it in EXPONENTIAL FORM by looking at the graph and using algebra. Label the graph to be appropriately. (20pts) my General Exponential Function:y:aC Linear Function of this Data:nytb Exponential Function of this Data: Ymxta What is the independent variable (T or In S)? (1pt) (Ipt) (Ipt) (1pt) T'emperature SolubilityX X- axis (Plotted) Y-axis (Plotted) C) (g KI/ 100g 1H20) (Ipt each) 20 50 70 90 100 (1pt each) 2.62 5.64 9.41 15.70 20.29 6.a63 1.73 2.21 2.75 3.01 10 I 00Explanation / Answer
General Exponential Function: y = aex
where a is a constant; y is the dependent variable and x is the independent variable.
Linear Function of this data: S = kT + C
where S denotes the solubility of KI, T is the temperature of the measurement and C is the constant. k is the proportionality constant or slope of the plot of S and t.
Exponential Function of this data: ln S = k’T + C’
where C’ is another constant and k’ is the proportionality constant between ln S and t.
T is the independent variable since the solubility, S increases with temperature T, i.e, S is a function of T.
Temperature (°C)
Solubility
(g KI/100 g H2O)
X-axis (Plotted)
T (K) =
(273 + temperature in °C)
Y-axis (Plotted)
S (g KI/100 g water)
Y-axis (Plotted)
ln S
20
2.62
293
2.62
0.963
50
5.64
323
5.64
1.730
70
9.41
343
9.41
2.242
90
15.70
363
15.70
2.754
100
20.29
373
20.29
3.010
Plot S vs T as below.
Plot of solubility S vs temperature T; the plot is exponential
Next plot ln S vs T.
Plot of ln S vs T; the plot is linear
A general characteristic of exponential functions is that the plot of the natural logarithm (ln function) vs the independent variable is linear.
Since the solubility varies exponentially with the temperature, hence, the solubility of KI in water is an exponential function of the temperature.
Temperature (°C)
Solubility
(g KI/100 g H2O)
X-axis (Plotted)
T (K) =
(273 + temperature in °C)
Y-axis (Plotted)
S (g KI/100 g water)
Y-axis (Plotted)
ln S
20
2.62
293
2.62
0.963
50
5.64
323
5.64
1.730
70
9.41
343
9.41
2.242
90
15.70
363
15.70
2.754
100
20.29
373
20.29
3.010
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