Answer:
Q = -811440 J
Explanation:
Given data:
Mass of oil = 2.76 Kg (2.76× 1000 = 2760 g)
Initial temperature = 191 °C
Final temperature = 23°C
Specific heat capacity of oil = 1.75 J/g.°C
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 23°C - 191 °C
ΔT = -168°C
Q = 2760 g ×1.75 J/g.°C ×-168°C
Q = -811440 J
Negative sign show heat is released.
Answer:
<h2>D</h2>
Explanation:
<em>According to Universe Today, The answer is D because Radiation from the Sun, which is more popularly known as sunlight, is a mixture of electromagnetic waves ranging from infrared (IR) to ultraviolet rays (UV). It of course includes visible light, which is in between IR and UV in the electromagnetic spectrum.</em>
<em>Hope this helps! <3</em>
良い日をお過ごしください！
The data set is missing in the question. The data set is given in the attachment.
Solution :
a). In the table, there are four positive examples and give number of negative examples.
Therefore,
and
The entropy of the training examples is given by :
= 0.9911
b). For the attribute all the associating increments and the probability are :
+ -
T 3 1
F 1 4
Th entropy for is given by :
![$\frac{4}{9}[ -\frac{3}{4}\log\left(\frac{3}{4}\right)-\frac{1}{4}\log\left(\frac{1}{4}\right)]+\frac{5}{9}[ -\frac{1}{5}\log\left(\frac{1}{5}\right)-\frac{4}{5}\log\left(\frac{4}{5}\right)]$](https://tex.z-dn.net/?f=%24%5Cfrac%7B4%7D%7B9%7D%5B%20-%5Cfrac%7B3%7D%7B4%7D%5Clog%5Cleft%28%5Cfrac%7B3%7D%7B4%7D%5Cright%29-%5Cfrac%7B1%7D%7B4%7D%5Clog%5Cleft%28%5Cfrac%7B1%7D%7B4%7D%5Cright%29%5D%2B%5Cfrac%7B5%7D%7B9%7D%5B%20-%5Cfrac%7B1%7D%7B5%7D%5Clog%5Cleft%28%5Cfrac%7B1%7D%7B5%7D%5Cright%29-%5Cfrac%7B4%7D%7B5%7D%5Clog%5Cleft%28%5Cfrac%7B4%7D%7B5%7D%5Cright%29%5D%24)
= 0.7616
Therefore, the information gain for is
0.9911 - 0.7616 = 0.2294
Similarly for the attribute 
the associating counts and the probabilities are :
+ -
T 2 3
F 2 2
Th entropy for is given by :
![$\frac{5}{9}[ -\frac{2}{5}\log\left(\frac{2}{5}\right)-\frac{3}{5}\log\left(\frac{3}{5}\right)]+\frac{4}{9}[ -\frac{2}{4}\log\left(\frac{2}{4}\right)-\frac{2}{4}\log\left(\frac{2}{4}\right)]$](https://tex.z-dn.net/?f=%24%5Cfrac%7B5%7D%7B9%7D%5B%20-%5Cfrac%7B2%7D%7B5%7D%5Clog%5Cleft%28%5Cfrac%7B2%7D%7B5%7D%5Cright%29-%5Cfrac%7B3%7D%7B5%7D%5Clog%5Cleft%28%5Cfrac%7B3%7D%7B5%7D%5Cright%29%5D%2B%5Cfrac%7B4%7D%7B9%7D%5B%20-%5Cfrac%7B2%7D%7B4%7D%5Clog%5Cleft%28%5Cfrac%7B2%7D%7B4%7D%5Cright%29-%5Cfrac%7B2%7D%7B4%7D%5Clog%5Cleft%28%5Cfrac%7B2%7D%7B4%7D%5Cright%29%5D%24)
= 0.9839
Therefore, the information gain for is
0.9911 - 0.9839 = 0.0072
Class label split point entropy Info gain
1.0 + 2.0 0.8484 0.1427
3.0 - 3.5 0.9885 0.0026
4.0 + 4.5 0.9183 0.0728
5.0 -
5.0 - 5.5 0.9839 0.0072
6.0 + 6.5 0.9728 0.0183
7.0 +
7.0 - 7.5 0.8889 0.1022
The best split for observed at split point which is equal to 2.
c). From the table mention in part (b) of the information gain, we can say that produces the best split.
You have 45 kJ of heat added and for every 40.6 kJ added, one mole is able to be vaporized. Find how many moles you can vaporize.
45 kJ / 40.6 kJ = 1.1 moles
Water is roughly 18 g per mole so 1.1 moles x 18 g per mol = 19.95 grams vaporized
Hope I helped!
Answer:
a) 24.7 mol
b) 790 g
Explanation:
Step 1: Given data
- Volume of the chamber (V): 200. L
- Room temperature (T): 23 °C
- Pressure of the gas (P): 3.00 atm
Step 2: Convert "T" to Kelvin
We will use the following expression.
K = °C + 273.15
K = 23°C + 273.15 = 296 K
Step 3: Calculate the moles (n) of oxygen
We will use the ideal gas equation.
P × V = n × R × T
n = P × V/R × T
n = 3.00 atm × 200. L/(0.0821 atm.L/mol.K) × 296 K = 24.7 mol
Step 4: Calculate the mass (m) corresponding to 24.7 moles of oxygen
The molar mass (M) of oxygen ga sis 32.00 g/mol. We will calculate the mass of oxygen using the following expression.
m = n × M
m = 24.7 mol × 32.00 g/mol = 790 g
