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Bezzdna [24]
2 years ago
10

A group of students measured the average monthly temperature of 1000 cities around the world and plotted the cumulative frequenc

y graph shown below. A cumulative frequency graph. The label for the x-axis reads 'Average monthly temperature (in degrees Celsius).' The label for the y-axis reads 'Cumulative Frequency (number of cities).' The coordinates that make up the shape of the line are as follows: (0,0), (10,50), (20,50), (30,400), (40,700), (50,950), (60,1000), (70,1000). Based on the graph, it can be concluded that (2 points) a about 300 cities have an average monthly temperature of 50°C b about 300 cities have an average monthly temperature of 20°C c almost 700 cities have an average monthly temperature less than or equal to 40°C d almost 1000 cities have an average monthly temperature less than or equal to 40°C
Chemistry
1 answer:
mr Goodwill [35]2 years ago
8 0

Answer:

The MAD of city 2 is less than the MAD for city 1, which means the average monthly temperature of city 2 vary less than the average monthly temperatures for City 1.

Explanation:

For comparing the mean absolute deviations of both data sets we have to calculate the mean absolute deviation for both data sets first,

So for city 1:

Now to calculate the mean deviations mean will be subtracted from each data value. (Note: The minus sign is ignored as the deviation is the distance of value from the mean and it cannot be negative. For this purpose absolute is used)

The deviations will be added then.

So the mean absolute deviation for city 1 is 24 ..

For city 2:

Now to calculate the mean deviations mean will be subtracted from each data value. (Note: The minus sign is ignored)

The deviations will be added then.

So the MAD for city 2 is 11.33 ..

So,

The MAD of city 2 is less than the MAD for city 1, which means the average monthly temperature of city 2 vary less than the average monthly temperatures for City 1.

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Calculate the cell potential E at 25°C for the reaction 2 Al(s) + 3 Fe2+(aq) → 2 Al3+(aq) + 3 Fe(s) given that [Fe 2+] = 0.020 M
Elodia [21]

Answer:

1.18 V

Explanation:

The given cell is:

Al(s)/Al^{3+}(0.10M)||Fe^{2+}(0.020M)/Fe(s)

Half reactions for the given cell follows:

Oxidation half reaction: Al(s)\rightarrow Al^{3+}(0.10M)+2e^-;E^o_{Al^{3+}/Al}=-1.66V

Reduction half reaction: Fe^{2+}(0.020M)+2e^-\rightarrow Fe(s);E^o_{Fe^{2+}/Fe}=-0.45V

Multiply Oxidation half reaction by 2 and Reduction half reaction by 3

Net reaction: 2Al(s)+3Fe^{2+}(0.020M)\rightarrow 2Al^{3+}(0.10M)+3Fe(s)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=-0.45-(-1.66)=1.21V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Al^{3+}]^2}{[Fe^{2+}]^3}

where,

E_{cell} = electrode potential of the cell = ?V

E^o_{cell} = standard electrode potential of the cell = +1.21 V

n = number of electrons exchanged = 6

Putting values in above equation, we get:

E_{cell}=1.21-\frac{0.059}{6}\times \log(\frac{0.10^2}{0.020^3})\\\\E_{cell}=1.18V

5 0
2 years ago
in a candy factory, the nutty chocolate bars contain 22.0% by mass pecans. If 5.0 kg of pecans were used for candy last Tuesday,
zysi [14]
22.0 is the same as saying that in 100 grams of a chocolate bar, there are 22.0 grams of pecans. or to make it easier because of this problem- 100 Kilograms of a chocolate bar, there is 22.0 Kg of pecans. we can use this as a conversion factor (what is used to convert a value to another value. 

conversion factor---> 22.0 kg of pecan= 100 kg of chocolate bar
Note: remember this, what you are converting from goes in the denominator, what you converting to goes in the numerator. 

5.0 Kg of pecan (100 Kg of chocolate bar/ 22.0 Kg of pecan)= 23 Kg of chocolate bar


7 0
2 years ago
Before landing, the brakes and the tires of an airliner have a temperature of 15.0∘C. Upon landing, the 90.7 kg carbon fiber bra
Goryan [66]

Answer:

0.921 J/g degrees C

Explanation:

Recall that the First Law of Thermodynamics demands that the total internal energy of an isolated system must remain constant. Any amount of energy lost by the brakes must be gained by the tires (in the form of heat in this situation).  Therefore, heat given off by the brakes = −heat taken in by tires, or:

−qbrakes=qtires

The equation used to calculate the quantity of heat energy exchanged in this process is:

−qbrakes=−cbrakes mbrakes ΔTbrakes=ctires mtires ΔTtires=qtires

First we must convert the mass of the tires and the brakes from  kg to  g.

massbrakes=90.7 kg×1,000. g1 kg=9.07×104 g

masstires=123 kg×1,000. g1 kg=1.23×105 g

Next, substitute in known values and rearrange to solve for ctires. Note that the final temperature for both the tires and the brakes is 172∘C, the initial temperature of the brakes is 312∘C and the initial temperature of the tires is 15∘C.

−(1.400Jg∘C)(9.07×104 g)(172∘C−312∘C)=(ctires)(1.23×105 g)(172∘C−15∘C)

ctires=−(1.400 Jg∘C)(9.07×104 g)(−140∘C)(1.23×105 g)(157∘C)=17,777,200 J19311000 g∘C=0.9206Jg∘C

The answer should have three significant figures, so round to 0.921Jg∘C.

6 0
2 years ago
Rank the formation of the solutions A, B, and C from the most exothermic to the most endothermic. Rank the enthalpy of solution
Elodia [21]

This is an incomplete question, the table is attached below.

Answer : The correct ranking of the solution from most exothermic to most endothermic will be: A, B and C.

Explanation :

As we know that the intermolecular force of attraction play an important role in the interaction of solute-solute, solute-solvent and solvent solvent solution.

In the solution A, the solute-solute and solvent-solvent interactions are weak. So, their solute-solvent interaction will be strong. That means, the solution will be more exothermic.

In the solution C, the solute-solute and solvent-solvent interactions are strong. So, their solute-solvent interaction will be weak. That means, the solution will be more endothermic.

Thus, the correct ranking of the solution from most exothermic to most endothermic will be: A, B and C.

4 0
2 years ago
The concentration of ozone in a sample of air that has a partial pressure of O3 of 0.33 torr and a total pressure of air of 695
Goryan [66]

Answer:

0.047 %

Explanation:

Step 1: Given data

  • Partial pressure of ozone (pO₃): 0.33 torr
  • Total pressure of air (P): 695 torr

Step 2: Calculate the %v/v of ozone in the air

Air is a mixture of gases. We can find the %v/v of ozone (a component) in the air (mixture) using the following expression.

<em>%v/v = pO₃/P × 100%</em>

%v/v = 0.33 torr/695 torr × 100%

%v/v = 0.047 %

8 0
2 years ago
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