Answer:
Variability of data around the mean is best expressed through the standard deviation.
Explanation:
We have some statistical metrics, and we need to find the best measure of the variability of data around the mean.
First variable is mode. Mode of a set of data values is the value that appears most often. It means the most frequent data.
Second variable is Median. The median is a simple measure of central tendency. It is located around 50% of data range.
Third variable is standard deviation. The standard deviation is found by taking the square root of the average of the squared deviations of the values subtracted from their average value. It means, it will show the dispersion of data around the average value.
According to the definition presents for the three variables, Standard deviation is the most appropriate measure of variability of data around the mean.
<span>You need information that you haven't provided in the question. Perhaps your text has a table of ΔGof for these compounds. In that case,
ΔGorxn = Sum of ΔGof products - ΔGof reactants
Just plug in the values that you find and calculate the free energy change of the reaction</span>
Answer:
665 g
Explanation:
Let's consider the following thermochemical equation.
2 C₄H₁₀(g) + 13 O₂(g) → 8 CO₂(g) + 10 H₂O(l), ΔH°rxn= –5,314 kJ/mol
According to this equation, 5,314 kJ are released per 8 moles of CO₂. The moles produced when 1.00 × 10⁴ kJ are released are:
-1.00 × 10⁴ kJ × (8 mol CO₂/-5,314 kJ) = 15.1 mol CO₂
The molar mass of CO₂ is 44.01 g/mol. The mass corresponding to 15.1 moles is:
15.1 mol × 44.01 g/mol = 665 g
Answer:
<h3>The answer is 2.50 g/mL</h3>
Explanation:
The density of a substance can be found by using the formula
![density = \frac{mass}{volume} \\](https://tex.z-dn.net/?f=density%20%3D%20%20%5Cfrac%7Bmass%7D%7Bvolume%7D%20%5C%5C)
From the question
mass = 25 g
volume = 10 mL
We have
![density = \frac{25}{10} = \frac{5}{2} \\](https://tex.z-dn.net/?f=density%20%3D%20%20%5Cfrac%7B25%7D%7B10%7D%20%20%3D%20%20%5Cfrac%7B5%7D%7B2%7D%20%20%5C%5C%20)
We have the final answer as
<h3>2.50 g/mL</h3>
Hope this helps you