Answer:
41.45 mL
Explanation:
Applying the general gas equation,
PV/T = P'V'/T'............... Equation 1
Where P = Initial pressure of hydrogen, V = Initial volume of hydrogen, T= Initial Temperature of hydrogen, P' = Final pressure of hydrogen, V' = Final Volume of Hydrogen, T' = Final Temperature.
make V' the subject of the equation
V' = PVT'/TP'................ Equation 2
Given: P = 718 torr = (718×133.322) N/m² = 95725.196 N/m², V = 47.9 mL = 0.0479 dm³, T = 26 °C = (26+273) = 299 K, T' = 273 K, P' = 101000 N/m²
Substitute these values into equation 2
V' = ( 95725.196×0.0479×273)/(299×101000)
V' = 0.04145 dm³
V' = 41.45 mL
Answer:
Option a. 0.5 m/s
Explanation:
This graph shows a straight line, where "Y" axis would be "Position" and "X" graph would be "Time". The ecuation that would describe this straight line is Y= aX + 1 , where "a" is the slope or inclination for this graph, and would give us the speed of the object
How do we find the slope (and hence, the speed)?: if you notice this graph, you will check that:
-When X (Time) is zero, Y (Position) is 1
-When X (Time) is 2, Y (Position) is 2
With these 4 points, you can calculate the slope (which will call "m") for this graph with:
m = (Y2-Y1)/(X2-X1) so: Y2=2, Y1=1, X2=2, X1=0
Which gives us: m=1/2 (0.5), the slope or speed of the object: 0.5 m/s
Answer:
Here's what I get
Explanation:
Assume the initial concentrations of H₂ and I₂ are 0.030 and 0.015 mol·L⁻¹, respectively.
We must calculate the initial concentration of HI.
1. We will need a chemical equation with concentrations, so let's gather all the information in one place.
H₂ + I₂ ⇌ 2HI
I/mol·L⁻¹: 0.30 0.15 x
2. Calculate the concentration of HI
![Q_{\text{c}} = \dfrac{\text{[HI]}^{2}} {\text{[H$_{2}$][I$_{2}$]}} =\dfrac{x^{2}}{0.30 \times 0.15} = 5.56\\\\x^{2} = 0.30 \times 0.15 \times 5.56 = 0.250\\x = \sqrt{0.250} = \textbf{0.50 mol/L}\\\text{The initial concentration of HI is $\large \boxed{\textbf{0.50 mol/L}}$}](https://tex.z-dn.net/?f=Q_%7B%5Ctext%7Bc%7D%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B%5BHI%5D%7D%5E%7B2%7D%7D%20%7B%5Ctext%7B%5BH%24_%7B2%7D%24%5D%5BI%24_%7B2%7D%24%5D%7D%7D%20%3D%5Cdfrac%7Bx%5E%7B2%7D%7D%7B0.30%20%5Ctimes%200.15%7D%20%3D%20%205.56%5C%5C%5C%5Cx%5E%7B2%7D%20%3D%200.30%20%5Ctimes%200.15%20%5Ctimes%205.56%20%3D%200.250%5C%5Cx%20%3D%20%5Csqrt%7B0.250%7D%20%3D%20%5Ctextbf%7B0.50%20mol%2FL%7D%5C%5C%5Ctext%7BThe%20initial%20concentration%20of%20HI%20is%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B0.50%20mol%2FL%7D%7D%24%7D)
3. Plot the initial points
The graph below shows the initial concentrations plotted on the vertical axis.
varying composition is not a property of a pure substance
Answer:
3.01 ·10↑22
Explanation:
First you want to convert the grams of Glucose to moles of Glucose.
Next find the formula units of glucose.
.008326Moles of Glucose · 6.022 · 10↑23Forumula Units*Moles↑-1 =
5.01 ·10↑21 Formula Units of Glucose
Now multiply the formula units of glucose by the amount of each element in the molecule.
So for Carbon:
6carbon · 5.01 · 10↑21 = 3.01 · 10↑22
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