Answer:
0.0277 M.
Explanation:
The integral rate law of a first order reaction:
<em>Kt = ln ([A₀]/[A]),</em>
where, k is the rate constant of the reaction <em>(k = 3.36 × 10⁻⁵ s⁻¹)</em>,
t is the time of the reaction <em>(t = 235.0 min = 14100 s)</em>,
[A₀] is the initial concentration of cyclopropane <em>([A₀] = 0.0445 M)</em>
<em>∵ Kt = ln ([A₀]/[A]),</em>
∴ (3.36 × 10⁻⁵ s⁻¹)(14100 s) = ln (0.0445 M)/[A]
Taking the exponential of both sides:
1.6 = (0.0445 M)/[A]
<em>∴ [A] = (0.0445 M)/1.6 = 0.0277 M.</em>
<em />
Answer:
1.5g/cm³
Explanation:
density=mass÷volume
mass= 1.5kg (<em>c</em><em>h</em><em>a</em><em>n</em><em>g</em><em>e</em><em> </em><em>i</em><em>n</em><em>t</em><em>o</em><em> </em><em>g</em>) = 1500g
volume of the cube = 10×10×10 = 1000cm³
density= divide 1500g÷1000cm = 1.5g/cm³
<h2>
Density= 1.5g/cm³</h2>
YOUR WELCOME!
<h3>
Answer: b) 0.250 mol</h3>
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Work Shown:
Using the periodic table, we see that
- 1 mole of carbon = 12 grams
- 1 mole of oxygen = 16 grams
These are approximations and these values are often found underneath the atomic symbol. For example, the atomic weight listed under carbon is roughly 12.011 grams. I'm rounding to 2 sig figs in those numbers listed above.
So 1 mole of CO2 is approximately 12+2*16 = 44 grams. The 2 is there since we have 2 oxygens attached to the carbon atom.
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Since 1 mole of CO2 is 44 grams, we can use that to convert from grams to moles.
11.0 grams of CO2 = (11.0 grams)*(1 mol/44 g) = (11.0/44) mol = 0.250 mol of CO2
In short,
11.0 grams of CO2 = 0.250 mol of CO2
This is approximate.
We don't need to use any of the information in the table.
Two German chemists, Justus von Liebig (1803–1873) and Friedrich Wöhler (1800–1882), were responsible for the emergence of organic chemistry in the early nineteenth century. Their quantitative analytical methods helped establish the constitution of newly isolated and synthesized carbon compounds.
