If you overheat copper sulfate higher of mass will be lost that is copper sulfate will loss sulfur and oxygen which led to a higher loss of mass than if you would have heated enough. This higher mass lost will be shown in calculation as percentage of water lost
Use the universal gas formula
PV=nRT
where
P=pressure ( 0.980 atm)
V=volume (L)
T=temperature ( 23 ° C = 23+273.15 = 296.15 ° K)
n=number of moles of ideal gas (0.485 mol)
R=universal gas constant = 0.08205 L atm / (mol·K)
Substitute values,
Volume, V (in litres)
=nRT/P
=0.485*0.08205*296.15/0.980
= 12.0256 L
= 12.0 L (to three significant figures)
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<em>Answer:</em>
- Element: Element are pure substance consist of same type of atoms
- Compound: Form by combination of atoms of different elements.
<em>Distinguish features:</em>
- Compounds represented by chemical formula while elements are represented by symbols.
- Compounds have fixed ratios of atoms of different elements while in elements are distinguished by atomic no.
- Compounds can be converted into simple substance by chemical methods but the elements cant be converted into simpler substance by means of chemical methods.
<em>Example:</em>
- Water (H2O) is a compounds.
- Iron (Fe) is an element.
<em>Summary:</em>
- When two or more elements are mixed then compounds are formed.
Answer:
d making models.
Explanation:
When scientists create a representation of a complex process, they are inferring that they are making models.
A model is an abstraction of the real world or a complex process. Models are very useful in developing solutions to processes that are not easily simplified.
- The models allow a part of a body to be simply studied.
- Through this simple abstraction, extrapolations to other parts of the system can be deduced.
- This can give very useful insights into the other parts of the system.
- The heterogeneity of complex processes is a huge limitation to understanding them.
- A homogenous part can be modelled and used to understand the system.
First i wanted to say you look preety LOL
1) Determine the rate constant for the reaction:
<span>k = (ln 2) / t_1/2 <--- I'll leave you to figure out how that came to be. Hint: use the integrate form of the first-order rate law </span>
<span>k = (ln 2) / 23.6 min </span>
<span>k = 0.02937 min^-1 <--- keep a few extra digits </span>
<span>1) use the integrated form of the first-order rate law: </span>
<span>ln A = -kt + ln A_o </span>
<span>ln A = - (0.02937 min^-1) (120 min) + ln 2.50 </span>
<span>ln A = -3.5244 + 0.91629 </span>
<span>ln A = -2.60811 </span>
<span>A = 0.07367 M <--- round off more as you see fit </span>
<span>Here's another way: </span>
<span>120 min / 23.6 min = 5.106383 half-lives </span>
<span>(0.5)^5.106383 = 0.02902856 <--- the decimal amount remaining after 5.106383 half-lives </span>
<span>0.02902856 x 2.50 M = the answer</span>
