Answer:
For temperatures higher than 533.49 K we will see a spontaneous reaction, and for temperatures lower than that the reaction will not be spontaneous.
Explanation:
When are chemical reactions spontaneous? To find out we need to look at the reaction's change in Gibbs Free energy:
When this is greater than zero, the reaction isn't spontaneous, when it is less than zero, we have a spontaneous reaction. The reaction must then change from spontaneous to non spontaneous when 
. If we insert that into our equation we get:
That is the temperature at which the reaction's spontaneity will change, plugging in our values we find:
At that temperature we have .
Now, at a temperature greater than this one, the entropy term in our equation for the Gibbs' free energy of reaction will take over, and make 
, thus the reaction will be spontaneous.
On the other hand, if we lower the temperature, we will have a smaller entropy term, and we will have: 
. That is, the reaction will not be spontaneous. Therefore for temperatures higher than 533.49 K we will see a spontaneous reaction, and for temperatures lower than that the reaction will not be spontaneous.
Mass = mass/molar mass of ch3ch2nh2
the molar mass of CH3CH2NH2 = 12 +(1x3)+12+(1 x2)+14+(1x2) =45 g/mol
moles is therefore = 1.50g /45g/mol = 0.033 moles
Answer:
A = 674.33mmHg
B = 0.385atm
Explanation:
Both question A and B requires the application of pressure law which states that the pressure of a fixed mass of gas is directly proportional to its temperature provided that volume is kept constant.
Mathematically,
P = kT, k = P / T
P1 / T1 = P2 / T2 = P3 / T3 =.......= Pn/Tn
A)
Data:
P1 = 799mmHg
T1 = 50°C = (50 + 273.15) = 323.15K
P2 = ?
T2 = 273.15K
P1 / T1 = P2 / T2
Solve for P2
P2 = (P1 × T2) / T1
P2 = (799 × 273.15) / 323.15
P2 = 674.37mmHg
The final pressure is 674.37mmHg
B)
P1 = 0.470atm
T1 = 60°C = (60 + 273.15)K = 333.15K
P2 = ?
T2 = 273.15K
P1 / T1 = P2 / T2
Solve for P2,
P2 = (P1 × T2) / T1
P2 = (0.470 × 273.15) / 333.15
P2 = 0.385atm
The final pressure is 0.385atm
Answer:
see attached
Explanation:
Dimensional analysis is useful whenever dimensions are involved. Unless it is quite clear that all of the problem dimensions are consistent (for example, all speeds in miles per hour, or all angles in degrees), dimensional analysis can be useful for keeping the math straight.
Only units of the same dimensions can be added or subtracted. When numbers are multiplied or divided or raised to a power, dimensional analysis can help ensure that the appropriate operations are being used on appropriate numbers. It can also help ensure that dimensions are being combined properly to give appropriate derived dimensions.
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Scientific notation is a way of writing very large or very small numbers compactly. It can also help with "order of magnitude" estimates. If an answer using SI prefixes is appropriate, or if a number can be conveniently expressed in standard form, then scientific notation is usually not required.
On the other hand, SI prefixes may not be appropriate in some cases, or a problem may specify that scientific notation be used for expressing results. In those instances, scientific notation should be used.
Some of the particles undergo a phase change and become a gas due to burning. An example would be CO2, which is lost in the air around/escapes into the atmosphere. Hope that helps!
