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andrey2020 [161]
3 years ago
10

Where does the energy required to break the interactions between butane molecules come from when butane boils?​

Chemistry
1 answer:
Arada [10]3 years ago
3 0

Answer:

The properties of liquids are intermediate between those of gases and solids, but are more similar to solids. In contrast to intramolecular forces, such as the covalent bonds that hold atoms together in molecules and polyatomic ions, intermolecular forces hold molecules together in a liquid or solid. Intermolecular forces are generally much weaker than covalent bonds. For example, it requires 927 kJ to overcome the intramolecular forces and break both O–H bonds in 1 mol of water, but it takes only about 41 kJ to overcome the intermolecular attractions and convert 1 mol of liquid water to water vapor at 100°C. (Despite this seemingly low value, the intermolecular forces in liquid water are among the strongest such forces known!) Given the large difference in the strengths of intra- and intermolecular forces, changes between the solid, liquid, and gaseous states almost invariably occur for molecular substances without breaking covalent bonds.

Explanation:

     im not sure this is what your looking for but i found this

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If a piece of cadmium with a mass of 37.60 g and a temperature of 100.0 oC is dropped into 25.00 cc of water at 23.0 oC, what wi
zlopas [31]

Answer:

T_{eq}=28.9\°C

Explanation:

Hello!

In this case, since it is observed that hot cadmium is placed in cold water, we can infer that the heat released due to the cooling of cadmium is gained by the water and therefore we can write:

Q_{Cd}+Q_{W}=0

Thus, we insert mass, specific heat and temperatures to obtain:

m_{Cd}C_{Cd}(T_{eq}-T_{Cd})+m_{W}C_{W}(T_{eq}-T_{W})=0

In such a way, since the specific heat of cadmium and water are respectively 0.232 and 4.184 J/(g °C), we can solve for the equilibrium temperature (the final one) as shown below:

T_{eq}=\frac{m_{Cd}C_{Cd}T_{Cd}+m_{W}C_{W}T_{W}}{m_{Cd}C_{Cd}+m_{W}C_{W}}

Now, we plug in to obtain:

T_{eq}=\frac{37.60g*0.232\frac{J}{g\°C}*100.00\°C+25.00g*4.184\frac{J}{g\°C}*23.0\°C}{37.60g*0.232\frac{J}{g\°C}+25.00g*4.184\frac{J}{g\°C}}\\\\T_{eq}=28.9\°C

NOTE: since the density of water is 1g/cc, we infer that 25.00 cc equals 25.00 g.

Best regards!

6 0
2 years ago
If i have an 86 and i don’t do a 50 point assignment, what will my grade be in the class?
nexus9112 [7]
Well you are at a 86% so t<span>his is considered a "B" grade on an average grade scale. If you take 50 points we would need to know the Total point you could have got in that class to be able to see how much a percent was the assignment work and then take that percentage of the assignment and subtract it from the 86% to see where on the scale you would fall in after the assignment points were taken off.   Hope this helps :)
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7 0
3 years ago
Add me on discord if u wanna be study buddies or just normal friends lol
Katyanochek1 [597]
Yesssdirrrrrr and yessss
7 0
3 years ago
Read 2 more answers
A 0.1326 g sample of magnesium was burned in an oxygen bomb calorimeter. the total heat capacity of the calorimeter plus water w
Sladkaya [172]

Answer: Th enthalpy of combustion for the given reaction is 594.244 kJ/mol

Explanation: Enthalpy of combustion is defined as the decomposition of a substance in the presence of oxygen gas.

W are given a chemical reaction:

Mg(s)+\frac{1}{2}O_2(g)\rightarrow MgO(s)

c=5760J/^oC

\Delta T=0.570^oC

To calculate the enthalpy change, we use the formula:

\Delta H=c\Delta T\\\\\Delta H=5760J/^oC\times 0.570^oC=3283.2J

This is the amount of energy released when 0.1326 grams of sample was burned.

So, energy released when 1 gram of sample was burned is = \frac{3283.2J}{0.1326g}=24760.181J/g

Energy 1 mole of magnesium is being combusted, so to calculate the energy released when 1 mole of magnesium ( that is 24 g/mol of magnesium) is being combusted will be:

\Delta H=24760.181J/g\times 24g/mol\\\\\Delta H=594244.3J/mol\\\\\Delta H=594.244kJ/mol

4 0
3 years ago
The molar mass of NH3 is 17.03 g/mol. The molar mass of H2 is 2.0158 g/mol. In a particular reaction, 0.575 g of NH3 forms. What
Arlecino [84]

The balanced chemical reaction is:

N2 + 3H2 = 2NH3

 

We are given the amount of ammonia formed from the reaction. This is where we start our calculations.

 

0.575 g NH3 (1 mol NH3 / 17.03 g NH3) (3 mol H2 / 2 mol NH3) ( 2.02 g H2 / 1 mol H2) = 0.10 g H2

3 0
3 years ago
Read 2 more answers
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