Answer:
The answer is C. Gas particles have no attractive forces between them.
Explanation:
"Physical changes occur when objects or substances undergo a change that does notchange their chemical composition. This contrasts with the concept of chemical change in which the composition of a substance changes or one or more substances combine or break up to form new substances."
Answer:
3Ba(OH)2 + 2H3PO4 —> Ba3(PO4)2 + 6H2O
Explanation:
Ba(OH)2 + H3PO4 —> Ba3(PO4)2 + H2O
There are 3 atoms of Ba on the right side and 1atom on the left side. It can be balance by putting 3 in front of Ba(OH)2 as shown below:
3Ba(OH)2 + H3PO4 —> Ba3(PO4)2 + H2O
There are 2 atoms of P on the right side and 1atom on the left. It can be balance by putting 2 in front of H3PO4 as shown below:
3Ba(OH)2 + 2H3PO4 —> Ba3(PO4)2 + H2O
Now, there are a total of 12 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 6 in front of H2O as shown below:
3Ba(OH)2 + 2H3PO4 —> Ba3(PO4)2 + 6H2O
Now the equation is balanced as the numbers of the atoms of the different elements present on both sides are equal
The reaction is:
Cl2 + 2 KBr --> 2 KCl + Br2
Moles of KCl is
n = m /M = 12 /74 = 0.16 mol
As, twice the moles of KCl is producing from 1 mol of chlorine
mole of Cl2 = 0.16 /2 = 0.08 mol
Mass of Cl2
m /70 = 0.08 = 5.6 g
Hence, 5.6 g mol Cl2 consumed to produce KCl
This problem is describing a gas mixture whose mole fraction of hexane in nitrogen is 0.58 and which is being fed to a condenser at 75 °C and 3.0 atm, obtaining a product at 3.0 atm and 20 °C, so that the removed heat from the system is required.
In this case, it is recommended to write the enthalpy for each substance as follows:
Whereas the specific heat of liquid and gaseous n-hexane are about 200 J/(mol*K) and 160 J/(mol*K) respectively, its condensation enthalpy is 31.5 kJ/mol, boiling point is 69 °C and the specific heat of gaseous nitrogen is about 29.1 J/(mol*K) according to the NIST data tables and 
and 
are the mole fractions in the gaseous mixture. Next, we proceed to the calculation of both heat terms as shown below:
It is seen that the heat released by the nitrogen is neglectable in comparison to n-hexanes, however, a rigorous calculation is being presented. Then, we add the previously calculated enthalpies to compute the amount of heat that is removed by the condenser:
Finally we convert this result to kJ:
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