Answer:
∆H > 0
∆Srxn <0
∆G >0
∆Suniverse <0
Explanation:
We are informed that the reaction is endothermic. An endothermic reaction is one in which energy is absorbed hence ∆H is positive at all temperatures.
Similarly, absorption of energy leads to a decrease in entropy of the reaction system. Hence the change in entropy of the reaction ∆Sreaction is negative at all temperatures.
The change in free energy for the reaction is positive at all temperatures since ∆S reaction is negative then from ∆G= ∆H - T∆S, we see that given the positive value of ∆H, ∆G must always return a positive value at all temperatures.
Since entropy of the surrounding= - ∆H/T, given that ∆H is positive, ∆S surrounding will be negative at all temperatures. This is so because an endothermic reaction causes the surrounding to cool down.
Answer:
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Explanation:
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Answer:
CaO + SO 3 → CaSO. 4
This is an acid-base reaction (neutralization): CaO is a base, SO 3 is an acid.
Answer : Hydrogen-bonding, Dipole-dipole attraction and London-dispersion force.
Explanation :
The given molecule is, 
Three types of inter-molecular forces are present in this molecule which are Hydrogen-bonding, Dipole-dipole attraction and London-dispersion force.
- Hydrogen-bonding : when the partial positive end of hydrogen is bonded with the partial negative end of another molecule like, oxygen, nitrogen, etc.
- Dipole-dipole attraction : When the partial positively charged part of the molecule is interact with the partial negatively charged part of the molecule. For example : In case of HCl.
- London-dispersion force : This force is present in all type of molecule whether it is a polar or non-polar, ionic or covalent. For example : In case of Br-Br , F-F, etc
Hydrogen-bonding is present between the oxygen and hydrogen molecule.
Dipole-dipole forces is present between the carbon and oxygen molecule.
London-dispersion forces is present between the carbon and carbon molecule.
Answer:
For the first oxide, 1 g gives 0.888 g of copper.
Dividing by 0.888 tells us that 1.126 g gives 1 g of copper so has 0.126 g of oxygen.
For the second oxide, 1 g gives 0.798 g of copper.
Dividing by 0.798 tells us that 1.253 g gives 1 g of copper so has 0.253 g of oxygen.
So 1 g of copper combines with either 0.126 g or 0.253 g of oxygen.
Within the limits of experimental error, 0.253 is twice 0.126, confirming the law of multiple proportion.
