Answer:
CaBr2 is a colorless solid and the charge on Calcium is +2
Explanation:
CaBr2 is a colorless solid at room temperature.
Other properties include -
a) It is a crystalline solid
b) Melting point of 765°C.
c) It is deliquescent
d) Soluble in water and absolute alcohol.
The charge of Br is -1. Here two atoms of Br attaches with one atom of calcium. Hence the charge of calcium is +2
<u>Answer:</u>
<em>B. A and B</em>
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<u>Explanation:</u>
1. Solids have a definite volume, definite size and definite shape
2. The particles present in a solid are very closely packed since the intermolecular forces between them are very strong. The molecules do not move apart.
3. Melting point is the temperature at which solid changes into a liquid.
4. When a solid is heated to the melting point the intermolecular forces are overcome by the energy and the molecules present in it moves from its fixed position to take its liquid state which is called as melting.
Answer:
−153.1 J / (K mol)
Explanation:
Calculate the standard entropy of reaction at 298 K for the reaction Hg(liq) + Cl2(g) → HgCl2(s) The standard molar entropies of the species at that temperature are: Sºm (Hg,liq) = 76.02 J / (K mol) ; Sºm (Cl2,g) = 223.07 J / (K mol) ; Sºm (HgCl2,s) = 146.0 J / (K mol)
Hg(liq) + Cl2(g) → HgCl2(s)
Given that;
The standard molar entropies of the species at that temperature are:
Sºm (Hg,liq) = 76.02 J / (K mol) ;
Sºm (Cl2,g) = 223.07 J / (K mol) ;
Sºm (HgCl2,s) = 146.0 J / (K mol)
The standard molar entropies of reaction = Sºm[products] - Sºm [ reactants]
= 146.0 J / (K mol) – [76.02 J / (K mol) +223.07 J / (K mol) ]
= -153.09 J / (K mol)
= or -153.1 J / (K mol)
Hence the answer is −153.1 J / (K mol)
Answer: The boiling point of a sample of Propanone at 180 kPa is 75 C. The correct answer is 2.
Explanation:
Propanone is more commonly known under the name "acetone." The boiling point of this compound at 180 kPa will be 75 C. It is an organic compound and its formula is (CH3)2CO.
This compound does not any color, it is highly flammable, and volatile. This compound does have a distinguished odor. It is used in labs, in industry settings, and in the home for various things. The human body does produce this compound naturally and the body disposes of this compound.
