Answer:
Au
Explanation:
For the density of a face-centered cubic:
where
= molar mass of the compound
avogadro's constant
the volume of a unit cell
Given that:
Density 
= 19.30 g/cm³
a = 0.408 nm
a = 
a = 
∴
Thus, the molar mass of 197.37 g/mol element is Gold (Au).
The balanced reaction is as follows;
BiCl₂ + Na₂SO₄ --> 2NaCl + BiSO₄
this is a double displacement reaction
the oxidation number of Bi is +2 in both BiCl₂ and BiSO₄
oxidation number of Cl is -1 in both BiCl₂ and NaCl
oxidation number of Na is +1 in both Na₂SO₄ and NaCl
oxidation numbers of elements in SO₄²⁻ remains the same in both compounds.Therefore the oxidation state in any of the elements in the reaction doesn't change. Neither of the elements show an increase or decrease in the oxidation numbers .
Answer for this question is no element decreases its oxidation number.
We observe that heat capacity of salted water we will find that it is less than pure water. We now that it takes less energy to increase the temperature of the salt water 1°C than pure water. Which means that the salted water heats up faster and eventually reaches to its boiling point first.
hope it helps
We need to know the relationship between atmospheric pressure and the density of gas particles in an area of increasing pressure.
The relationship is: As air pressure in an area increases, the density of the gas particles in that area increases.
For any gaseous substance, density of gas is directly proportional to pressure of gas.
This can be explained from idial gas edquation:
PV=nRT
PV=
RT [where, w= mass of substance, M=molar mass of substance]
PM=
RT
PM=dRT [where, d=density of thesubstance]
So, for a particular gaseous substance (whose molar mass is known), at particular temperature, pressure is directly related to density of gaseous substance.
Therefore, as air pressure in an area increases, the density of the gas particles in that area increases.
Answer:
0.508 mole
Explanation:
NOTE: Since no hydrogen is attached to the compound given in question above, it means the compound is CCl₄.
The number of mole present in 78.2 g of CCl₄ can be obtained as follow:
Mass of CCl₄ = 78.2 g
Molar mass of CCl₄ = 12 + (35.5×4)
= 12 + 142
= 154 g/mol
Mole of CCl₄ =?
Mole = mass / molar mass
Mole of CCl₄ = 78.2 / 154
Mole of CCl₄ = 0.508 mole
Therefore, 0.508 mole is present in 78.2 g of CCl₄
