Answer:
it's a precipitation reaction.
Explanation:
since a solid is produced, one of the elements are insoluble with one another–making a precipitate.
Answer:
Temperature at which molybdenum becomes superconducting is-272.25°C
Explanation:
Conductor are those hard substances which allows path of electric current through them. And super conductors are those hard substances which have resistance against the flow of electric current through them.
As given, molybdenum becomes superconducting at temperatures below 0.90 K.
Temperature in Kelvins can be converted in °C by relation:
T(°C)=273.15-T(K)
Molybdenum becomes superconducting in degrees Celsius.
T(°C)=273.15-0.90= -272.25 °C
Temperature at which molybdenum becomes superconducting is -272.25 °C
<u>Answer:</u> The experimental van't Hoff factor is 1.21
<u>Explanation:</u>
The expression for the depression in freezing point is given as:
where,
i = van't Hoff factor = ?
= depression in freezing point = 0.225°C
= Cryoscopic constant = 1.86°C/m
m = molality of the solution = 0.100 m
Putting values in above equation, we get:
Hence, the experimental van't Hoff factor is 1.21
We can write the balanced equation for the synthesis reaction as
H2(g) + Cl2(g) → 2HCl(g)
We use the molar masses of hydrogen chloride gas HCl and hydrogen gas H2 to calculate for the mass of hydrogen gas H2 needed:
mass of H2 = 146.4 g HCl *(1 mol HCl / 36.46 g HCl) * (1 mol H2 / 2 mol HCl) *
(2.02 g H2 / 1 mol H2)
= 4.056 g H2
We also use the molar masses of hydrogen chloride gas HCl and chlorine gas CL2 to calculate for the mass of hydrogen gas H2:
mass of CL2 = 146.4 g HCl *(1 mol HCl / 36.46 g HCl) * (1 mol Cl2 / 2 mol HCl) *
(70.91 g Cl2 / 1 mol Cl2)
= 142.4 g Cl2
Therefore, we need 4.056 grams of hydrogen gas and 142.4 grams of chlorine gas to produce 146.4 grams of hydrogen chloride gas.
