Na₂S(aq) + Cd(NO₃)₂(aq) = CdS(s) + 2NaNO₃(aq)
v=25.00 mL
c=0.0100 mmol/mL
M(Na₂S)=78.046 mg/mmol
n(Na₂S)=n{Cd(NO₃)₂}=cv
m(Na₂S)=M(Na₂S)n(Na₂S)=M(Na₂S)cv
m(Na₂S)=78.046*0.0100*25.00≈19.5 mg
First we need to know that the boiling point of water in C is 100 and we just need to solve for x in the equation:
-33.75-(-77.75) / 100 = 100-(-77.75) / x
44.4/100 = 177.75 / x
x = 177.75*100/44.4 = 400.33
The boiling point of water in ∘a would be 400.33∘a.
Answer:-A. It is less than 890 kJ/mol because the amount of energy required to break bonds is less than the amount of energy released in forming bonds.
Explanation: Endothermic reactions are defined as the reactions in which energy of the product is greater than the energy of the reactants. The total energy is absorbed in the form of heat and 
for the reaction comes out to be positive.
Exothermic reactions are defined as the reactions in which energy of the product is lesser than the energy of the reactants. The total energy is released in the form of heat and for the reaction comes out to be negative.
In the formation of new bonds more energy is released than is required to break the existing bonds, heat is released.
In the formation of bonds less energy is released than is required to break the existing bonds, heat is absorbed.
First, let us find the corresponding amount of moles H₂ assuming ideal gas behavior.
PV = nRT
Solving for n,
n = PV/RT
n = (6.46 atm)(0.579 L)/(0.0821 L-atm/mol-K)(45 + 273 K)
n = 0.143 mol H₂
The stoichiometric calculations is as follows (MW for XeF₆ = 245.28 g/mol)
Mass XeF₆ = (0.143 mol H₂)(1 mol XeF₆/3 mol H₂)(245.28 g/mol) = <em>11.69 g</em>
Answer:
- Molar mass = 608.36 g/mol
Explanation:
It seems the question is incomplete. However a web search us shows this data:
" Reserpine is a natural product isolated from the roots of the shrub Rauwolfia serpentina. It was first synthesized in 1956 by Nobel Prize winner R. B. Woodward. It is used as a tranquilizer and sedative. When 1.00 g reserpine is dissolved in 25.0 g camphor, the freezing-point depression is 2.63 °C (Kf for camphor is 40 °C·kg/mol). Calculate the molality of the solution and the molar mass of reserpine. "
The <em>freezing-point depression</em> is expressed by:
We put the data given by the problem and <u>solve for m</u>:
- 2.63 °C = 40°C·kg/mol * m
For the calculation of the molar mass:<em> Molality</em> is defined as moles of solute per kilogram of solvent:
- 0.06575 m = Moles reserpine / kg camphor
- 25.0 g camphor ⇒ 25.0/1000 = 0.025 kg camphor
We<u> calculate moles of reserpine:</u>
- 0.06575 m = Moles reserpine / 0.025 kg camphor
- Moles reserpine = 1.64x10⁻³ mol
Finally we use the mass of reserpine and the moles to calculate <u>the molar mass</u>:
- 1.00 g reserpine / 1.64x10⁻³ mol = 608.36 g/mol
<em>Keep in mind that if the data in your problem is different, the results will be different. But the solving method remains the same.</em>
