Answer:
d. 127 g/mol.
Explanation:
Hello!
In this case, since we have the amount of molecules of this this compound, we are able to compute the moles out there by using the Avogadro's number:
Which correspond to the moles of X2. Then, by using the mass we are able to compute the molar mass of X2:
It means that the atomic mass of X halves the molar mass of X2, which is then d. 127 g/mol.
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We will use boiling point formula:
ΔT = i Kb m
when ΔT is the temperature change from the pure solvent's boiling point to the boiling point of the solution = 77.85 °C - 76.5 °C = 1.35
and Kb is the boiling point constant =5.03
and m = molality
i = vant's Hoff factor
so by substitution, we can get the molality:
1.35 = 1 * 5.03 * m
∴ m = 0.27
when molality = moles / mass Kg
0.27 = moles / 0.015Kg
∴ moles = 0.00405 moles
∴ The molar mass = mass / moles
= 2 g / 0.00405 moles
= 493.8 g /mol
Beta particle in the reaction
Answer:
1.17 mol
Explanation:
Step 1: Write the balanced equation
2 Al + 6 HCl → 2 AlCl₃ + 3 H₂
Step 2: Calculate the moles corresponding to 85.0 g of HCl
The molar mass of HCl is 36.46 g/mol.
85.0 g × 1 mol/36.46 g = 2.33 mol
Step 3: Calculate the number of moles of H₂ produced from 2.33 moles of HCl
The molar ratio of HCl to H₂ is 6:3.
2.33 mol HCl × 3 mol H₂/6 mol H₂ = 1.17 mol H₂
Answer:
1.72x10⁻⁵ g
Explanation:
To solve this problem we use the PV=nRT equation, where:
- R = 0.082 atm·L·mol⁻¹·K⁻¹
- T = 25 °C ⇒ (25+273.16) = 298.16 K
And we <u>solve for n</u>:
- 1 atm * 5.7x10⁶ L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 298.16 K
Finally we <u>convert moles of helium to grams</u>, using its <em>molar mass</em>:
- 4.29x10⁻⁶ mol * 4 g/mol = 1.72x10⁻⁵ g
