Answer:
The density of the metal is 0.561 g/mL
Explanation:
The computation of the density of the metal is shown below;
As we know that
The Density of the metal is
where,
Mass = 4.9g
Change in volume = 6.9 mL
Now place these values to the above formula
So, the density of the metal is
= 0.561 g/mL
Hence, the density of the metal is 0.561 g/mL
We simply applied the above formula so that the correct density could arrive
It’s an equipment in a home , work place , many other places.
A metalloid can be:
- Boron (B)
- Silicon (Si)
- Germanium ( Ge)
- Arsenic (As)
- Antimony ( Sb)
- Tellurium (Te)
- Polonium (Po)
Hope this helps :)
Answer:
The freezing point of the solution is 78.71 °C
Explanation:
Step 1: Data given
Freezing point for the pure solvent is 79.4 °C.
Molality = 0.1 molal
Freezing point depression = 6.9 °C/m
Step 2: Calculate freezing point of the solution
ΔT = i*Kb*m
⇒ with i = the van't Hoff factor: naphthalene = a non-electrolyte, which means that the van't Hoff factor for this solution will be 1
⇒ with Kb = the freezing point depression constant = 6.9 °C/m
⇒ m = the molality = 0.1 molal
ΔT = 1 * 6.9 * 0.1
ΔT = 0.69
Freezing point of the solution = 79.4 °C - 0.69 = 78.71 °C
The freezing point of the solution is 78.71 °C
Answer:
A) 8.00 mol NH₃
B) 137 g NH₃
C) 2.30 g H₂
D) 1.53 x 10²⁰ molecules NH₃
Explanation:
Let us consider the balanced equation:
N₂(g) + 3 H₂(g) ⇄ 2 NH₃(g)
Part A
3 moles of H₂ form 2 moles of NH₃. So, for 12.0 moles of H₂:
Part B:
1 mole of N₂ forms 2 moles of NH₃. And each mole of NH₃ has a mass of 17.0 g (molar mass). So, for 4.04 moles of N₂:
Part C:
According to the <em>balanced equation</em> 6.00 g of H₂ form 34.0 g of NH₃. So, for 13.02g of NH₃:
Part D:
6.00 g of H₂ form 2 moles of NH₃. An each mole of NH₃ has 6.02 x 10²³ molecules of NH₃ (Avogadro number). So, for 7.62×10⁻⁴ g of H₂: