A certain mass of gàs occupied 5.0dm cube at 2 atm and 10°C .calculate d no of moles present
1 answer:
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Answer:
a) volume of ammonium iodide required =349 mL
b) the moles of lead iodide formed = 0.0436 mol
Explanation:
The reaction is:
It shows that one mole of lead nitrate will react with two moles of ammonium iodide to give one mole of lead iodide.
Let us calculate the moles of lead nitrate taken in the solution.
Moles=molarityX volume (L)
Moles of lead nitrate = 0.360 X 0.121 =0.0436 mol
the moles of ammonium iodide required = 2 X0.0436 = 0.0872 mol
The volume of ammonium iodide required will be:
the moles of lead iodide formed = moles of lead nitrate taken = 0.0436 mol
Answer:
5.41 ×10⁻²²
Explanation:
We were told right from the question that both the Zinc ions and the Zinc oxide adopts a face-centered cubic arrangement.
Then, the number of ZnO molecule in one unit cell = 4
The standard molar mass of ZnO = 81.38g
Avogadro's constant = 6.023 × 10²³ mole
∴
The mass of one unit cell of zinc oxide can be calculated as:
=
= 5.40461564×10⁻²²
≅ 5.41 ×10⁻²²
∴ The mass of one unit cell of zinc oxide = 5.41 ×10⁻²²
Answer:
See solution.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to set up the formula for the calculation of the by-mass percentage of the metal:
Thus, we solve for the molar mass of the metal to obtain:
For the subsequent problems, we proceed as follows:
a.
b.
c.
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