Explanation:
Formula to calculate osmotic pressure is as follows.
Osmotic pressure = concentration × gas constant × temperature( in K)
Temperature =
= (25 + 273) K
= 298.15 K
Osmotic pressure = 531 mm Hg or 0.698 atm (as 1 mm Hg = 0.00131)
Putting the given values into the above formula as follows.
0.698 =
C = 0.0285
This also means that,
= 0.0285
So, moles = 0.0285 × volume (in L)
= 0.0285 × 0.100
=
Now, let us assume that mass of = x grams
And, mass of = (1.00 - x)
So, moles of
=
Now, moles of
=
=
= x = 0.346
Therefore, we can conclude that amount of present is 0.346 g and amount of present is (1 - 0.346) g = 0.654 g.
Answer:
Atoms He (Avogadro’s number) → Moles of He (molar mass of He) → Mass of He
• molar mass of He (from the periodic table) = 4.003 g/mol
• Avogadro’s Number: Avogadro’s number gives us the number of entities present in 1 mole: 6.022 × 1023 He atoms in 1 mole of He
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Answer:
0. 414
Explanation:
Octahedral interstitial lattice sites.
Octahedral interstitial lattice sites are in a plane parallel to the base plane between two compact planes and project to the center of an elementary triangle of the base plane.
The octahedral sites are located halfway between the two planes. They are vertical to the locations of the spheres of a possible plane. There are, therefore, as many octahedral sites as there are atoms in a compact network.
The Octahedral interstitial void ratio range is 0.414 to 0.732. Thus, the minimum cation-to-anion radius ratio for an octahedral interstitial lattice site is 0. 414.
Answer:
The atomic number is the number of protons in the nucleus