P₁ = 0.90 atm
V₁ = 50.0 mL
T₁ = 298 K
P₂ = 1 atm
T₂ = 273 K
V₂ = P₁ x V₁ x T₂ / T₁ x P₂
V₂ = 0.90 x 50.0 x 273 / 298 x 1
V₂ = 12285 / 298
V₂ = 41 mL
Answer (1)
hope this helps!
Answer:
Total mass of original sample = 0.38 g
Explanation:
Percentage of chloride = 74.5%
Mass of AgCl precipitate = 1.115 g
Mass of original sample = ?
Solution:
Mass of chloride:
1.115 g/ 143.3 g/mol × 35.5 g/mol
0.0078 g × 35.5 = 0.28 g
Total mass of compound:
Total mass = mass of chloride / %of Cl × 100%
Total mass = 0.28 g/ 74.5% × 100%
Total mass = 0.0038 g× 100
Total mass = 0.38 g
Answer:
8.937g/cm³
Explanation:
To answer this question we need to know that, in 1 unit FCC cell you have:
Edge length = √8 * R
<em>Volume = 8√8 * R³</em>
<em>And there are 4 atoms per unit cell</em>
<em />
<em>Mass of 4 atoms in g:</em>
4 atom * (1mol / 6.022x10²³atom) * (63.55g / mol) = 4.221x10⁻²²g
<em />
<em>Volume in cm³:</em>
0.1278nm * (1x10⁻⁷cm / 1nm) = 1.278x10⁻⁸cm
Volume = 8√8 * (1.278x10⁻⁸cm)³
Volume = 4.723x10⁻²³cm³
And density is:
4.221x10⁻²²g / 4.723x10⁻²³cm³ =
<h3>8.937g/cm³</h3>
