Removing salt from water will reduce it's density
Answer:
B
Explanation:
i think leter B it because
Answer:
given statement is completely true
e.g . water H2O
Answer:
MoO₃
Explanation:
To solve this question we must find the moles of molybdenum in Mo2O3. The moles of Mo remain constant in the new oxide. With the differences in masses we can find the mass of oxygen and its moles obtaining the empirical formula as follows:
<em>Moles Mo2O3 -Molar mass: 239,878g/mol-</em>
11.79g * (1mol / 239.878g) = 0.04915 moles Mo2O3 * (2mol Mo / 1mol Mo2O3) = 0.09830 moles Mo
<em>Mass Mo in the oxides:</em>
0.09830 moles Mo * (95.95g/mol) = 9.432g Mo
<em>Mass oxygen in the new oxide:</em>
14.151g - 9.432g = 4.719g oxygen
<em>Moles Oxygen:</em>
4.719g oxygen * (1mol/16g) = 0.2949 moles O
The ratio of moles of O/Mo:
0.2949molO / 0.09830mol Mo = 3
That means there are 3 moles of oxygen per mole of Molybdenum and the empirical formula is:
<h3>MoO₃</h3>
Answer:
4.9 x 10⁻⁶.
Explanation:
<em>2H₂O(g) ⇄ 2H₂(g) + O₂(g),</em>
'
P of H₂ = 0.002 atm, P of O₂ = 0.006 atm, and P of H₂O = 0.07 atm.
∴ The equilibrium constant (Kp) = (P of H₂)²(P of O₂)/(P of H₂O)² = (0.002 atm)²(0.006 atm)/(0.07 atm)² = 4.9 x 10⁻⁶.
