Answer:
- The room mantained at a lower temperature will contain more air molecules.
Explanation:
1) Since the two rooms are <em>connected by an open door</em>, you assume pressure equilibrium: the pressure on the two rooms is the same.
2) Since you consider <em>two equal size rooms</em>, both volumes are equal.
3) Assuming ideal gas behavior, pressure (P), temperature (T), volume (V) and number of moles (n) are related by the equation PV = nRT
4) Naming T₁ the lower temperature, T₂ the higher temperature, n₁ the number of moles of air in the room at lower temperature, and n₂ the number of moles of air in the room at higher temperature, you get:
- n₁ T₁ = n₂ T₂, or n₁ / n₂ = T₂ / T₁
5) That means that the amount of molecules (number of moles) is inversely related to the temperature: the higher the temperature the lower the number of moles, and the lower the temperature the greater the number of moles.
Hence, the answer is that <em>the room that contains more air molecules is the room mantained at a lower temperature.</em>
20.06 g of Hg and 1.6 g of O₂
<u>Explanation:</u>
To Find:
Number of Mercury and oxygen that can be obtained from 21.7 g of HgO
First we have to write the balanced equation for the decomposition reaction of Mercury(II) oxide as,
2 HgO (s) → 2Hg(l) + O₂ (g)
21.7 g of HgO = 
= 0.1 mol of HgO.
As per the above equation, we can find the mole ratio between HgO and Hg is 1: 1 and that of HgO and oxygen is 2:1 .
So amount of Hg produced = 0.1 mol × 200.59 g / mol ( molar mass of Hg)
= 20.06 g of Hg
Amount of oxygen produced = 0.05 mol × 32 g/ mol = 1.6 g of O₂
Thus it is clear that 20.06 g of Hg and 1.6 g of O₂ is obtained from 21.7 g of HgO
Answer:
my define it will be turst me is c
Answer:
Explanation:
1)<u><em> Ionization equilibrium equation: given</em></u>
- H₂O(l) + H₂O(l) ⇌ H₃O⁺(aq) + OH⁻(aq)
2) <em><u>Ionization equilibrium constant, at 25°C, Kw: given</u></em>
<u>3) Stoichiometric mole ratio:</u>
As from the ionization equilibrium equation, as from the fact it is stated, the concentration of both ions, at 25°C, are equal:
- [H₃O⁺(aq)] = [OH⁻(aq)] = 1.0 × 10⁻⁷ M
- ⇒ Kw = [H3O⁺] [OH⁻] = 1.0 × 10⁻⁷ × 1.0 × 10⁻⁷ = 1.0 × 10⁻¹⁴ M
<u><em>4) A solution has a [OH⁻] = 3.4 × 10⁻⁵ M at 25 °C </em></u><em><u>and you need to calculate what the [H₃O⁺(aq)] is.</u></em>
Since the temperature is 25°, yet the value of Kw is the same, andy you can use these conditions:
Then you can substitute the known values and solve for the unknown:
- 1.0 × 10⁻¹⁴ M² = [H₃O⁺] × 3.4 × 10⁻⁵ M
- ⇒ [H₃O⁺] = 1.0 × 10⁻¹⁴ M² / ( 3.4 × 10⁻⁵ M ) = 2.9⁻¹⁰ M
As you see, the increase in the molar concentration of the ion [OH⁻] has caused the decrease in the molar concentration of the ion [H₃O⁺], to keep the equilibrium law valid.
The mass of electrons , i just did the same test and got it right
