Cl is stable as a diatomic molecule where the 2 Cl atoms are held together by a covalent bond
molar mass of the diatomic molecule is 70.9 g/mol
therefore 70.9 g of Cl₂ is - 1 mol
then 140 g of Cl₂ is - 1/70.9 x 140 = 1.97 mol
there are 1.97 mol of Cl₂ present
It's 95 F
The formula is the following:
T
Answer:
<u>225.6 kJ</u>, <em>assuming the water is already at 100 °C</em>
Explanation:
The correct answer to this question will depend on the initial temperature of the water to which heat is added to produce steam. Energy is required to raise the water temperature to 100°C. At that point, an energy of vaporization is needed to convert liquid water at 100 °C to water vapor at 100°C. The heat of vaporization for water is 2256.4 kJ/kg. The energy required to bring 100g of water from a lower temperature to 100°C is calculated at 4.186 J/g°C. We don't know the starting temperature, so this step cannot be calculated.
<em><u>Assuming</u></em> that we are already at 100 °C, we can calculate the heat required for vaporization:
(100.0g)(1000.0g/1 kg)(2256.4 kJ/kg) = 225.6 kJ for 100 grams water.
<span> </span>
Answer
is: volume is 20 mL.<span>
c</span>₁(CH₃COOH) = 2,5 M.<span>
c</span>₂(CH₃COOH) = 0,5 M.<span>
V</span>₂(CH₃COOH) = 100 mL.<span>
V</span>₁(CH₃COOH) = ?<span>
c</span>₁(CH₃COOH) · V₁(CH₃COOH)
= c₂(CH₃COOH) · V₂(CH₃COOH).<span>
2,5 M · V</span>₁(CH₃COOH)
= 0,5 M · 100 mL.<span>
V</span>₁(CH₃COOH) = 0,5 M · 100 mL ÷ 2,5 M.<span>
V</span>₁(CH₃COOH) = 20 mL ÷ 1000 mL/L =0,02 L.
