Answer:
You will have 19.9L of Cl2
Explanation:
We can solve this question using:
PV = nRT; V = nRT/P
<em>Where V is the volume of the gas</em>
<em>n the moles of Cl2</em>
<em>R is gas constant = 0.082atmL/molK</em>
<em>T is 273.15K assuming STP conditions</em>
<em>P is 1atm at STP</em>
The moles of 63g of Cl2 gas are -molar mass: 70.906g/mol:
63g * (1mol / 70.906g) = 0.8885 moles
Replacing:
V = 0.8885mol*0.082atmL/molK*273.15K/1atm
V = You will have 19.9L of Cl2
From the information given, the total volume of rubbing alcohol is 88.2 ml
68.6 % of this volume is isopropanol.
We will assume 88.2 ml represents 100% volume, so the volume of water will be 31.4 %
The volume of isopropanol is
68.6/100 x 88.2 → 0.686 × 88.2 = 60.505 ml
The volume of isopropanol is 60.5 ml.
Volume of water will be 88.20 - 60.5 = 27.7 ml
(27.7 / 88.2 × 100 = 31.4% )
Adding 60.5 ml of isopropanol to 27.7 ml of water to make up 88.2 ml will give 68.6 % v/v isopropanol to water solution.
Answer:
heat energy is released into the surrounding
In order to answer this question, the units of volume must be consistent. In this problem, we decide the unit m3 to be uniform. Option A is equal to 12 m3, option b is equal to 1.2x10^8/100^3 or 120 m3. Option C is 2.0 x10^4/ 10^3 or 20 m3. Option D is 1.2x10^8/ 1000^3 or 0.12 m3. The greatest volume is option b. 120 m3.
Limestone (CaCO₃) is the second most abundant mineral on Earth after SiO₂. For many uses, it is first decomposed thermally to quicklime (CaO). MgO is prepared similarly from MgCO₃.
AT 871°C CaCO 3 needs about 1 hour for complete decomposition.
<h3>
At what temperature Caco3 decompose to Cao?</h3>
At any temperature higher than 835°C, the value of Δ G ∘ will be negative and the decomposition reaction will be spontaneous.
AT 871°C CaCO 3 needs about 1 hour for complete decomposition.
Calcium carbonate decomposes on heating to give calcium oxide and carbon dioxide.
To learn more about Caco3, refer
brainly.com/question/26187973
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