The answer is c ( to disturbance heat
Answer:
Question 1: Conversion of SO3 into concentrated sulphuric acid.
Question 2: It is necessary to provide heat to the reaction mixture.
Explanation:
Question 1: The first thing you do is dissolve the sulfur trioxide in concentrated sulphuric acid, which produces disulfuric acid. It cannot be carried out by a simple reaction of sulfur trioxide with water. Disulfuric acid does react with water safely, breaking down into 97-99% concentrated sulfuric acid.
Question 2: One reason is due to kinetics (low temperatures will cause the equilibrium to be displaced, but the reaction will be slower) and the other is due to the performance of the catalyst. Thus, the temperature used must be high, since the catalyst works well in this way, and produces a sufficiently high proportion of SO3 in the equilibrium mixture and at a sufficiently high speed.
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hallo! heres you answer
^Extrusive igneous rocks are formed from the cooling lava that flows above the crust and Intrusive Igneous rocks are formed within the earth's interior.^
is the correct one because
-Intrusive, or plutonic, igneous rock forms when magma is trapped deep inside the Earth. ... Intrusive rocks have a coarse grained texture.
- Extrusive Igneous Rocks: Extrusive, or volcanic, igneous rock is produced when magma exits and cools above (or very near) the Earth's surface.
Answer:
41.9(w/w) %
Explanation:
Based on the reaction:
Na₂C₂O₄(s) + 2HCl(aq) → H₂C₂O₄(aq) + 2NaCl(aq)
<em>Where 1 mole of sodium oxalate reacts with 2 moles of HCl</em>
Moles of HCl solution to reach end point are:
44.15mL = 0.04415L ₓ (0.250mol / L) = 0.01104 moles of HCl
As 2 moles of HCl reacts per mole of Na₂C₂O₄:
0.01104mol HCl ₓ (1 mol Na₂C₂O₄ / 2 mol HCl) = <em>5.519x10⁻³ moles Na₂C₂O₄</em> are in the sample.
Molar mass of Na₂C₂O₄ is 134g/mol; thus, mass of 5.519x10⁻³ moles Na₂C₂O₄ is:
5.519x10⁻³ moles Na₂C₂O₄ ₓ (134g / mol) = <em>0.740g of Na₂C₂O₄</em> in the sample.
Thus, percent by mass of sodium oxalate in the sample is:
0.740g of Na₂C₂O₄ / 1.766g ₓ 100 =
<h3>41.9(w/w) %</h3>