<span> the </span>vapor pressure<span> of the liquid at a temperature T</span>2<span> ... Now, </span>it's<span> important to realize that the </span>normal boiling point<span> of a substance is measured at an atmoshperic ... ΔHvap=−ln(</span>134mmHg760mmHg<span> )⋅8.314J mol−1K−1 (1(273.15+</span>0)−1(273.15+40))K−1 ... Give equations that can be used tocalculate<span> the .
Now try it yourself :)</span>
Answer:
A. endothermic.
A. Yes, absorbed.
Explanation:
Let's consider the following thermochemical equation.
2 HgO(s) ⇒ 2 Hg(l) + O₂(g) ΔH = 182 kJ
The enthalpy of the reaction is positive (ΔH > 0), which means that the reaction is endothermic.
182 kJ are absorbed when 2 moles of HgO react (molar mass 216.59 g/mol). The heat absorbed when 72.8 g of HgO react is:
4.48 mol Cl2. A reaction that produces 0.35 kg of BCl3 will use 4.48 mol of Cl2.
(a) The <em>balanced chemical equation </em>is
2B + 3Cl2 → 2BCl3
(b) Convert kilograms of BCl3 to moles of BCl3
MM: B = 10.81; Cl = 35.45; BCl3 = 117.16
Moles of BCl3 = 350 g BCl3 x (1 mol BCl3/117.16 g BCl3) = 2.987 mol BCl3
(c) Use the <em>molar ratio</em> of Cl2:BCl3 to calculate the moles of Cl2.
Moles of Cl2 = 2.987 mol BCl3 x (3 mol Cl2/2 mol BCl3) = 4.48 mol Cl2
Volume well in the picture the fish tank seems to be a rectangle, so you would do V=length x width x height. So V=5x3x3=45
Answer:
And we have to calculate the number of moles of sucrose present in a lb mass of sucrose: Moles of sucrose=454⋅g342.30⋅g⋅mol−1=1.33⋅mol .
Explanation:
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