Answer:
120.0 kJ
Explanation:
- First we convert the given mass of C to moles:
30.00 g C ÷ 12g/mol = 2.5 mol
- The ΔH° value given by the problem, is the heat absorbed when 5 moles of C react.
So when we have<u> half the moles of C</u> (2.5 instead of 5.0), t<u>he heat absorbed will also be half</u>, thus the answer is:
Las vitaminas liposolubles (A, D, E, y K) son lípidos a base de isopreno que se almacenan en el hígado y la grasa.
Allene (1,2-propadiene) has point group D2d, itself is achiral because it has two planes of symmetry. ... An allene with substituents on one terminal carbon atom are unlike and substituent on other terminal carbon atoms are same, allene will be achiral. It will have one symmetry plane.
Hope this helped :)
Answer:
94.325 g
Explanation:
We'll begin by converting 350 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
350 mL = 350 mL × 1 L /1000 mL
350 mL = 0.35 L
Next, we shall determine the number of mole of KC₂H₃O₂ in the solution. This can be obtained as follow:
Volume = 0.35 L
Molarity of KC₂H₃O₂ = 2.75 M
Mole of KC₂H₃O₂ =?
Molarity = mole /Volume
2.75 = Mole of KC₂H₃O₂ / 0.35
Cross multiply
Mole of KC₂H₃O₂ = 2.75 × 0.35
Mole of KC₂H₃O₂ = 0.9625 mole
Finally, we shall determine the mass of KC₂H₃O₂ needed to prepare the solution. This can be obtained as illustrated below:
Mole of KC₂H₃O₂ = 0.9625 mole
Molar mass of KC₂H₃O₂ = 39 + (12×2) +(3×1) + (16×2)
= 39 + 24 + 3 + 32
= 98 g/mol
Mass of KC₂H₃O₂ =?
Mass = mole × molar mass
Mass of KC₂H₃O₂ = 0.9625 × 98
Mass of KC₂H₃O₂ = 94.325 g
Thus, the mass of KC₂H₃O₂ needed to prepare the solution is 94.325 g
