Answer:
295.7 mL of 24% trichloroacetic acid (tca) is needed .
Explanation:
Let the volume of 24% trichloroacetic acid solution be x
Volume of required 10% trichloroacetic acid solution =8 bottles of 3 ounces
= 24 ounces = 709.68 mL
(1 ounces = 29.57 mL)
Amount of trichloroacetic acid in 24% solution of x volume of solution will be equal to amount of trichloroacetic acid in 10% solution of volume 709.68 mL.
x = 295.7 mL
295.7 mL of 24% trichloroacetic acid (tca) is needed .
Answer:
Assume that 100 grams of C2H4 is present. This means that there are 85.7 grams of carbon and 14.3 grams of hydrogen.
Convert these weights to moles of each element:
85.7 grams carbon/12 grams per mole = 7 moles of carbon.
14.3 grams hydrogen/1 gram per mole = 14 moles of hydrogen.
Divide by the lowest number of moles to obtain one mole of carbon and two moles of hydrogen.
Since we know that there cannot be a stable CH2 molecule, multiply by two and you have C2H4 which is ethylene - a known molecule.
The secret is to convert the percentages to moles and find the ration of the constituents.
Answer:
Pb3O4 + 4H2 → 3Pb + 4H2O
Explanation:
Pb3O4
Tritium - H2
Molar Mass of H2 Bond Polarity H-3 Hydrogen-3 3H T
Products
Lead - Pb
Molar Mass of Pb Plumbum Element 82 Bulk Lead
Water - H2O
Molar Mass of H2O Oxidation Numbers of H2O Dihydrogen Monoxide Dihydridooxygen Hoh Hydrogen Hydroxide Dihydrogen Oxide Oxidane Hydrogen Oxide Pure Water
We want to solve Q = mcΔT for the liquid water; its change in temperature will tell us the amount of thermal energy that flowed out of the reaction. The specific heat, c, of water is 4.184 J/g °C.
Q = (72.0 g)(4.184 J/g °C)(100 °C - 25 °C) = 22593.6 J
Q ≈ 2.26 × 10⁴ J or 22.6 kJ (three significant figures).
