Each half-life results in ~50% (1/2) of the original element remaining.
7500/1250 = 6 half-lives, so 100(1/2)^6
= 100(0.015625)
= 1.5625% of the original element would remain
Answer:Use an excess of ethane
Explanation:
The halogenation of alkanes is a substitution reaction. All the hydrogen atoms in the alkanes could be potentially substituted. How ever the reaction can be controlled by using an excess of either the alkane or the halogen. If the aim (as it is in this question) is to minimize the yield of halogenated alkanes, an excess of the alkane (in this case, ethane) is used.
Answer:
-431.5 J/g
Explanation:
Mass of solution = Mass of solute + mass of solvent
Solute is KOH while solvent is water.
Mass of KOH = 16.9 g
Mass of water = 90.8 g
Mass of solution = 16.9 + 90.8
= 107.7 g
Change in temperature (Δt) = 34.27 - 18.5
= 16.2 °C
Heat required to raise the temperature of water is released by dissolving KOH.
Therefore,
Heat released by KOH = m × s× Δt
= 107.7 × 4.18 × 16.2
= 7293 J
Heat released by per g KOH = 7293 J/16.9 g
= 431.5 J/g
As heat is released therefore, enthalpy change would be negative.
Enthalpy change of KOH = -431.5 J/g
Answer:
83.20 g of Na3PO4
Explanation:
1 mole of Na3PO4 contains 3 moles of Na+.
Mole of Na ion to be prepared = Molarity x volume
= 0.700 x 725/1000
= 0.5075 mole
If 1 mole of Na3PO4 contains 3 moles of Na ion, then 0.5075 Na ion will be contained in:
0.5075/3 x 1 = 0.1692 mole of Na3PO4
mole of Na3PO4 = mass/molar mass = 0.1692
Hence, mass of Na3PO4 = 0.1692 x molar mass
= 0.1692 x 163.94
= 83.20 g.
83.20 g of Na3PO4 will be needed.
Answer:
Swan received a formal education, while Edison did not.
Explanation:
got it right
