Answer:
cell :—
•They lack a well-defined nucleus, have a nucleoid instead.
•Usually have double-stranded, circular DNA.
•Do not have mitochondria.
Eukaryotic cell :—
•Have a well-defined nucleus enclosed in the nuclear membrane.
•Have a linear double-stranded nucleus.
•Mitochondria are present.
The balanced equation is 2HgO --> 2Hg + O2
Not sure but I think it’s B try it out and double check it
(4 mol H2O) x (112 kJ / 3 mol H2O) = 149 kJ
<span>(14.5 g HCl) / (36.4611 g HCl/mol) x (112 kJ / 3 mol HCl) = 14.9 kJ </span>
<span>(475 kJ) / (181 kJ / 2 mol HgO) x (216.5894 g HgO/mol) = 1137 g HgO </span>
<span>(179 kJ) / (181 kJ / 1 mol O2) x (31.99886 g O2/mol) = 31.6 g O2 </span>
<span>(145 kJ) / (112 kJ / 3 mol Cl2) x (70.9064 g Cl2/mol) = 275 g Cl2 </span>
<span>(14.5 g S2Cl2) / (135.0360 g S2Cl2/mol) x (112 kJ / 1 mol S2Cl2) = 12.0 kJ </span>
<span>CaCO3 + 2 NH3 → CaCN2 + 3 H2O; ∆H = –90.0 kJ </span>
<span>(798 kJ) / (90.0 kJ / 2 mol HN3) x (17.03056 g NH3/mol) = 302 g NH3 </span>
<span>(19.7 g H2O) / (18.01532 g H2O/mol) x (90.0 kJ / 3 mol H2O) = 32.8 kJ</span>
Answer:
TRIAL 1:
For “Event 0”, put 100 pennies in a large plastic or cardboard container.
For “Event 1”, shake the container 10 times. This represents a radioactive decay event.
Open the lid. Remove all the pennies that have turned up tails. Record the number removed.
Record the number of radioactive pennies remaining.
For “Event 2”, replace the lid and repeat steps 2 to 4.
Repeat for Events 3, 4, 5 … until no pennies remain in the container.
TRIAL 2:
Repeat Trial 1, starting anew with 100 pennies.
Calculate for each event the average number of radioactive pennies that remain after shaking.
Plot the average number of radioactive pennies after shaking vs. the Event Number. Start with Event 0, when all the pennies are radioactive. Estimate the half-life — the number of events required for half of the pennies to decay.
Explanation:
