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Sauron [17]
3 years ago
15

A wooden boat discovered just south of the great pyramid in egypt has a carbon-14>carbon-12 ratio that is 72.5% of that found

in living organisms. how old is the boat
Chemistry
1 answer:
Kazeer [188]3 years ago
7 0
The half life for C14 is 5730 years.

We assume that Carbon 14/ Carbon 12 ratio was steady for living organisms over time, the problem is actually telling us that
\frac{Nt}{N0} = 0.0725 = \frac{1}{2}ˣ

Take the natural logarithm and In on both sides.
ln(0.725) = ln\frac{1}{2}ˣ
= - 0.3216 = xln (\frac{1}{2} = -0.6931x.
So x = (-.3216) / (-0.6931) = 0.464
or

t/t₁/₂ = 0.464
So t = 0.464 x t₁/₂ = 0.464 * 5730 yrs = 2660 years.
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Which were Lenin's actions before and during the Russian Revolution? Check all that apply.
Stolb23 [73]

Answer:

He supported the ideology of Marxism.

He opposed the tsar and was exiled.

He led the Bolsheviks.

Explanation:

Lenin was Russian Revolution political theorist. He served as first founding head of government of Soviet Russia. He supported the ideology of Marxism and opposed Tsar.

7 0
3 years ago
How many grams of FeCl 3 are in 250. mL of a 0.100 M solution?
Naddika [18.5K]

Answer:

The correct answer is option B

Explanation:

$Molarity=\frac{Weight}{Molecular \,weight} \frac{1000}{V(in \, ml)}

Given values,

Molarity of $FeCl_3=0.100M$

Volume of solution, $V=250ml$

Molecular weight of $FeCl_3=162.2$

Substituting this values in Molarity formula, we get

$0.1=\frac{weight}{162.2} \times\frac{1000}{250} $\\$\Rightarrow 16.22=weight\times4$\\$\Rightarrow weight=\frac{16.22}{4} $\\$\therefore weight=4.06g$

5 0
3 years ago
The standard enthalpy of formation for glucose [c6h12o6(s)] is −1273.3 kj/mol. what is the correct formation equation correspond
balu736 [363]
The standard formation equation for glucose C6H12O6(s) that corresponds to the standard enthalpy of formation or enthalpy change ΔH°f = -1273.3 kJ/mol is 
     C(s) + H2(g) + O2(g) → C6H12O6(s)
and the balanced chemical equation is 
     6C(s) + 6H2(g) + 3O2(g) → C6H12O6(s)

Using the equation for the standard enthalpy change of formation 
     ΔHoreaction = ∑ΔHof(products)−∑ΔHof(Reactants)
     ΔHoreaction = ΔHfo[C6H12O6(s)] - {ΔHfo[C(s, graphite) + ΔHfo[H2(g)] + ΔHfo[O2(g)]}

C(s), H2(g), and O2(g) each have a standard enthalpy of formation equal to 0 since they are in their most stable forms:
     ΔHoreaction = [1*-1273.3] - [(6*0) + (6*0) + (3*0)]
                           = -1273.3 - (0 + 0 + 0)
                           = -1273.3
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