Which statement describes iron?

A wooden artifact from an ancient tomb contains 60% of the carbon-14 that is present in living trees. How long ago was the artif

noname [10]

Answer:

The age of the sample is 4224 years.

Explanation:

Let the age of the sample be t years old.

Initial mass percentage of carbon-14 in an artifact = 100%

Initial mass of carbon-14 in an artifact = $[A_o]$

Final mass percentage of carbon-14 in an artifact t years = 60%

Final mass of carbon-14 in an artifact = $[A]=0.06[A_o]$

Half life of the carbon-14 = $t_{1/2}=5730 years$

$k=\frac{0.693}{t_{1/2}}$

$[A]=[A_o]\times e^{-kt}$

$[A]=[A_o]\times e^{-\frac{0.693}{t_{1/2}}\times t}$

$0.60[A_o]=[A_o]\times e^{-\frac{0.693}{5730 year}\times t}$

Solving for t:

t = 4223.71 years ≈ 4224 years

The age of the sample is 4224 years.

8 0

4 years ago

Which of the following would release the most heat? Assume the same mass of in each case. Specific heats of ice, liquid water, a

lesya692 [45]

Answer:

The process which releases most heat is E)

Explanation:

As we know that water freezes at 0ºC and vaporizes at 100ºC, we calculate the heat as follows:

Processes with temperatures < 0ºC : by using specific heat of ice (Sh ice) multiplied by the change in temperature (ΔT= Final Temperature - Initial Temperature)⇒ Sh ice x ΔT

Processes of ice melting (at 0ºC): by using heat of fusion of ice (ΔH fus) multiplied by a conversor factor (1 mol H20= 18 g)⇒ ΔHfus x 1mol/18g

Processes between 0ºC and 100ºC: by using specific heat of liquid water (Sh liq) multiplied by change in temperature ⇒ Sh liq x ΔT

Processes of water evaporation (at 100ºC): by using heat of vaporization (ΔH vap) multiplied by the conversor factor ⇒ ΔH vap x 1mol/18 g

Processes at a temperature >100ºC: by using specific heat of water vapor (Sh vap) multiplied by the change in temperature ⇒ Sh vap x ΔT

A) Water at -25ºC is ice. Ice is heated from -25ºC to 0ºC, then it melts at 0ºC (ice became liquid water) and then liquid water is heated from 0ºC to 70ºC. T

This is the only process in with the heat is absorbed (not releases), so it cannot be the right answer, but we calculate the heat involved to practice:

Heat= (Sh ice x ΔT) + (ΔH fus x 1/18 g) + Sh liq x ΔT

Heat= (2.05 J/g ºC x (0ºC -(-25ºC) ) + (6.01 x 10³ J/mol x 1 mol/18 g) + (4.18 J/g ºC x (70ºC-0ºC)

Heat= 51.25 J + 333,8 J +292.6 J

Heat= 677.65 J (heat is absorbed)

B) Water is cooled from 13ºC to 0ºC, then it is freezed at 0ºC and then the ice is cooled from 0ºC to -2.6 ºC

Heat= (Sh liq x ΔT) + (-ΔH melt x 1/18 g) + (Sh ice x ΔT)

Heat= 4.18 J/ºC x (0ºC- 13ºC) + (-6.01 x 10³ J/mol x 1mol/18 g) + (2.05 J/ºC x (-2.5ºc - 0ºC)

Heat= -54.34 J - 333.8 J + 5.33 J

Heat= -393.47 J (heat is released)

C) Liquid water is cooled from 74ºC to 95ºC

Heat= Sh liq x ΔT

Heat= 4.18 J/ºC x (74ºC - 95ºC)

Heat = -87.78 J (heat is released)

D) Water at 140ºC is in vapor state. Vapor at 140ºC is cooled to 110ºC (still vapor).

Heat = Sh vap x ΔT

Heat= 2.01 J/ºC x (110ºC - 140ºC)

Heat= -60.3 J (heat is released)

E) Vapor at 106ºC is cooled to 100ºC, then it condenses at 100ºC (convertion from gas to liquid), and liquid water is cooled from 100ºC to 88ºC.

Heat= (Sh vap x ΔT) + (-ΔHvap x 1mol/18g) + (Sh liq x ΔT)

Heat= (2.01 J/ºC x (100ºC-106ºC)) - (40.7 x 10³ J/mol x 1mol/18 g) + (4.18 J/ºC x (88ºC -100ºC)

Heat= -2323.32 J (heat is released) <u>THIS IS THE RIGHT ANSWER</u> (the more negative= the more released)

7 0

3 years ago

How many moles are in 7.46 x 1025 particles of iron

sladkih [1.3K]

Answer:

1 mole of iron =6.023×10^23 particles

1 particles of iron=1/6.023×10^23 mole

7.46×10^25 particles =1/6.023×10^23×7.46×10^25

=1.238×10^48 mole is a required answer.

4 0

3 years ago

Use the data given below to construct a Born-Haber cycle to determine the heat of formation of KCl. Δ H°(kJ) K(s) → K(g) 89 K(g)

AURORKA [14]

Explanation:

The net equation will be as follows.

$K(s) + Cl_{2}(g) \rightarrow KCl(s)$

So, we are required to find $\Delta H_{formation}$ for this reaction.

Therefore, steps involved for the above process are as follows.

Step 1: Convert K from solid state to gaseous state

$K(s) \rightarrow K(g)$ , $\Delta H_{1}$ = 89 kJ

Step 2: Ionization of gaseous K

$K(g) \rightarrow K^{+}(g) + e^{-}$ , $H_{2}$ = 418 KJ

Step 3: Dissociation of $Cl_{2}$ gas into chlorine atom .

$\frac{1}{2} Cl_{2}(g) \rightarrow Cl(g)$ , $\Delta H_{3} = \frac{244}{2}$ = 122 KJ

Step 4: Iozination of chlorine atom.

$Cl(g) + e^{-} \rightarro Cl^{-}(g)$ , $H_{4}$ = -349 KJ

Step 5: Add $K^{+}$ ion and $Cl^{-}$ ion formed above to get KCl .

$K^{+}(g) + Cl^{-}(g) \rightarrow KCl(s)$ , $H_{5}$ = -717 KJ

Now, using Born-Haber cycle, value of enthalpy of the formation is calculated as follows.

$\Delta H_{f} = \DeltaH_{1} + \Delta H_{2} + \Delta H_{3} + \Delta H_{4} + \Delta H_{5}$

= 89 + 418 + 122 - 349 - 717

= - 437 KJ/mol

Thus, we can conclude that the heat of formation of KCl is - 437 KJ/mol.

5 0

4 years ago

Why does the boiling point of a liquid vary with atmospheric pressure?

tigry1 [53]

Atmospheric pressure decreases because air is less dense at higher altitudes. Because the atmospheric pressure is lower, the vapour pressure of the liquid needs to be lower to reach boiling point.

8 0

3 years ago