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2 years ago

What is radioactive dating in chemistry? PLEASE HELP!!!

1 answer:

7 0

Radioactive dating uses the concept of half life to determine the age of something.
The general idea is that elements exist naturally as a combination of isotopes. Some isotopes are stable while others decay radioactively. The half-life of a radioactive isotope is the time required to reduce the initial quantity of the isotope by 50% through the process of radioactive decay.
Carbon 14 dating works this way. A tree growing in the past would have an equilibrium mixture of C-12 and C-14 atoms in proportions consistent with living matter today. We know this value.
When that tree dies, it no longer accumulates new carbon atoms from carbon dioxide in the atmosphere (through photosynthesis). At this point, the proportion of C-14 in the organic matter that was the tree decreases due to the decay of C-14 to an isotope of nitrogen.
When we dig up the remnants of the tree today, we can measure the proportion of C-14 to C-12 and see how much it has been reduced. From this we can calculate backwards using the half life of C-14 (5730 years) to estimate how old the remains of the tree are.

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Which plant characteristic is the least important with regard to attracting bats? (1 point)

Sav [38]

Answer:

O flower shape

Explanation:

Bats are just as important for pollination as insects and just like insects, plants use artifices to attract bats to their flowers, allowing them to have access to the pollen that will be taken elsewhere.

Bats are mainly attracted by the color and smell of flowers, the size and shape also play a role in attracting these animals, but the shape is less influential.

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3 years ago

Complete the chemical change. _______ + e - → Cl-

Sedaia [141]

You are left with a chloride ion, so you started off with a chlorine atom.
Cl + e- → Cl-

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2 years ago

Read 2 more answers

Is backbonding is possible in BF4- ? If no then why?

Rudik [331]

Answer:

No, because Flourine can only form 1 bond, thus backbonding is not obtainable

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2 years ago

Dominic made the table below to organize his notes about mixtures. A 1-column table. The first column labeled properties of mixt

ankoles [38]

Answer:

Exxplanation:

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2 years ago

Read 2 more answers

Determine Z and V for steam at 250°C and 1800 kPa by the following: (a) The truncated virial equation [Eq. (3.38)] with the foll

makvit [3.9K]

Answer:

Explanation:

Given that:

the temperature $T_1$ = 250 °C= ( 250+ 273.15 ) K = 523.15 K

Pressure = 1800 kPa

The truncated viral equation is expressed as:

$\frac{PV}{RT} = 1 + \frac{B}{V} + \frac{C}{V^2}$

where; B = - $152.5 \ cm^3 /mol$ C = -5800 $cm^6/mol^2$

R = 8.314 × 10³ cm³ kPa. K⁻¹.mol⁻¹

Plugging all our values; we have

$\frac{1800*V}{8.314*10^3*523.15} = 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

$4.138*10^{-4} \ V= 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

Multiplying through with V² ; we have

$4.138*10^4 \ V ^3 = V^2 - 152.5 V - 5800 = 0$

$4.138*10^4 \ V ^3 - V^2 + 152.5 V + 5800 = 0$

V = 2250.06 cm³ mol⁻¹

Z = $\frac{PV}{RT}$

Z = $\frac{1800*2250.06}{8.314*10^3*523.15}$

Z = 0.931

b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)].

The generalized Pitzer correlation is :

$T_c = 647.1 \ K \\ \\ P_c = 22055 \ kPa \\ \\ \omega = 0.345$

$T__{\gamma}} = \frac{T}{T_c}$

$T__{\gamma}} = \frac{523.15}{647.1}$

$T__{\gamma}} = 0.808$

$P__{\gamma}} = \frac{P}{P_c}$

$P__{\gamma}} = \frac{1800}{22055}$

$P__{\gamma}} = 0.0816$

$B_o = 0.083 - \frac{0.422}{T__{\gamma}}^{1.6}}$

$B_o = 0.083 - \frac{0.422}{0.808^{1.6}}$

$B_o = 0.51$

$B_1 = 0.139 - \frac{0.172}{T__{\gamma}}^{ \ 4.2}}$

$B_1 = -0.282$

The compressibility is calculated as:

$Z = 1+ (B_o + \omega B_1 ) \frac{P__{\gamma}}{T__{\gamma}}$

$Z = 1+ (-0.51 +(0.345* - 0.282) ) \frac{0.0816}{0.808}$

Z = 0.9386

$V= \frac{ZRT}{P}$

$V= \frac{0.9386*8.314*10^3*523.15}{1800}$

V = 2268.01 cm³ mol⁻¹

c) From the steam tables (App. E).

At $T_1 = 523.15 \ K \ and \ P = 1800 \ k Pa$

V = 0.1249 m³/ kg

M (molecular weight) = 18.015 gm/mol

V = 0.1249 × 10³ × 18.015

V = 2250.07 cm³/mol⁻¹

R = 729.77 J/kg.K

Z = $\frac{PV}{RT}$

Z = $\frac{1800*10^3 *0.1249}{729.77*523.15}$

Z = 0.588

3 0

2 years ago