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3 years ago

What process uses analysis of the decay of atomic particles to determine exact age?

Chemistry

2 answers:

padilas [110]3 years ago

4 0

Answer:

Radioactive dating

Explanation:

Radioactive dating measures the ratio of C¹²(normal) and C¹⁴ (radioactive) in fossils to determine the exact age of the specimen. C¹⁴ is unstable and has a half life of 5730 years, so as the organism ages the amount of C¹⁴ diminishes but the amount of C¹² remains constant. The ratio of C¹⁴ to C¹² is used to analyze the decay of atomic particles to determine exact age.

Lesechka [4]3 years ago

3 0

Answer:

Explanation:

Because carbon-14 and uranium-235 are used in radioactive dating.

Hope it helped!

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8. Consider the following reaction, for which ΔG° = −21.7 kJ/mol at 25°C: A2B(g) ↔ 2 A(g) + B(g) At the moment that 0.10 mol A2B

Mrac [35]

Answer:

(2) b, c, f

Explanation:

Hello,

In this case, for the undergoing chemical reaction:

$A_2B(g) \rightleftharpoons 2 A(g) + B(g)$

The Gibbs free energy of reaction at 25 °C is related with an equilibrium constant of:

$K=exp(-\frac{\Delta G^0 }{RT} )=exp[-\frac{-21700J/mol}{8.314J/(mol*K)*298.15K}]=1.58x10^{-4}$

Now, at the given moment, the reaction quotient turns out:

$Q=\frac{(\frac{0.10mol}{10.0L})^2(\frac{0.050mol}{10.0L})}{(\frac{0.10mol}{10.0L})} =5x10^{-5}$

Thus, since Q<K, the reaction will have too much reactants that will produce more products, shifting the reaction rightwards to them.

Hence, with the given information the statements b, c and f are correct, so the answer is: (2) b, c, f.

Best regards.

5 0

3 years ago

Find the density of a piece of chocolate with these measurements: 2.4 g and 5.12 mL

Dahasolnce [82]

Answer:

0.46875g/ml

Explanation:

Density(p) = m / v unit - g/ml or Kg/m^3

Given

mass = 2.4g

volume = 5.12ml

p = m / v

= 2.4g / 5.12ml

= 0.46875g/ml

5 0

2 years ago

Calculate the number of moles and the mass of the solute in each of the following solutions:

frosja888 [35]

Answer:

For a: The number of moles of KI are $2.7\times 10^{-5}$ and mass is $4.482\times 10^{-3}g$

For b: The number of moles of sulfuric acid are $1.65\times 10^{-5}$ and mass is $1.617\times 10^{-3}g$

For c: The number of moles of potassium chromate are $2.84\times 10^{-2}$ and mass is 5.51 g.

For d: The number of moles of ammonium sulfate are 39.018 moles and mass is 5155.84 grams.

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

To calculate the number of moles of a substance, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(2)

For a:

Molarity of KI = $8.23\times 10^{-5}M$

Volume of solution = 325 mL = 0.325 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$8.25\times 10^{-5}mol/L=\frac{\text{Moles of KI}}{0.325L}\\\\\text{Moles of KI}=2.7\times 10^{-5}mol$

Now, using equation 2, we get:

Moles of KI = $2.7\times 10^{-5}mol$

Molar mass of KI = 166 g/mol

Putting values in equation 2, we get:

$2.7\times 10^{-5}mol=\frac{\text{Mass of KI}}{166g/mol}\\\\\text{Mass of KI}=4.482\times 10^{-3}g$

Hence, the number of moles of KI are $2.7\times 10^{-5}$ and mass is $4.482\times 10^{-3}g$

For b:

Molarity of sulfuric acid = $22\times 10^{-5}M$

Volume of solution = 75 mL = 0.075 L

Putting values in equation 1, we get:

$22\times 10^{-5}mol/L=\frac{\text{Moles of sulfuric acid}}{0.075L}\\\\\text{Moles of }H_2SO_4=1.65\times 10^{-5}mol$

Now, using equation 2, we get:

Moles of sulfuric acid = $1.65\times 10^{-5}mol$

Molar mass of sulfuric acid = 98 g/mol

Putting values in equation 2, we get:

$1.65\times 10^{-5}mol=\frac{\text{Mass of }H_2SO_4}{98g/mol}\\\\\text{Mass of }H_2SO_4=1.617\times 10^{-3}g$

Hence, the number of moles of sulfuric acid are $1.65\times 10^{-5}$ and mass is $1.617\times 10^{-3}g$

For c:

Molarity of potassium chromate = $0.1135M$

Volume of solution = 0.250 L

Putting values in equation 1, we get:

$0.1135mol/L=\frac{\text{Moles of }K_2CrO_4}{0.250L}\\\\\text{Moles of }K_2CrO_4=2.84\times 10^{-2}mol$

Now, using equation 2, we get:

Moles of potassium chromate = $2.84\times 10^{-2}mol$

Molar mass of potassium chromate = 194.2 g/mol

Putting values in equation 2, we get:

$2.84\times 10^{-2}mol=\frac{\text{Mass of }K_2CrO_4}{194.2g/mol}\\\\\text{Mass of }K_2CrO_4=5.51g$

Hence, the number of moles of potassium chromate are $2.84\times 10^{-2}$ and mass is 5.51 g.

For d:

Molarity of ammonium sulfate = 3.716 M

Volume of solution = 10.5 L

Putting values in equation 1, we get:

$3.716mol/L=\frac{\text{Moles of }(NH_4)_2SO_4}{10.5L}\\\\\text{Moles of }(NH_4)_2SO_4=39.018mol$

Now, using equation 2, we get:

Moles of ammonium sulfate = 39.018 mol

Molar mass of ammonium sulfate = 132.14 g/mol

Putting values in equation 2, we get:

$39.018mol=\frac{\text{Mass of }(NH_4)_2SO_4}{132.14g/mol}\\\\\text{Mass of }(NH_4)_2SO_4=5155.84g$

Hence, the number of moles of ammonium sulfate are 39.018 moles and mass is 5155.84 grams.

3 0

3 years ago

Giving brainliest and thanks to best answer <333 :>

valkas [14]

Answer:

Explanation:

radiation from the sun first warms the outer atmosphere (trophosphere)

convection(aka just heat moving through gas or liquid) brings the warmth down lower

conduction heats the ground

gl lol :))

7 0

3 years ago

Read 2 more answers

Determine the standard enthalpy of formation in kJ/mol for NO given the following information about the formation of NO2 under s

Zarrin [17]

Answer:

90.3 kJ/mol

Explanation:

Let's consider the following thermochemical equation.

2 NO(g) + O₂(g) → 2 NO₂(g) ∆H°rxn = –114.2 kJ

We can find the standard enthalpy of formation for NO using the following expression.

∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × ΔH°f(O₂(g))

∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × 0 kJ/mol

∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g))

ΔH°f(NO(g)) = (2 mol × ΔH°f(NO₂(g)) - ∆H°rxn) / 2 mol

ΔH°f(NO(g)) = (2 mol × 33.2 kJ/mol + 114.2 kJ) / 2 mol

ΔH°f(NO(g)) = 90.3 kJ/mol

8 0

3 years ago