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4 years ago

Which of the following is a technique used in relative dating?

Chemistry

2 answers:

olga_2 [115]4 years ago

6 0

Answer: Index fossil correlation

Relative dating refers to the age estimation method in which the order of the past events is determined to compare the age of an object, rock or organism with the respect to the other. This method gives the estimated or approximate age of the entity under consideration. It does not give the absolute age. Index fossils are the fossils which are found in a particular geological time. They have a short life span and wide geographical distribution. These fossils can be used to determine the relative age. As, these fossils can be compared with the layers of strata and rocks, where they are found, to know their approx age.

lesantik [10]4 years ago

6 0

Radiocarbon dating
By checking the amount of radioactive carbon in a sample, its relative age can be estimated.

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MaRussiya [10]

The answer is tsunami

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4 years ago

Consider the decomposition of a metal oxide to its elements, where M represents a generic metal.

VladimirAG [237]

Answer:

For 1: The standard Gibbs free energy change of the reaction is 10.60 kJ/mol

For 2: The equilibrium constant for the given reaction at 298 K is $1.386\times 10^{-2}$

For 3: The equilibrium pressure of oxygen gas is 0.0577 atm

Explanation:

For 1:

The equation used to calculate standard Gibbs free energy change of a reaction is:

$\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]$

For the given chemical reaction:

$M_2O_3(s)\rightarrow 2M(s)+\frac{3}{2}O_2(g)$

The equation for the standard Gibbs free energy change of the above reaction is:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(M(s))})+(\frac{3}{2}\times \Delta G^o_f_{(O_2(g))})]-[(1\times \Delta G^o_f_{(M_2O_3(s))})]$

We are given:

$\Delta G^o_f_{(M_2O_3(s))}=-10.60kJ/mol\\\Delta G^o_f_{(M(s))}=0kJ/mol\\\Delta G^o_f_{(O_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (0))+(\frac{3}{2}\times (0))]-[(1\times (-10.60))]\\\\\Delta G^o_{rxn}=10.60kJ/mol$

Hence, the standard Gibbs free energy change of the reaction is 10.60 kJ/mol

For 2:

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G^o=-RT\ln K_{eq}$

where,

$\Delta G^o$ = Standard Gibbs free energy = 10.60 kJ/mol = 10600 J/mol (Conversion factor: 1 kJ = 1000 J )

R = Gas constant = 8.314 J/K mol

T = temperature = 298 K

$K_{eq}$ = equilibrium constant = ?

Putting values in above equation, we get:

$10600J/mol=-(8.314J/Kmol)\times 298K\times \ln (K_{eq})\\\\K_{eq}=1.386\times 10^{-2}$

Hence, the equilibrium constant for the given reaction at 298 K is $1.386\times 10^{-2}$

For 3:

The expression of $K_{eq}$ for above equation follows:

$K_{eq}=p_{O_2}^{3/2}$

The concentration of pure solids and pure liquids are taken as 1 in the expression. That is why, the concentration of metal and metal oxide is taken as 1 in the expression.

Putting values in above expression, we get:

$1.386\times 10^{-2}=p_{O_2}^{3/2}\\\\p_{O_2}=0.0577atm$

Hence, the equilibrium pressure of oxygen gas is 0.0577 atm

6 0

3 years ago

6c.Calculate the maximum volume, in dm3, of chlorine gas at Stp that can be obtained from 23.4 tonnes of molten sodium chloride.

sweet [91]

Answer:

4.48×10⁶ dm³

Explanation:

We'll begin by converting 23.4 tonnes to grams (g). This can be obtained as follow:

1 tonne = 10⁶ g

Therefore,

23.4 tonnes = 23.4 × 10⁶

23.4 tonnes = 2.34×10⁷ g

Thus, 23.4 tonnes is equivalent to 2.34×10⁷ g

Next, we shall determine the number of mole in 2.34×10⁷ g of NaCl. This can be obtained as follow:

Mass NaCl = 2.34×10⁷ g

Molar mass of NaCl = 58.5 g/mol

Mole of NaCl =?

Mole = mass / molar mass

Mole of NaCl = 2.34×10⁷ / 58.5

Mole of NaCl = 4×10⁵ moles

Next, we shall determine the number of mole of chlorine, Cl₂ produced from the reaction. This can be obtained as follow:

2NaCl —> 2Na + Cl₂

From the balanced equation above,

2 moles of NaCl reacted to produce 1 mole of Cl₂.

Therefore, 4×10⁵ moles of NaCl will react to produce = (4×10⁵ × 1)/2 = 2×10⁵ moles of Cl₂.

Thus, 2×10⁵ moles of Cl₂ were obtained from the reaction.

Finally, we shall determine the volume of Cl₂ produced. This can be obtained as follow:

1 mole of Cl₂ at stp = 22.4 dm³

Therefore,

2×10⁵ moles of Cl₂ at stp = 2×10⁵ 22.4

2×10⁵ moles of Cl₂ at stp = 4.48×10⁶ dm³

Thus, the volume of chlorine obtained from the reaction is 4.48×10⁶ dm³

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