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3 years ago

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Relative Dating Using Index Fossils

1 answer:

3241004551 [841]3 years ago

3 0

Index fossils can help geologists compare rock Layers at Distance locations.

Index fossils help identify where Unconformities may have occurred.

Then that layer can be Compared to other locations.

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A 60g sample of tetraethyl lead, a gasoline additive, is found to contain 38.43g lead, 17.83g carbon and 3.74g hydrogen. Determi

anastassius [24]

Answer:

Empirical formula (which matches the molecular formula) is = PbC₈H₂₀

Explanation:

Our sample: 60 g of tetraethyl lead

In order to determine the compound empirical formula we need the centesimal composition:

(Mass of element / Total mass) . 100 =

(38.43 g lead / 60g ) . 100 = 64.05%

(17.83 g C / 60g) . 100 = 29.72%

(3.74 g H / 60g) . 100 = 6.23 %

These % are the mass of the elements in 100 g of compound. Let's find out the moles of them:

64.05 g / 207.2 g/mol = 0.309 moles

29.72 g / 12 g/mol = 2.48 moles

6.23 g/ 1 g/mol = 6.23 moles

Next, we divide the moles, by the lowest value of them (0.309)

0.309 / 0.309 = 1 mol Pb

2.48 / 0.309 = 8 mol C

6.23 / 0.309 = 20 mol H

There, we have our formula PbC₈H₂₀

6 0

3 years ago

Which of the following elements belong to the same pair?

butalik [34]

Answer:

B. Na and Li

Both are group I elements.

5 0

3 years ago

The name of an alkyl group that contains THREE carbon atoms is____

tester [92]

Answer:

The answer is "Propyl"

Explanation:

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3 years ago

An atom's valence electrons are located in the atom's outermost energy level, true or false?

Sedaia [141]

Answer: The answer is True

Explanation: I hope my answer helps :)

3 0

3 years ago

Read 2 more answers

Given these reactions, where X represents a generic metal or metalloid 1 ) H 2 ( g ) + 1 2 O 2 ( g ) ⟶ H 2 O ( g ) Δ H 1 = − 241

anygoal [31]

Answer : The enthalpy of the given reaction will be, -1048.6 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

The main reaction is:

$XCl_4(s)+2H_2O(l)\rightarrow XO_2(s)+4HCl(g)$ $\Delta H=?$

The intermediate balanced chemical reactions are:

(1) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(g)$ $\Delta H_1=-241.8kJ$

(2) $X(s)+2Cl_2(g)\rightarrow XCl_4(s)$ $\Delta H_2=+461.9kJ$

(3) $\frac{1}{2}H_2(g)+\frac{1}{2}Cl_2(g)\rightarrow HCl(g)$ $\Delta H_3=-92.3kJ$

(4) $X(s)+O_2(g)\rightarrow XO_2(s)$ $\Delta H_4=-789.1kJ$

(5) $H_2O(g)\rightarrow H_2O(l)$ $\Delta H_5=-44.0kJ$

Now reversing reaction 2, multiplying reaction 3 by 4, reversing reaction 1 and multiplying by 2, reversing reaction 5 and multiplying by 2 and then adding all the equations, we get :

(1) $2H_2O(g)\rightarrow 2H_2(g)+O_2(g)$ $\Delta H_1=2\times 241.8kJ=483.6kJ$

(2) $XCl_4(s)\rightarrow X(s)+2Cl_2(g)$ $\Delta H_2=-461.9kJ$

(3) $2H_2(g)+2Cl_2(g)\rightarrow 4HCl(g)$ $\Delta H_3=4\times -92.3kJ=-369.2kJ$

(4) $X(s)+O_2(g)\rightarrow XO_2(s)$ $\Delta H_4=-789.1kJ$

(5) $2H_2O(l)\rightarrow 2H_2O(g)$ $\Delta H_5=2\times 44.0kJ=88.0kJ$

The expression for enthalpy of main reaction will be:

$\Delta H=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4+\Delta H_5$

$\Delta H=(483.6)+(-461.9)+(-369.2)+(-789.1)+(88.0)$

$\Delta H=-1048.6kJ$

Therefore, the enthalpy of the given reaction will be, -1048.6 kJ

4 0

3 years ago