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2 years ago

How can you use a fossil to indemnify the environment where the organism lived

1 answer:

8 0

Climate is one of the factors that determines where different species of plants and animals can live, so paleontologists look for clues to a location's ancient climate in the types of fossil plants and animals they find there. For example, no modern crocodile species lives in a climate with long periods of freezing temperatures, so scientists hypothesize that ancient crocodiles had the same requirement for year round warmth. That leads them to consider the 110-million-year-old crocodile fossils from the Washington, D.C. to be part of a large body of circumstantial evidence that temperatures there were warm year round during the Early Cretaceous. Similarly, coal beds and fossil trees in the Arctic Slope of Alaska are among the many clues that Alaskan temperatures were very warm during the Late Cretaceous.

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Given the following heats of combustion. CH3OH(l) + 3/2 O2(g) CO2(g) + 2 H2O(l) ΔH°rxn = -726.4 kJ C(graphite) + O2(g) CO2(g) ΔH

sergeinik [125]

Answer:

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation:

The formation reaction of CH_3OH will be,

$C(s)+2H_2(g)+\frac{1}{2}O_2\rightarrow CH_3OH(g),\Delta H_{formation}=?$

The intermediate balanced chemical reaction will be,

$C(graphite)+O_2(g)\rightarrow CO_2(g), \Delta H_1=-393.5kJ/mole$ ..[1]

$H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l), \Delta H_2=-285.8kJ/mole$ ..[2]

$CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l) , \Delta H_3=-726.4kJ/mole$ ..[3]

Now we will reverse the reaction 3, multiply reaction 2 by 2 then adding all the equations, Using Hess's law:

We get :

$C(graphite)+O_2(g)\rightarrow CO_2(g) , \Delta H_1=-393.5kJ/mole$ ..[1]

$2H_2(g)+2O_2(g)\rightarrow 2H_2O(l) ,\Delta H_2=2\times (-285.8kJ/mole)=-571.6kJ/mol$ ..[2]

$CO_2(g)+2H_2O(l)\rightarrow CH_3OH(g)+\frac{3}{2}O_2(g) ,\Delta H_3=726.4kJ/mole$ [3]

The expression for enthalpy of formation of $C_2H_4$ will be,

$\Delta H_{formation}=\Delta H_1+2\times \Delta H_2+\Delta H_3$

$\Delta H=(-393.5kJ/mole)+(-571.6kJ/mole)+(726.4kJ/mole)$

$\Delta H=-238.7kJ/mole$

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

4 0

2 years ago

When the name of an anion that is part of an acid ends in -ite, the acid name includes the suffix _____.

Marina86 [1]

(B) suffixous

Hope It Helps

3 0

2 years ago

Read 2 more answers

A scientist did an investigation on the growth of koi fish. He selected forty newly hatched koi from the same clutch of eggs. He

Luda [366]

D have a good day my dude

8 0

3 years ago

How much heat energy is needed to raise the temperature of 59.7g of cadmium from 25°C to 100°C? The specific heat of cadmium is

Ilia_Sergeevich [38]

Heat energy can be calculated by using the specific heat of a substance multiplying it to the mass of the sample and the change in temperature. It is expressed as:

Energy = mCΔT
Energy = 59.7 (0.231) (100-25)
Energy = 1034.30 J

3 0

3 years ago

A sample of g of pure aluminum metal is added to mL of M hydrochloric acid. The volume of hydrogen gas produced at standard temp

MArishka [77]

Answer:

V = 11.2L are produced

Explanation:

... Sample of 27g of pure aluminium, 3added to 333 mL of 3.0 M HCl..

Based on the chemical reaction:

2Al(s) + 6HCl(aq) → 2AlC₃(aq) + 3H₂(g)

Where 3 moles of hydrogen are produced when 6 moles of hydrochloric acid reacts with 2 moles of Al.

To solve this question, we need to determine limiting reactant converting each reactant to moles. With limiting reactant and the chemical reaction we can find moles of hydrogen and its volume at STP (T=273.15K; P=1atm), thus:

Moles Al-Molar mass: 26.98g/mol-:

27g * (1mol / 26.98g) = 1mol of Al

Moles HCl:

333mL = 0.333L * (3mol/L) = 1mol HCl

For a complete reaction of 1 mole of HCl are required:

1mol HCl * (2mol Al / 6mol HCl) = 0.333 moles of Al. As there is 1 mole of Al, Al is in excess and HCl is limiting reactant.

Moles of Hydrogen produced are:

1mol HCl * (3 moles H₂ / 6 mol HCl) = 0.5moles H₂ are produced.

Using ideal gas law:

PV = nRT

V = nRT/P

Where V is volume

n are moles: 0.5mol

R is gas constant: 0.082atmL/molK

T is absolute temperature: 273.15K

P is pressure: 1atm.

Solving for V:

V = 0.5mol*0.082atmL/molK*273.15K / 1atm

<h3>V = 11.2L are produced</h3>

3 0

2 years ago