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3 years ago

Which shows the correct order of steps in the formation of coal?

Chemistry

2 answers:

tamaranim1 [39]3 years ago

8 0

Answer:

it is Dead plants fall in swamp, Layers of dead plants pile up, Plant matter is buried by rocks and dirt, Plant matter is compressed and is exposed to heat, and Coal forms.

if iam right please make this into brainlest

EleoNora [17]3 years ago

5 0

Answer:

it is Dead plants fall in swamp, Layers of dead plants pile up, Plant matter is buried by rocks and dirt, Plant matter is compressed and is exposed to heat, and Coal forms.

Explanation:

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¿Cuál de las siguientes NO es una característica de los ácidos?

Serga [27]

Answer: B

Explanation: La respuesta es b porque su electricidad y no es un ácido.

4 0

3 years ago

A solution containing 8.3 g of a nonvolatile, nondissociating substance dissolved in 1.00 mol of chloroform, CHCl3, has a vapor

Margaret [11]

Answer:

a. Xm = 0.0229

b. 0.0234 moles

c. 354.1 g/mol

Explanation:

ΔP = P° . Xm

ΔP = P° - P', where P° is vapor pressure of pure solvent and P', vapor pressure of solution-

This is the formula for lowering vapor pressure.

If we apply the data given: 523 Torr - 511 Torr = 523 . Xm

Xm = ( 523 Torr - 511 Torr) / 523 Torr → 0.0229

Xm = Mole fraction of solute → Moles of solute / Total moles (sv + solute)

We can make this equation to determine moles of solute

0.0229 = Moles of solute / Moles of solute + 1

0.0229 (Moles of solute + 1) = Moles of solute

0.0229 = Moles of solute - 0.0229 moles of solute

0.0229 = 0.9771 moles of solute → 0.0229 / 0.9971 = 0.0234 moles

Molecular mass of solute → g/mol → 8.3 g / 0.0234 mol = 354.1 g/mol

7 0

3 years ago

What are the two different Energetics that relate to Environmental Science

Taya2010 [7]

Answer:

Organisms inhabit nearly every environment on Earth, from hot vents deep in the ocean floor to the icy reaches of the Arctic. Each environment offers both resources and constraints that shape the appearance of the species that inhabit it, and the strategies these species use to survive and reproduce. Some of the broadest patterns of environmental difference arise from the way our planet orbits the Sun and the resulting global distribution of sunlight (Chapin et al. 2002).

Explanation:

In the tropics, where solar radiation is plentiful year-round, temperatures are warm, and plants may photosynthesize continuously as long as water and nutrients are available. In polar regions, where solar radiation is seasonally limited, mean temperatures are much lower, and organisms must cope with extended periods when photosynthesis ceases.

3 0

3 years ago

Does this make a reaction? Because both products are soluble.

Crank

Answer:

Yes.

Explanation:

It will react. Because they are compounds

3 0

3 years ago

i am begging anyone to help me with this! (all tutors i've asked said they can't solve it but i need someone to help me out) - i

9966 [12]

First, we need to calculate how much energy we will get from this combustion.

Assuming the combustion is complete, we have the octane reacting with O₂ to form only water and CO₂, so:

$C_8H_{18}+O_2\to CO_2+H_2O$

We need to balance the reaction. Carbon only appear on two parts, so, we can start by it:

$C_8H_{18}+O_2\to8CO_2+H_2O$

Now, we balance the hydrogen:

$C_8H_{18}+O_2\to8CO_2+9H_2O$

And in the end, the oxygen:

$C_8H_{18}+\frac{25}{2}O_2\to8CO_2+9H_2O$

We can multiply all coefficients by 2 to get integer ones:

$2C_8H_{18}+25O_2\to16CO_2+18H_2O$

Now, we need to use the enthalpies of formation to get the enthalpy of reaction of this reaction.

The enthalpy of reaction can be calculated by adding the enthalpies of formation of the products multiplied by their stoichiometric coefficients and substracting the sum of enthalpies of formation of the reactants multiplied by their stoichiometric coefficients.

For the reactants, we have (the enthalpy of formation of pure compounds is zero, which is the case for O₂):

$\begin{gathered} \Delta H\mleft\lbrace reactants\mright\rbrace=2\cdot\Delta H\mleft\lbrace C_8H_{18}\mright\rbrace+25\cdot\Delta H\mleft\lbrace O_2\mright\rbrace \\ \Delta H\lbrace reactants\rbrace=2\cdot(-250.1kJ)+25\cdot0kJ \\ \Delta H\lbrace reactants\rbrace=-500.2kJ+0kJ \\ \Delta H\lbrace reactants\rbrace=-500.2kJ \end{gathered}$

For the products, we have:

$\begin{gathered} \Delta H_{}\mleft\lbrace product\mright\rbrace=16\cdot\Delta H\lbrace CO_2\rbrace+18\cdot\Delta H\lbrace H_2O\rbrace \\ \Delta H_{}\lbrace product\rbrace=16\cdot(-393.5kJ)+18\cdot(-285.5kJ) \\ \Delta H_{}\lbrace product\rbrace=-6296kJ-5139kJ \\ \Delta H_{}\lbrace product\rbrace=-11435kJ \end{gathered}$

Now, we substract the rectants from the produtcs:

$\begin{gathered} \Delta H_r=\Delta H_{}\lbrace product\rbrace-\Delta H\lbrace reactants\rbrace \\ \Delta H_r=-11435kJ-(-500.2kJ) \\ \Delta H_r=-10934.8kJ \end{gathered}$

Now, this enthalpy of reaction is for 2 moles of C₈H₁₈, so for 1 mol of C₈H₁₈ we have half this value:

$\Delta H_c=\frac{1}{2}\Delta H_r=\frac{1}{2}\cdot(-10934.8kJ)=-5467.4kJ$

Now, we have 100 g of C₈H₁₈, and its molar weight is approximately 114.22852 g/mol, so the number of moles in 100 g of C₈H₁₈ is:

$\begin{gathered} M_{C_8H_{18}}=\frac{m_{C_8H_{18}}}{n_{C_8H_{18}}} \\ n_{C_8H_{18}}=\frac{m_{C_8H_{18}}}{M_{C_8H_{18}}}=\frac{100g}{114.22852g/mol}\approx0.875438mol \end{gathered}$

Since we have approximately 0.875438 mol, and 1 mol releases -5467.4kJ when combusted, we have:

$Q=-5467.4kJ/mol\cdot0.875438mol\approx-4786.37kJ$

Now, for the other part, we need to calculate how much heat it is necessary to melt a mass, <em>m</em>.

First, we have to heat the ice to 0 °C, so:

$\begin{gathered} Q_1=m\cdot2.010J/g.\degree C\cdot(0-(-10))\degree C \\ Q_1=m\cdot2.010J/g\cdot10 \\ Q_1=m\cdot20.10J/g \end{gathered}$

Then, we need to melt all this mass, so we use the latent heat now:

$Q_2=n\cdot6.03kJ/mol$

Converting mass to number of moles of water we have:

$\begin{gathered} M=\frac{m}{n} \\ n=\frac{m}{M}=\frac{m}{18.01528g/mol} \end{gathered}$

So:

$Q_2=\frac{m}{18.01528g/mol}_{}\cdot6.03kJ/mol\approx m\cdot0.334716kJ/g$

Adding them, we have a total heat of:

$\begin{gathered} Q_T=m\cdot20.10J/g+m\cdot0.334716kJ/g \\ Q_T=m\cdot0.02010kJ/g+m\cdot0.334716kJ/g \\ Q_T=m\cdot0.354816kJ/g \end{gathered}$

Since we have a heat of 4786.37 kJ form the combustion, we input that to get the mass (the negative sign is removed because it only means that the heat is released from the reaction, but now it is absorbed by the ice):

$\begin{gathered} 4786.37kJ=m\cdot0.354816kJ/g \\ m=\frac{4786.37kJ}{0.354816kJ/g}\approx13489g\approx13.5\operatorname{kg} \end{gathered}$

Since we have a total of 20kg of ice, we can clculate the percent using it:

$P=\frac{13.5\operatorname{kg}}{20\operatorname{kg}}=0.675=67.5\%$

5 0

11 months ago