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3 years ago

Over the past 150 years, how has fossil fuel consumption affected the carbon dioxide concentrations in the atmosphere?

2 answers:

6 0

Hello!

Fossil fuels are Carbon-based compounds that when heated in the presence of oxygen undergo a combustion reaction to produce thermal energy and gases. The chemical equation for the combustion of a generic Fossil Fuel (C) is the following:

C + O₂ →CO₂

From this chemical reaction, we can see that CO₂ (Carbon Dioxide) is produced, so the consumption of fossil fuels has increased the Carbon Dioxide concentrations in the atmosphere.

Have a nice day!

jolli1 [7]3 years ago

6 0

Answer: Increased fossil fuel use has increased atmospheric carbon dioxide concentrations.

Explanation: Gradpoint

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A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli

Diano4ka-milaya [45]

Answer:

$\large \boxed{109.17 \, ^{\circ}\text{C}}$

Explanation:

Data:

50/50 ethylene glycol (EG):water

V = 4.70 gal

ρ(EG) = 1.11 g/mL

ρ(water) = 0.988 g/mL

Calculations:

The formula for the boiling point elevation ΔTb is

$\Delta T_{b} = iK_{b}b$

i is the van’t Hoff factor — the number of moles of particles you get from 1 mol of solute. For EG, i = 1.

1. Moles of EG

$\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}$

2. Kilograms of water

$m = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{998 g}}{\text{1 L}} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{8.88 kg}$

3. Molal concentration of EG

$b = \dfrac{\text{159 mol}}{\text{8.88 kg}} = \text{17.9 mol/kg}$

4. Increase in boiling point

$\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 17.9 \cdot mol \cdot kg^{-1} = 9.17 \, ^{\circ}\text{C}$

5. Boiling point

$\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 9.17 \, ^{\circ}\text{C} = \mathbf{109.17 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}$

7 0

3 years ago

How does 0.5 m sucrose (molecular mass 342) solution compare to 0.5 m glucose (molecular mass 180) solution?

mash [69]

Answer : Both solutions contain $3.011 X 10^{23}$ molecules.

Explanation : The number of molecules of 0.5 M of sucrose is equal to the number of molecules in 0.5 M of glucose. Both solutions contain $3.011 x 10^{23}$ molecules.

Avogadro's Number is $N_{A}$ = $6.022 X 10^{23}$ which represents particles per mole and particles may be typically molecules, atoms, ions, electrons, etc.

Here, only molarity values are given; where molarity is a measurement of concentration in terms of moles of the solute per liter of solvent.

Since each substance has the same concentration, 0.5 M, each will have the same number of molecules present per liter of solution.

Addition of molar mass for individual substance is not needed. As if both are considered in 1 Liter they would have same moles which is 0.5.

We can calculate the number of molecules for each;

Number of molecules = $N_{A} X M$ ;

∴ Number of molecules = $6.022 X 10^{23} X 0.5 mol/L X 1 L$ which will be = $3.011 X 10^{23}$

Thus, these solutions compare to each other in that they have not only the same concentration, but they will have the same number of solvated sugar molecules. But the mass of glucose dissolved will be less than the mass of sucrose.

7 0

3 years ago

Read 2 more answers

8. How many grams are there in 5 moles of lithium?

nikitadnepr [17]

Answer: m = n·M = 34.7 g

Explanation:

M(Li) = 6.941 g/mol, n = 5 mol

6 0

3 years ago

Consider this chemical reaction, where moving from left to right represents moving forward in time. A five panel comic strip. In

yulyashka [42]

Answer:

The reaction reaches equilibrium at the fourth panel.

Explanation:

Chemical Equilibrium is achieved when the overall properties the system seem to be constant, that is, stop changing.

Although, for chemical equilibrium, the right term for this equilibrium is dynamic equilibrium; the rate of forward reaction balances the rate of backward reaction, but concentrations can keep changing.

The point where equilibrium is achieved is when exactly when we reach the panel where the spheres that make up this panel is the same as the next panel and the next, that is, the specific colour and number of spheres start to become unchanged.

And from the description given in the question,

- In the first panel, there are ten large red spheres.

- In the second panel, there are 8 large red spheres and two small blue spheres.

- In the third panel, there are six large red spheres and four small blue spheres.

- In the fourth panel, there are four large red spheres and six small blue spheres.

- In the fifth panel, there are four large red spheres and six small blue spheres.

It is evident that the make-up of the spheres have become the same as at the fourth and fifth panel. This means that the first point where this final configuration of spheres first appeared is the fourth panel.

Hence, equilibrium is first reached at the fourth panel.

Hope this Helps!!!

6 0

3 years ago

Pls help me with this

Zina [86]

Answer:

$[I_2]=[Br]=0.31M$

Explanation:

Hello there!

In this case, according to the given information, it is possible for us to set up the following chemical equation at equilibrium:

$I_2+Br_2\rightleftharpoons 2IBr$

Now, we can set up the equilibrium expression in terms of x (reaction extent) whereas the initial concentration of both iodine and bromine is 0.5mol/0.250L=2.0M:

$K=\frac{[IBr]^2}{[I_2][Br_2]} \\\\1.2x10^2=\frac{(2x)^2}{(2.0-x)^2}$

Thus, we solve for x as show below:

$\sqrt{1.2x10^2} =\sqrt{\frac{(2x)^2}{(2.0-x)^2}} \\\\10.95=\frac{2x}{2.0-x}\\\\21.91-10.95x=2x\\\\21.91=12.95x\\\\x=\frac{21.91}{12.95} \\\\x=1.69M$

Therefore, the concentrations of both bromine and iodine are:

$[I_2]=[Br]=2.0M-1.69M=0.31M$

Regards!

8 0

3 years ago