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3 years ago

All scientists try to base their conclusions on

2 answers:

6 0

<span>All scientists try to base their conclusions on the experiments they have conducted and they have repeated them to be sure of it. </span>

Inessa05 [86]3 years ago

3 0

The information drawn from completing lab test and proven data from tests and hypothesis

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A civil engineering student is considering buying an electric car. However, the conscious student wants to make sure her carbon

svetlana [45]

Answer:

$\text{An gasoline-powered vehicle has }\\\boxed{\text{five times the carbon footprint}}\text{ of an electric vehicle.}$

Explanation:

1. Miles travelled in an average month

$\text{Miles} = \text{1 mo} \times \dfrac{\text{52 wk}}{\text{12 mo}} \times \dfrac{\text{5 da}}{\text{1 wk}} \times \dfrac{\text{16 mi}}{\text{1 da}} = \text{347 mi}$

2. Using a gasoline powered vehicle

(a) Moles of heptane used

$n = \text{347 mi} \times \dfrac{\text{1 gal}}{\text{36.5 mi}}\times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{679.5 g}}{\text{1L}} \times \dfrac{ \text{1 mol}}{\text{100.20 g}}= \text{244 mol}$

(b) Equation for combustion

C₇H₁₆ + O₂ ⟶ 7CO₂ + 8H₂O

(c) Moles of CO₂ formed

$n = \text{244 mol heptane} \times \dfrac{\text{7 mol CO$_{2}$}}{\text{1 mol heptane}} = \text{1710 mol CO$_{2}$}$

(d) Volume of CO₂ formed

At 20 °C and 1 atm, the molar volume of a gas is 24.0 L.

$V = \text{1710 mol } \times \dfrac{\text{24.0 L}}{\text{1 mol}} \times \dfrac{\text{1 gal}}{\text{3.785 L}} = \textbf{10 800 gal}$

3. Using an electric vehicle

(a) Theoretical energy used

$\text{Theor. Energy} = \text{347 mi} \times \dfrac{\text{1 kWh}}{\text{5.2 mi}} = \text{66.7 kWh theor.}$

(b) Actual energy used

The power station is only 85 % efficient.

$\text{Actual energy used} = \text{66.7 kWh theor.} \times \dfrac{\text{100 kWh actual}}{\text{ 85 kWh theor.}}\times \dfrac{\text{3600 kJ}}{\text{1 kWh}}\\\\ = 2.82\times 10^{5} \text{ kJ}\$

(c) Combustion of CH₄

CH₄ + 2O₂ ⟶ CO₂ +2 H₂O

(d) Equivalent volume of CO₂

The heat of combustion of methane is -802.3 kJ·mol⁻¹

$V= 2.82\times 10^{5}\text{ kJ} \times \dfrac{\text{1 mol methane}}{\text{802.3 kJ}} \times \dfrac{\text{1 mol CO$_{2}$} }{\text{1 mol methane}} \times \dfrac{ \text{24.0 L}}{ \text{1 mol CO$_{2}$}}\\\\ \times \dfrac{\text{1 gal}}{\text{3.875 L}} = \textbf{2180 gal}$

4. Comparison

$\dfrac{V_{\text{gasoline}}}{V_{\text{electric}}} = \dfrac{10800}{2180} = 5.0\\\\ \text{An gasoline-powered vehicle has }\\ \boxed{\textbf{ five times the carbon footprint}}\text{ of an electric vehicle.}$

6 0

3 years ago

Suppose that, from measurements in a microscope, you determine that a certain layer of graphene covers an area of 1.50μm2. Conve

Alex73 [517]

The answer is 1.5*10^-12 square meters. For a detailed calculation, please refer to the attachment.

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7 0

3 years ago

The graph shows the volume of a gaseous product formed during two trials of a reaction. A different concentration of reactant wa

solniwko [45]

A: Trial 1, because the average rate of the reaction is lower.

The rate of reaction is the speed with which reactants are converted into products. It is also the rate at which reactants disappear and products appear. The higher the rate of reaction, the greater the amount of product formed in a reaction.

If we look at the graph, we will realize that trial 1 produces a lesser amount of product than trial 2. This implies that the average rate of the reaction in trial 1 is lower than in trial 2.

Lower average rate of reaction implies lower concentration of the reactants since the rate of reaction depends on the concentration of reactants.

Hence trial 1 has a lower concentration of reactants because the average rate of the reaction is lower.

8 0

3 years ago

Read 2 more answers

A 250ml glass of orange juice contains 22 grams of sugar. how much sugar is in a 2 liter (2.500ml) bottle of orange juice?

Maslowich

220 grams of sugar would be in 2 liters of orange juice

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3 years ago

The SI unit of pressure is the ______________.

steposvetlana [31]