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lesya [120]
3 years ago
9

Using your understanding of environmental science, make a hypothesis about how these six vehicles compare with regard to total l

ifetime greenhouse gas emissions. Predict which ranking reflects least CO2 emitted to most CO2 emitted. (Note that all responses will be marked as "correct" at this point.)
Chemistry
1 answer:
gtnhenbr [62]3 years ago
5 0

Answer:

Answer is option (d) (lowest to highest)

We have to rank the six vehicles according to the CO2 emission causing an impact on global warming. We know that burning of coal has the highest emission of all fossil fuel so EV powered by electricity from it will be the highest impact on global warming with highest CO2 emission, next is burning of gasoline, but the CO2 emission can be varied by the process of the burning in engines obviously a high-efficiency burning will have less impact than a regular standard engine. So, after ICEV standard will have high CO2 emission than the ICEV (high efficiency), now the next will be EV which is having its electricity supplied by a mix of resources, the production involves both fossil and non-fossil fuel which make this vehicle more environmentally friendly than the full fossil fuel-based. The ICEV with less driving has the second least emission because you are not burning the fuel like other vehicles because of not driving the vehicle. In the last will be the EV having powered by electricity of wind it will be the least emission almost zero because it using wind power to generate electricity. Hence arrangement comes like this, EV(wind), ICEV(less driving), EV (mix of sources), ICEV (high-efficiency), standard ICEV, EV (coal) from lowest to highest CO2 emission.

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Given stock solutions of glucose (1M), asparagine (100 mM), and NaH2PO4 (50 mM), how much of each solution would you need to pre
WITCHER [35]

Answer:

25 mL,50 mL, 20 mL

Explanation:

Molarity = numbers of mole / volume in liters

Glucose

stock solutions concentration = 1M

concentration needed = 0.05 M

volume of the preparation = 500ml = 500 / 1000 = 0.5l

number of moles of glucose in the solution = Molarity × molar mass = 0.05 × 0.5 L = 0.025 moles

volume of  the stock needed = number of moles / molarity of the stock solution = 0.025 moles / 1 M = 0.025L = 25 mL

Asparagine

Molarity of the stock solution = 100mM = 0.1 M

Molarity needed = 10mM = 0.01 M

volume of the prepared solution = 0.5 L

number of moles in the prepared solution = 0.01 M × 0.5 L = 0.005 moles

volume of the stock solution needed = number of moles of the prepared solution / Molarity of the stock = 0.005 / 0.1 = 0.05L = 50 mL

NaH₂PO₄

Molarity of the stock solution = 50 mM = 0.05 M

Molarity of the needed = 2 mM = 0.002 M

Volume of the prepared solution = 0.5 L

number of moles in the prepared solution = 0.002 M × 0.5 L = 0.002 × 0.5 L = 0.001 moles

volume of the stock needed = 0.001 moles / 0.05 M = 0.02 L = 20 mL

4 0
3 years ago
If a buffer solution is 0.120 M in a weak acid (Ka = 2.0 × 10-5) and 0.440 M in its conjugate base, what is the pH?
Helen [10]
-log ka=pH - log ([A-]/[HA])

3 0
3 years ago
In h2o, the type of bond that holds one of the hydrogen atoms to the oxygen atom is a
Murrr4er [49]
The hydrogen and oxygen<span> atoms from H</span>₂O are <span>bonded together through covalent </span>bonding. 
3 0
3 years ago
7.
pickupchik [31]

Answer:

higher infant mortality

Explanation:

3 0
3 years ago
Calculate the volume in milliliters of a iron(II) bromide solution that contains of iron(II) bromide . Round your answer to sign
miv72 [106K]

This is an incomplete question, here is a complete question.

Calculate the volume in milliliters of a 1.29 mol/L iron(II) bromide solution that contains 275 mmol of iron(II) bromide . Round your answer to significant 3 digits.

Answer  : The volume of iron(II) bromide solution is, 2.13\times 10^2mL

Explanation : Given,

Concentration of iron(II) bromide = 1.29 mo/L

Moles of iron(II) bromide = 275 mmol = 0.275 mol

conversion used : 1 mmol = 0.001 mol

Now we have to calculate the volume of iron(II) bromide.

\text{Volume of iron(II) bromide}=\frac{Moles of iron(II) bromide}}{\text{Concentration of iron(II) bromide}}

Now put all the given values in this formula, we get:

\text{Volume of iron(II) bromide}=\frac{0.275mol}{1.29mol/L}=0.213L=2.13\times 10^2mL

Thus, the volume of iron(II) bromide solution is, 2.13\times 10^2mL

5 0
3 years ago
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