John W replied: "As an electrical engineer, I love electric cars but I don't think it's the right answer. It may become part of the answer in the long run but I am against the mass promotion of electric or fuel cell vehicles. Think of it this way, the concept of using electric vehicles is to generate energy centrally from clean sources and to store the energy chemically for use in vehicles for transportation. The concept of using a hydrogen fuel cell car is to generate energy centrally from clean sources and to store the energy chemically for use in vehicles for transportation. The current concept of gasoline cars is to refine chemically stored energy collected by nature over millions of years and to use this chemically stored energy in vehicles for transportation. The only difference is that we are just finding the fossil fuels instead of making them. It's a bit counter-intuitive for most people but hydrocarbon fuels such as gasoline and diesel are no different than hydrogen and batteries except that gasoline and diesel have much higher volumetric energy densities and we already have the infrastructure and vehicles to use them. Also, just as we can make hydrogen from H2O, we can make gasoline from CO2 and H2O. Sandia National Labs has done this, they were researching more efficient ways to produce hydrogen from H2O and concluded that it made more sense to continue the process and synthesize liquid hydrocarbon fuels for use in existing infrastructure and vehicles. There's already 1 billion gas and diesel powered vehicles out there. If we wanted to benefit the environment, we would be simply changing how we make gasoline and diesel such that they become sustainable and perhaps even carbon negative. Promoting electric and fuel cell vehicles are just a matter of increasing consumerism so that manufacturers can sustain a growth rate that's higher than what's justified by population growth and end of life replacement alone. Indeed an artificially inflated economic growth rate has been sustained via designed obsolescence such as model years, and tail fins. There's probably more of an economic boost from electric and hydrogen vehicles but this is at the cost of increased consumerism and hence increased environmental impact. Hopefully there will be more electric vehicles in the future but there's no need to switch en masse to them. We already have a very good battery called gasoline and diesel, we just need to produce them from clean energy sources instead of drilling for them."

Megan W replied: "This can be solved by using a system of equations. Let x represent the percentage of the smaller isotope, and y represent the percent of the larger isotope. (I'll keep the percents as decimals to start). The two percentages must add up to 100%, or simply 1 in the case of decimals. This makes the first equation: x+y = 1 The atomic mass of an atom is the weighted average of its isotopes, in which the fractions of each isotope are multiplied by the isotope mass and then added together to give the official mass. For example, say element Z has two isotopes, one with a weight of 4 and the other with a weight of 5. 60% (.6) has weight 4 and thus 40% (.4) has weight five. The weighted average gives: (.6*4) + (.4*5) = 4.4 4.4 would be elements Z's official atomic weight. Now back to the original equation. The same concept applies for lithium. X and y, multiplied by their respective isotope masses, will result in the official atomic mass given. This gives the equation: 6.10512x + 7.01600y = 6.941 This can now be solved as a system of equations. The two equations are: 6.10512x + 7.01600y = 6.941 x+y = 1 Upon solving these two equations, we get x = .0823 and y = .9177. Therefore: 8.23% of lithium has the atomic mass 6.10512 91.77% of lithium has the atomic mass of 7.01600 Hope it helps!"

klinky86 replied: "you can use algebra to do this question. with the info above, you can set up the equation: (6.0151x + 7.0160y)/(x + y) = 6.941 Where x is number of Li 6 and y is number of Li 7, x + y is total number. you can move x+y to the other side by multiplying both sides by x+y, then put all the x values on the left, and y values on the right. after that you can get the value for y as a function of x, and simply substitute for y. After you solve for the values of x and y, their abundances are correlated to x/(x+y) and y/(x+y), converted to percent."

smilam replied: "Since 6.0151 (x) + 7.0160 (1-x) = 6.941 So we get 6.0151x - 7.0160x = 6.941 -7.0160 1.0009x = 0.075 x = .0749 So the 6.0151 amu has a 7.49 % abundance and the 7.0160 amu has a 92.51 % abundance"

Dr.A replied: "let x = % Li-6 abundance Li-7 = 100 -x 6.941 = 6.0151 x + 7.0160 ( 100-x)/ 100 694.1 = 6.0151 x + 701.60 - 7.0160 x - 7.5 = - 1.0009 x x = 7.49 % = Li-6 100 - 7.49=92.51 %= Li-7"