Recent content by clintyip

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    Unit Conversions: kJ/mol to kJ/litre

    Don't I just multiply it by 1000? So I get 31645.4 kJ/L. But the thing is, this substance is Hexane, and according to Wikipedia, it has a heat of combustion of 29.3 kJ/L. The two numbers aren't even remotely close.
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    Unit Conversions: kJ/mol to kJ/litre

    That means, I can multiply 4163kj/mol by 1/86.14g, which then I get 4163kJ/86.14g. Then I multiply this number by the density. So 4163kJ/86.14g x 0.6578g/mL which equals to 31.64 kJ/mL Is that right?
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    Unit Conversions: kJ/mol to kJ/litre

    But I don't have a number with mols on top o_o
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    Unit Conversions: kJ/mol to kJ/litre

    The density of this substance is 0.6548 g/mL.
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    Unit Conversions: kJ/mol to kJ/litre

    Let's say I have 4163 kJ/mol of energy coming from a substance with molar mass 86.14g/mol. How would I convert the 4163 kJ/mol of energy to kJ/litre? Thanks.
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    How Do You Convert Energy from J/kg to kJ/mol?

    Step One: 120g/mol => 0.12kg/mol Step Two: mol = mass/molecular weight = kg/0.12 = mol kg = 0.12mol So 55300kJ/0.12mol = 460833 kJ/mol Wow, that's a big number, did I do it right?
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    How Do You Convert Energy from J/kg to kJ/mol?

    So I have 55300 kJ/kg Assuming that my Molecular weight is 120g/mol... Yeah, I have no idea. It seems like there's a missing variable to me :S
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    How Do You Convert Energy from J/kg to kJ/mol?

    Well, this is actually a fuel, Kerosene, with a molecular weight of 120...
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    How Do You Convert Energy from J/kg to kJ/mol?

    so that means if I had 55300000 j/kg, that would be 55300 kJ/kg. How would I make it into kJ/mol... (sorry, I'm not very good at converting...)
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    How Do You Convert Energy from J/kg to kJ/mol?

    Let's say I have a value of: 553 x 10^5 j/kg How would I convert that to kJ/mol?
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    Distance Between Lenses for a Magnification of 10

    Homework Statement A convex lens and a concave lens both of 6cm focal length are placed to give an object which is 5cm from the convex lens of magnification of 10. How far are the lenses apart? Homework Equations 1/f = 1/u + 1/v Magnification=v/u The Attempt at a Solution Lens...
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    Question - Circular motion amusement park physics

    Definitely the most challenging one I've done yet. Thanks again!
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    Question - Circular motion amusement park physics

    When I check the validity of the 21 metres, I used Kinetic energy formulas - one for the gravitron itself, and the other one for the gravitational potential energy from the mass. For Gravitron- KE = 1/2 I (w)^2 KE = 0.5 x 19440 x 2.094^2 KE = 42620.6 Joules For Mass- GPE = mgh GPE =...
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    Question - Circular motion amusement park physics

    So, what I got from that is: a = r(alpha), which is 2x0.2094 = 0.4188m/s^2 S = ut + 1/2 at^2 u = 0, therefore cancel out ut S = 1/2 at^2 S = 0.5 x 0.4188 x 10^2 S = 20.94metres. Thing is, would a mass of 216.92kg that falls 20.94metres really be able to accelerate the gravitron of mass 1500kg...
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