# how do I find 0.74 kcal/min to cal/sec using deminsional analysis when I don’t have the conversion for kcal to cal?

how do I find 0.74 kcal/min to cal/sec using deminsional analysis when I don’t have the conversion for kcal to cal?​

Inot sure what it is i sorry : ((

i’m pretty sure its b i might be wrong though!

The answer to your question really depends on three things: how much water you have, how cold it is to start out, and how cold the things around it are. water actually freezes when it gets to 32 degrees fahrenheit (0 degrees celsius), but the time it takes to get there may be different.  let’s start with the first. if you take two glasses, and fill one with a tiny bit of water, and the other about halfway, then put them both in the freezer, the one with less water will freeze first (you can try this at home, but i recommend using plastic cups and not glass ones).  now let’s move on to the second part. let’s say you have two glasses, and you fill one with really cold water that has been in the refrigerator, and the other with really hot water from the sink. if you put both of them in the freezer, the one that started out colder will freeze first.  for the third part, let’s imagine that you have two glasses with the same amount of water in them, and the water is at the same temperature. imagine putting one outside on a really really cold day in georgia, and having a friend in alaska put one outside on the same day. since it would be so much colder in alaska, the glass of water there would freeze before yours.  so, if you took a tiny bit of really cold water in a glass, and put it outside on a cold day in alaska, it would freeze a lot faster than a big glass of hot water outside on a cold day in georgia.

Also Read :   How many moles of pbr3 contain 3.68 ã 1025 bromine atoms?

answer is: empirical formula is tl₂o. if we use 100 grams of compound, tha: m(tl) = 96.2%  · 100 g ÷ 100%. m(tl) = 96.2 g. n(tl) = m(tl)  ÷ m(tl). n(tl) = 96.2 g  ÷ 204.4 g/mol. n(tl) = 0.47 mol. m(o) = 3.77 g. n(o) = m(o)  ÷ m(o). n(o) = 3.77 g  ÷ 16 g/mol. n(ag) = 0.235 mol. n(tl) . n(o) = 0.47 mol : 0.235 mol /0.235. n(tl) . n(o) = 2 : 1. m(tl₂o) = 204.4 g/mol  · 2 + 16 g/mol = 424.8 g/mol. m(compound) = 456. 8 g/mol – 424.8 g/mol = 32 g/mol ( 2 oxygens); tl₂o₃.