Chem ? i dont understand ...?

Question

calculate the percentage of Ca in a mixture that contains 14.090g of cadmium, 15.050g calcium, and 17.500g sand.........

Answer 1

32.268%

1) Divide the amount of Ca by the sum of the amounts of all three materials (.322684391).
2) Express as a percentage by shifting the decimal point two positions to the right (32.3684391%).
3) Reduce to five significant digits since the material amounts are all given to five significant digits (32.268%).

And you're done!

Answer 2

in case you have an outstanding extensive form m, that is represented via 10^-b if m is under a million or 10^b if m is larger than a million. Then we can write that -b = Log m m is under a million b = Log m m is larger than a million. We use the convention that "Log" is alway a base-10 logarithm. Logs are (or have been) very functional in math operations, because of the fact once you upload logs, that's the comparable as multiplying the corresponding numbers. In different words, in case you had 2 numbers, for the sake of dialogue, the two better than a million, and that they have been m and n, Log m + Log n = Log (mn). so which you would be able to desire to take the numbers m and n, fifnd their logarithms, upload them jointly, and then discover the "antilog" of the sum, which gave you the product mn. Doing powers grew to grow to be much less complicated; as quickly as switched over to logs, the logs could desire to be elevated or divided, and the resultant log switched over back to the respond. Chemists, being detrimental math human beings, and managing small numbers, used pH to describe the hydrogen ion content textile in water. It runs inversely to the hydrogen ion content textile. for this reason, a pH of two potential a [H+] of 10-2 and pH of three potential a [H+] of 10-3. If the pH is between 2 and 3, the [H+] is larger than 10-3 yet no longer as intense as 10-2. with a bit of luck, it rather is sufficient to get into pH conversions to [H+] and vice versa. PROB a million. seems alright to me. PROB 2. You lost a decimal there, [H+]=10^-2.7 At this element we do somewhat trick, and rewrite this as [H+]= 10^-3 x 10^0.3. Cranking this via your calculator, you will get an [H+] of two x 10^-3 PROB 3. This seems ok. PROB 4. With this, you employ the relation [H+][OH-] = 10x10^-15. it rather is fairly much less complicated to artwork with than 1x10^-14 for issues like this. [H+] = 3.5x10-5 so [OH-] = 2.8x10^-10. it rather is advantageous.

Answer 3

by weight?
15.050 / (14.090+15.050+17.500) = .32268
= 32.268%

Answer 4

As I said............