A 30A is a rather rare bra size, as it means the ribcage equals 30 inches and the bust equals 31 inches, since an "A cup" only means there is a one inch difference between the ribcage and the bust. Because of that, an A cup on different bands will be a different size-a 36A would be for someone whose ribcage equals 36 inches and their bust 37, a 26A would be for someone whose ribcage equals 26 inches and their bust 27, et al. Since an A cup marks a one inch difference, yes, a 30A is a small bra size. There is a very good change you have been fitted incorrectly though, possibly with the "Plus 4" method, which spawned "The War on Plus 4". You may be more like a 26DD, which means your ribcage equals 26 inches and your bust equals 31 inches, marking a 5 inch difference, which equals a DD.
161.5 is 85% of 190. Al you do is divide 190 by 100 which equals 1.9 and then multiply by 85 which gives 161.5.
First you should combine equal terms. For example, 2x2 and -4x2 can be combined as -2x2. Then you can use the quadratic formula.Additonal Answer:-2x2+5x+3-4x2+22 = 0Collect like terms: -2x2+5x+25 = 0Divide al terms by -1: 2x2-5x-25 = 0Factorise: (2x+5)(x-5) = 0Solutions are: x = -5/2 or x = 5
it's Al-Masjed Al Haram In Mekka Saudi Arabia
Al. Example: Makiah Al Amani
Al2S3. Obviously, the 2 and 3 are subscript. The reason this happens is because of the bonding ratio: there are 3 S ions required to balance 2 Al ions.
S2Ur5 NO this is wrong! There isn't even an element that is Ur wow! The answer is Al2S3
Each molecule of Al2S3 is composed of one Al3+ ion. Since a mole is equal to 6.022 x 1023 molecules, there would be 6.022 x 1023 Al3+ ions in one mole of Al2S3.
The oxidation number of Al is +III and S is -II.
S has an oxidation number of -2. Al is +3.
Al + O2 will never yield Fe2O3.
Al and AgNO3. Al has a 3plus charge plus 3e negative. 3(Ag positive plus le negative equals Ag). Al plus 3Ag plus is equal to Al with a 3 plus charge plus 3Ag.
The chemical equation is:2 Al +3 CuCl2 = 3 Cu + 2 AlCl3
The answer i got is 12.33 grams of Al2S3. Below i will try to show the steps i used: n=moles m=mass (grams) M= molecular weight (from periodic table) 2Al + 3S --> Al2S3 nAl= m/M = 9/13 = 0.692307 moles of Al nS= m/M = 8/16 = 0.5 moles of S Limiting Reagent: 1.5 moles of S required for every mole of Al (3:2 ratio) This means that there should be about 1.04 moles of S to completely use up all the Al. Since there is less than this amount of S present (only 0.5 moles), S is the limiting reagent and should be used in the mole calculation of the product Amount of product: 0.5 moles S x (1 mol Al2S3/ 3 mol S) = 0.16666 moles of Al2S3 nAl2S3 = m/M Molecular weight of Al2S3 = (2 x 13) + (3 x 16) = 74 0.16666 moles Al2S3 = m/74 m = 74 x 0.16666 = 12.33 grams of Al2S3 Therefore, 12.33 grams of Al2S3 is formed in this reaction ( hopefully this is right :P )
The balanced chemical equation for the reaction between Al and O2 is: 4Al + 3O2 -> 2Al2O3 Therefore, the coefficient needed for Al to balance the equation is 4.
Moles Al2S3 = 14.2 g / 150.16 g/mol =0.9456 the rario between Al2S3 and Al(OH)3 is 1 : 2 moles Al(OH)3 = 2 x 0.9456 =0.1891 Mass Al(OH)3 = 0.1891 mol x 78 g/mol =14.75 g