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
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The chemical formula for the ionic compound formed from aluminum and sulfur is Al2S3. This compound is formed by the transfer of electrons between aluminum (Al) and sulfur (S) atoms, resulting in the formation of Al3+ and S2- ions which combine in a 2:3 ratio to create Al2S3.
The oxidation number of Sulfur (S) in Al2S3 is -2. This is because aluminum (Al) typically has an oxidation state of +3, and in Al2S3, the overall compound is neutral, so the oxidation state of sulfur must be -2 to balance the charges.
The formula for the ionic compound formed from Al^3+ and S^2- ions will be Al2S3. This is because the charges must balance out, requiring two Al^3+ ions for every three S^2- ions to cancel out the charges and form a neutral compound.
The chemical formula for a compound between Al and S is Al2S3. Aluminum typically forms a 3+ cation and sulfur typically forms a 2- anion. To balance the charges, you need two aluminum atoms for every three sulfur atoms.
The oxidation number of aluminum (Al) in Al2S3 is +3, while the oxidation number of sulfur (S) is -2. Each aluminum atom has an oxidation number of +3, and each sulfur atom has an oxidation number of -2 in order to balance the charge in the compound.
There are 2 aluminum atoms (Al) in each molecule of Al2S3 and 3 sulfur atoms (S), making a total of 5 metal atoms in one molecule of Al2S3.
The chemical formula for aluminum sulfide is Al2S3.
The oxidation number of sulfur in Al2S3 is -2. This is because aluminum has an oxidation number of +3, and the overall compound is neutral, so the total oxidation number contribution from sulfur must be -6 to balance the charge.
To determine the limiting reactant and calculate the grams of Al2S3 formed, we need to first convert the grams of each reactant (Al and S) to moles. Then, we find the mole ratio of Al to Al2S3 and S to Al2S3 according to the balanced chemical equation. After that, we compare the moles of Al2S3 that can be formed from Al and S, choose the smaller value, and convert it back to grams of Al2S3.
The ionic formula for aluminum (Al) and sulfur (S) is Al2S3. This is because aluminum has a 3+ charge and sulfur has a 2- charge, so the formula needs two aluminum ions to balance out the three sulfur ions.
The ionic compound formed between Al^3+ and S^2- ions would have the formula Al2S3. This formula ensures that the positive and negative charges balance out in the compound, with two aluminum ions combining with three sulfide ions.
The oxidation half-reaction is: 3Ag --> 3Ag^+ + 3e^-