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Is 30a a small bra size?
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.
What are the solutions for 2x squared plus 5x plus 3 minus 4x squared plus 22 equals 0?
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
85 percent of 190 is what number?
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.
What is the second qibla?
it's Al-Masjed Al Haram In Mekka Saudi Arabia
What is a good middle for the fitst name makiah?
Al. Example: Makiah Al Amani
What is the chemical formula for the ionic compound formed from aluminum and sulfur?
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.
What is the oxidation number fo S in 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.
What will the formula for the ionic compound formed from al3 and s2?
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.
Predict the following chemical formula for a compound between Al and S?
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.
What is the oxidation number of Al2S3?
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.
How many metal atoms are in 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.
Chemical formula for aluminum sulfide?
S2Ur5 NO this is wrong! There isn't even an element that is Ur wow! The answer is Al2S3
What is the mass of Al formed when 0.500 moles Al2S3 is reduced completely with excess H2?
To determine the mass of aluminum (Al) formed from the complete reduction of 0.500 moles of aluminum sulfide (Al2S3), we first need to establish the stoichiometry of the reaction. The balanced equation for the reduction of Al2S3 is: [ Al2S3 + 6H2 \rightarrow 2Al + 3H2S. ] From this equation, one mole of Al2S3 produces two moles of Al. Therefore, 0.500 moles of Al2S3 will yield (0.500 \times 2 = 1.000) moles of Al. The molar mass of aluminum is approximately 27.0 g/mol, so the mass of Al produced is (1.000 , \text{mol} \times 27.0 , \text{g/mol} = 27.0 , \text{g}).
What is the oxidation number of sulfur in 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.
How many grams of Al2S3 can be formed from the reaction of 108.00 grams of Al with 5.00 grams of S?
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 )
What is the formula for ionic compound for Al 3 and S2-?
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.
What is he ionic formula for Al and S?
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.
