3.21 moles HBr (6.022 X 10^23/1mole HBr)
= 1.93 X 10^24 molecules of HBr
Given the balanced equation2Al + 6HBr --> 2AlBr3 + 3H2In order to find how many grams of HBr are required to produce 150g AlBr3, we must convert from mass to mass (mass --> mass conversion).150g AlBr3 * 1 mol AlBr3 * 6 molecules HBr = 136.52 or 137g HBr----------- 266.6g AlBr3 * 2 molecules AlBr3
yes HBr is an electrolyte
C18H25NO•HBr
The answer is 0,3422 grams.
HCl HBr HF
2.01 mol
For this you need the atomic (molecular) mass of HBr. Take the number of grams and divide it by the atomic mass. Multiply by one mole for units to cancel. HBr= 81.0 grams186 grams HBr / (81.0 grams) =2.30 moles HBr
In the English language hydrobromic acid is the solution of HBr in water; the name of the acid HBr is hydrogen bromide. 135 g HBr are equivalent to 1,67 moles.
-73
Balanced equation. KOH + HBr -> KBr + H2O everything is one to one, so... Molarity = moles of solute/liters of solution ( change ml to liters ) 0.25 M KOH = moles KOH/0.015 liters = 0.00375 moles of KOH this is as many moles that you have of HBr, so... Molarity of HBr = 0.00375 moles/0.012 liters = a concentration of HBr that is 0.31 M
A mole is the quantity of any molecule, atom, etc that has the same number of ... If I have 6.022×1023 H2 molecules, I have a mass of 2 gram of hydrogen molecules. ... How many moles are present in 1.21 X 10^24 molecules of HBr? ... How do you convert the amount of atoms in each battery to moles of lithium atoms?
The energy is -72,26 kJ.
This is a titration question: we want to have the same number of hydroxide ions as hydroxide ions so that they will form water and the pH will be neutral. In chemistry, we count atoms and molecules in moles, and we can calculate how many moles of HBr we have, because concentration in molarity is the number of moles divided by the volume in liters... M = moles/V. We plug in what we got: 1.45M = moles/0.0350L, and solve for moles: 0.0508 moles. Now we know we need 0.0508 moles of NaOH, whose molecular weight is 40g/mole. MW x moles = grams, so (40g/mole)(0.0508 moles) = 2.03 g of NaOH.
Given the balanced equation2Al + 6HBr --> 2AlBr3 + 3H2In order to find how many grams of HBr are required to produce 150g AlBr3, we must convert from mass to mass (mass --> mass conversion).150g AlBr3 * 1 mol AlBr3 * 6 molecules HBr = 136.52 or 137g HBr----------- 266.6g AlBr3 * 2 molecules AlBr3
Ther answer is none! ammonium bromide is made from hydrogen bromide and ammonia NH3 + HBr = NH4Br i mole of each makes 1mole of ammonium salt.
Ka = [H+].[Br-] / [HBr] However the value of this expression is very high, because HBr is a STRONG acid, meaning that much more than 99.9% of the HBr molecules in water are protolized (ionized), making [H+] and [Br-] equal to the original (added) HBr amount, and the [HBr]-value nearly zero.
Some gases have polar molecules as HCl, HBr, SO2, but not all gases have polar molecules.