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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 grams
186 grams HBr / (81.0 grams) =2.30 moles HBr
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How many grams of solid sodium hydroxide would need to be added to completely neutralize 35.0 mL of 1.45 M HBr?
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.
How many kilojoules are associated with the formation of 2 moles of HBr?
The standard enthalpy of formation of HBr is -36.3 kJ/mol. For 2 moles of HBr, the total energy associated with the formation would be: -36.3 kJ/mol * 2 mol = -72.6 kJ.
What is the concentration of a HBr solution in 12.0 mL of the solution is neutralized by 15.0 mL of a 0.25 M KOH solution?
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
What is the concentration of a HBr if 12.0 mL of the solution is neutralized by 15.0 mL of a 0.25 M KOH solution?
To find the concentration of HBr, you first need to determine the number of moles of KOH that react with the HBr. This can be done using the volume and concentration of KOH solution. Then, using the stoichiometry of the neutralization reaction between HBr and KOH, you can find the number of moles of HBr present in the sample. Finally, divide the moles of HBr by the volume of the sample (12.0 mL) to obtain the concentration of HBr.
How many kilo-joules are associated with the formation of 2 moles of HBr(g)?
The standard enthalpy of formation of HBr(g) is -36.2 kJ/mol. For 2 moles of HBr(g), the total energy associated with its formation would be 2 * -36.2 kJ/mol = -72.4 kJ.
How many grams of solid sodium hydroxide would need to be added to completely neutralize 35.0 mL of 1.45 M HBr?
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.
How many molecules are in 3.21 moles of HBr?
3.21 moles HBr (6.022 X 10^23/1mole HBr) = 1.93 X 10^24 molecules of HBr
How many atoms are in 2.13 grams of HBr?
To find the number of atoms in 2.13 grams of HBr, you first need to determine the number of moles using the molar mass of HBr (molar mass = 80.91 g/mol). Then, you can use Avogadro's number (6.022 × 10^23 atoms/mol) to convert moles to atoms.
How many grams in 4.23 m mol of HBr?
The answer is 0,3422 grams.
How many moles are present in 1.21 molecules of HBr?
To find the number of moles in 1.21 molecules of HBr, divide the number of molecules by Avogadro's number (6.022 x 10^23 molecules/mol). Thus, 1.21 molecules of HBr is approximately 2.01 x 10^-24 moles.
How many kilojoules are associated with the formation of 2 moles of HBr?
The standard enthalpy of formation of HBr is -36.3 kJ/mol. For 2 moles of HBr, the total energy associated with the formation would be: -36.3 kJ/mol * 2 mol = -72.6 kJ.
How many moles are in 135 g of hydrobromic acid?
To calculate moles use the eq'n moles = mass(g) / Mr (Relative molceular Mass). We have a mass of 135 g The Mr is calcualted from the Periodic Table, using atomic weights/masses. HBr = 1 x H = 1 x 1 = 1 1 x Br = 1 x 79.9 = 79.9 79.9 + 1 = 80.9 Substituting into the eq'n moles(HBr) = 135 / 80.9 Moles = 1.66872..... ~ 1.67 moles.
What is the concentration of a HBr solution in 12.0 mL of the solution is neutralized by 15.0 mL of a 0.25 M KOH solution?
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
How many milliliters of 0.305M NaOH would be required to titrate a 5.00mL solution of 0.616M HBr?
Since NaOH and HBr react in a 1:1 ratio, the moles of NaOH needed to titrate HBr can be calculated. Moles of NaOH = moles of HBr. Next, use the concentration and volume of HBr to find the moles present. Finally, use the concentration of NaOH to calculate the volume needed. In this case, approximately 2.41 mL of 0.305M NaOH would be needed.
What is the concentration of a HBr if 12.0 mL of the solution is neutralized by 15.0 mL of a 0.25 M KOH solution?
To find the concentration of HBr, you first need to determine the number of moles of KOH that react with the HBr. This can be done using the volume and concentration of KOH solution. Then, using the stoichiometry of the neutralization reaction between HBr and KOH, you can find the number of moles of HBr present in the sample. Finally, divide the moles of HBr by the volume of the sample (12.0 mL) to obtain the concentration of HBr.
How many kilo-joules are associated with the formation of 2 moles of HBr(g)?
The standard enthalpy of formation of HBr(g) is -36.2 kJ/mol. For 2 moles of HBr(g), the total energy associated with its formation would be 2 * -36.2 kJ/mol = -72.4 kJ.
A 25.00 mL sample of HBr is titrated with 0.150 M standardized sodium hydoxide solution the endpoint was reached when 18.80 mL of titrant had been added calculate the molar concentration of the HBr?
Using the balanced chemical equation for the reaction between HBr and NaOH (1:1 ratio), we can determine the moles of NaOH used. From the volume of NaOH used, we can then calculate the moles of HBr present in the sample. Finally, dividing moles of HBr by the initial volume of the sample (in liters) gives the molar concentration of HBr.
