Need moles sodium bromide first.
18.7 grams NaBr (1 mole NaBr/102.89 grams)
= 0.1817 moles NaBr
=====================Now,
Molarity = moles of solute/Liters of solution
0.256 M NaBr = 0.1817 moles NaBr/X Liters
Liters = 0.1817/0.256
= 0.7098 Liters
-------------------------( you do sigi figis )
Sodium Bromide is of course very soluble in water. In fact 116 grams of NaBr will dissolve in 100 cc of H2O at 50 degrees C.
All you really need here is a ratio of molar masses, ie molar mass of Na/molar mass of NaBr So look at your periodic table and is says: Na = 23g/mol Br = 80g/mol therefore NaBr = 103g/mol so from this, you actually know a percentage of Sodium to Bromine, correct me if I'm wrong, someone else please but just take: (23g/mol)/103g/mol = 22.3% of you NaBr is actually Na, so take 5.35g * 0.223 and you get: 1.19g of Na in your NaBr Hope that helps. Cheers
To calculate the amount of potassium bromide needed, use the formula: moles = molarity x volume (in liters). First, convert the volume to liters by dividing 0.50L by 1000. Then, multiply the molarity (0.125M) by the volume in liters to find the moles of potassium bromide needed. Finally, convert moles to grams using the molar mass of potassium bromide (KBr).
117 grams of sodium chloride (NaCl) is equivalent to 117 grams of chlorine gas because each molecule of NaCl contains one sodium atom and one chlorine atom.
The molar volume of a gas at STP is 22.4 liters/mol. The molar mass of hydrogen bromide is 80.9 g/mol. Therefore, the density of hydrogen bromide at STP is 80.9 g/mol / 22.4 L/mol = 3.61 g/L.
From the definition of molarity, one liter of the stated solution contains 0.256 mole of sodium bromide. The gram formula weight of sodium bromide is 102.89; therefore, one liter of the solution contains 0.256 X 102.89 or 26.34 grams. 18.7/26.34 = 0.710, the fraction of a liter that contains 18.7 grams of sodium bromide.
Sodium Bromide is of course very soluble in water. In fact 116 grams of NaBr will dissolve in 100 cc of H2O at 50 degrees C.
All you really need here is a ratio of molar masses, ie molar mass of Na/molar mass of NaBr So look at your periodic table and is says: Na = 23g/mol Br = 80g/mol therefore NaBr = 103g/mol so from this, you actually know a percentage of Sodium to Bromine, correct me if I'm wrong, someone else please but just take: (23g/mol)/103g/mol = 22.3% of you NaBr is actually Na, so take 5.35g * 0.223 and you get: 1.19g of Na in your NaBr Hope that helps. Cheers
To calculate the grams of lithium bromide present in the solution, you would first determine the moles of lithium bromide using the formula: moles = Molarity x Volume (L). Once you have the moles, you can convert it to grams using the molar mass of lithium bromide (86.85 g/mol).
1.6 grams.
Normal saline is 0.9% weight/volume sodium chloride to water. This is 9 grams per litre. NaCl has a molecular weight of 58.5, sodium (Na) has a weight of 23, which is 39.3% of the molecular weight. So sodium is 39.3% of the weight. 1 litre of saline has 9 grams, 250ml is a quarter of a litre, so has 9/4 grams = 2.25 grams. 39.3% of 2.25 g is 0.884 grams of sodium.
The first step is to convert the mass of sodium chloride to moles using its molar mass. Then, divide the moles of sodium chloride by the volume of the solution in liters to calculate the molarity.
To calculate the amount of potassium bromide needed, use the formula: moles = molarity x volume (in liters). First, convert the volume to liters by dividing 0.50L by 1000. Then, multiply the molarity (0.125M) by the volume in liters to find the moles of potassium bromide needed. Finally, convert moles to grams using the molar mass of potassium bromide (KBr).
To determine the number of grams of pure sodium hydroxide present in a solution with a known volume, you need to know the concentration of the solution in g/ml. Then you can use the formula: grams = concentration (g/ml) x volume (ml)
That depends on how much solvent you are dissolving the 9 grams of salt in.
117 grams of sodium chloride (NaCl) is equivalent to 117 grams of chlorine gas because each molecule of NaCl contains one sodium atom and one chlorine atom.
At 10°C, the solubility of strontium bromide (SrBr₂) in water is approximately 3.5 grams per 100 grams of water. This means that you can dissolve about 3.5 grams of strontium bromide in 100 grams of water at this temperature.