To find the volume of the solution containing 18.7 grams of bromide, first calculate the moles of bromide using the molar mass. Then, use the molarity to determine the volume of the solution. The volume of 0.256 M sodium bromide solution that contains 18.7 grams of bromide would be approximately 369.1 mL.
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.
To find out how many grams of sodium are in the sample of sodium bromide, we need to determine the percentage of sodium in sodium bromide. Sodium bromide is composed of one sodium atom for every one bromine atom. Sodium's atomic mass is approximately 23 g/mol, while bromine's is approximately 80 g/mol. Thus, the molar mass of sodium bromide is around 102 g/mol. Since the molar mass of sodium is about 23 g/mol, sodium makes up around 23/102 = 0.2255 or 22.55% of sodium bromide's mass. Therefore, a 5.35 g sample of sodium bromide contains around 5.35 g * 0.2255 ≈ 1.21 g of sodium.
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.
To find the volume of 0.256 M sodium bromide that contains 18.7 grams, you first need to calculate the number of moles of sodium bromide using its molar mass. Then, you can use the formula: moles = molarity x volume (in liters) to find the volume of the solution in liters. Finally, convert the volume to the desired unit if needed.
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.
To find out how many grams of sodium are in the sample of sodium bromide, we need to determine the percentage of sodium in sodium bromide. Sodium bromide is composed of one sodium atom for every one bromine atom. Sodium's atomic mass is approximately 23 g/mol, while bromine's is approximately 80 g/mol. Thus, the molar mass of sodium bromide is around 102 g/mol. Since the molar mass of sodium is about 23 g/mol, sodium makes up around 23/102 = 0.2255 or 22.55% of sodium bromide's mass. Therefore, a 5.35 g sample of sodium bromide contains around 5.35 g * 0.2255 ≈ 1.21 g of sodium.
1.6 grams.
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).
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.
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.