If you think to a simple binary bromide as NaBr: 166,57.10e23 atoms.
To find the volume of a 0.185 M solution of sodium bromide (NaBr) needed to obtain 7.41 grams, first calculate the number of moles of NaBr: Molar mass of NaBr = 22.99 g/mol (Na) + 79.90 g/mol (Br) = 102.89 g/mol. Moles of NaBr = 7.41 g / 102.89 g/mol ≈ 0.0720 moles. Now, using the molarity formula (M = moles/volume in liters), rearrange to find volume: Volume (L) = moles / M = 0.0720 moles / 0.185 M ≈ 0.3892 L, or 389.2 mL. Therefore, you need approximately 389.2 mL of the solution.
To find the molarity (M) of a solution, you use the formula: M = moles of solute / liters of solution. Here, you have 4.50 moles of NaBr and 3.00 L of solution. Therefore, the molarity is M = 4.50 moles / 3.00 L = 1.50 M. Thus, the molarity of the NaBr solution is 1.50 M.
The melting point of NaBr is 747 oC.
2 Na + Br2 --> 2 NaBr
Molarity = moles of solute/Liters of solution ( 75.0 ml = 0.075 Liters ) Get moles NaBr 1.5 M NaBr = moles NaBr/0.075 Liters = 0.1125 moles NaBr (102.89 grams/1 mole NaBr) = 11.575 grams NaBr ( call it 12 grams ) ----------------------------------------------------
2 moles, if you can find the proper catalyst, or set of reactions to complete the reaction.
Cl2 + 2NaBr => Br2 + 2NaCl One mole Cl2 reacts with 2 moles NaBr Cl2 = 71 NaBr = 102.9 Molar volume = 22.414 L/mole for ideal gas @STP 3L Cl2 = 3/22.414 = 0.1338 mole 25g NaBr = 25/102.9 = 0.2430 mole 0.1338 moles Cl2 requires 0.2676 moles NaBr for complete reaction The NaBr is the limiting reagent
If you think to a simple binary bromide as NaBr: 166,57.10e23 atoms.
You would need 50 mL of the 2.0 M NaBr solution to make 200 mL of 0.50 M NaBr solution. This can be calculated using the formula: (C1V1) = (C2V2), where C1 = concentration of stock solution, V1 = volume of stock solution, C2 = final concentration, and V2 = final volume.
To find the volume of a 0.185 M solution of sodium bromide (NaBr) needed to obtain 7.41 grams, first calculate the number of moles of NaBr: Molar mass of NaBr = 22.99 g/mol (Na) + 79.90 g/mol (Br) = 102.89 g/mol. Moles of NaBr = 7.41 g / 102.89 g/mol ≈ 0.0720 moles. Now, using the molarity formula (M = moles/volume in liters), rearrange to find volume: Volume (L) = moles / M = 0.0720 moles / 0.185 M ≈ 0.3892 L, or 389.2 mL. Therefore, you need approximately 389.2 mL of the solution.
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 )
To find the molarity (M) of a solution, you use the formula: M = moles of solute / liters of solution. Here, you have 4.50 moles of NaBr and 3.00 L of solution. Therefore, the molarity is M = 4.50 moles / 3.00 L = 1.50 M. Thus, the molarity of the NaBr solution is 1.50 M.
0.758 moles of NH3 is the amount of moles in 50 grams of NH42SO4.
Sodium and bromine are the elements in sodium bromide (NaBr) compound.
.15/1000 * 50 = 0.0075 moles or 7.5mmol
A formula unit of NaBr contains 2 atoms: 1 sodium and 1 bromine.