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How many atoms are there in 13 moles of bromide?
If you think to a simple binary bromide as NaBr: 166,57.10e23 atoms.
How many ml of a 0.185 M aqueous solution of sodium bromide must be taken to obtain 7.41 grams of the salt?
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.
What is the melting point of NaBr?
The melting point of NaBr is 747 oC.
What is a balance chemical equation for the synthesis of NaBr from Na and Br2?
2 Na + Br2 --> 2 NaBr
If 25 moles of sucrose are mixed into 50 liters of water what is the molarity?
The molarity can be calculated using the formula: moles of solute divided by liters of solution. In this case, the moles of sucrose is 25, and the liters of solution is 50. This gives a molarity of 0.5 M.
How many grams of NaBr are contained in 75.0 mL of a 1.5 M NaBr solution?
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 ) ----------------------------------------------------
How many moles of NaBr can be produced from 2 moles of NaF?
2 moles, if you can find the proper catalyst, or set of reactions to complete the reaction.
What reactant is limiting if 3000 cm3 of Cl2 at STP react with a solution containing 25 grams of NaBr?
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
How many atoms are there in 13 moles of bromide?
If you think to a simple binary bromide as NaBr: 166,57.10e23 atoms.
How many mL of a 2.0 m NaBr solution are needed to make 200 mL of 0.50 m NaBr?
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.
How many ml of a 0.185 M aqueous solution of sodium bromide must be taken to obtain 7.41 grams of the salt?
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.
What volume of 0.256 M sodium bromide contains 18.7 grams of bromide?
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 )
How many moles of (NH4)2SO4 do you have?
0.758 moles of NH3 is the amount of moles in 50 grams of NH42SO4.
How many elements NaBr?
Sodium and bromine are the elements in sodium bromide (NaBr) compound.
How many moles of ammonium hydroxide are contained in 50 cubic centimetre of 0.15 molar ammoniumhydroxide?
.15/1000 * 50 = 0.0075 moles or 7.5mmol
How many atoms is in nabr?
A formula unit of NaBr contains 2 atoms: 1 sodium and 1 bromine.
How many moles of HCl are found in 50 mL of a 6.0 M solution?
To find the number of moles, first calculate the number of moles of HCl in the 50 mL solution by multiplying the volume (in liters) by the molarity. Volume in liters = 50 mL / 1000 mL/L = 0.05 L Moles = 0.05 L * 6.0 mol/L = 0.3 moles of HCl.
