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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 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 ) ----------------------------------------------------
What is the melting point of NaBr?
The melting point of NaBr is 747 oC.
How many elements NaBr?
Sodium and bromine are the elements in sodium bromide (NaBr) compound.
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 is the chemical symbol for sodium bromide?
The chemical symbol for sodium bromide is NaBr.
Is NaBr a acid or base or salt?
NaBr is a salt composed of sodium (Na+) cations and bromide (Br-) anions. It is neither an acid nor a base.
Is nabr a acid?
No, NaBr is not an acid. It is the chemical formula for sodium bromide, which is a salt composed of sodium cations (Na+) and bromide anions (Br-).
What is a balance chemical equation for the synthesis of NaBr from Na and Br2?
2 Na + Br2 --> 2 NaBr
How many atoms is in nabr?
A formula unit of NaBr contains 2 atoms: 1 sodium and 1 bromine.
Which is the percent composition of bromine in the compound NaBr?
The percent composition of bromine in NaBr is approximately 77.7%. This is calculated by dividing the molar mass of bromine by the molar mass of NaBr and then multiplying by 100.
What is the mass of NaBr that will be produced from 42.7 g of AgbR?
Na2S2O3 + AgBr → NaBr + Na3[Ag(S2O3)2] First check that the given equation is balanced ... it isn't ... so the first thing to do is balance the equation: balancing Na: 2Na2S2O3 + AgBr → NaBr + Na3[Ag(S2O3)2] and everything is now balanced so we've got the balanced equation molar mass AgBr = 107.87 + 79.90 = 187.77 g/mol mol AgBr available = 42.7 g AgBr x [1 mol / 187.77 g] = 0.2274 mol AgBr from the balanced equation the mole ratio AgBr : Na2S2O3 = 1 : 2 so mol Na2S2O3 required = 0.2274 mol AgBr x [ 2 mol Na2S2O3 / mol AgBr] = 0.455 mol Na2S2O3 (to 3 sig figs)
