The melting point of NaBr is 747 oC.
2 Na + Br2 --> 2 NaBr
The binary name for KBr is Potassium Bromide.
yes it is because kbr is just one word not 2.
There are two elements that make up the compound NaBr, or sodium bromide. These two elements are sodium and bromine.
KOH, NaOH, KCl, KBr, NaBr
NaCl, NaOH, NaBr, KBr, KOH
Salts of bromine are called bromides and many are known: CaBr2, KBr, NaBr, LiBr etc.
Well, honey, technically speaking, a buffer is a solution that can resist changes in pH when small amounts of acid or base are added. So, if you mix hydrobromic acid (HBr) and sodium bromide (NaBr) together, you could potentially have a buffer solution if the concentrations are right. But hey, don't get too excited, it's not as simple as just throwing them together and calling it a day.
The reaction between NaOH (sodium hydroxide) and KBr (potassium bromide) would result in the formation of NaBr (sodium bromide) and KOH (potassium hydroxide) as products. This is a double displacement reaction where the cations and anions switch partners.
When you flame test the two solutions, any Na solution burns yellow, while any K solution burns violet/purple. So both the KCl and the KBr will burn purple, while both NaBr and NaCl will burn yellow.
Bromine is found in a variety of compounds, such as sodium bromide (NaBr), potassium bromide (KBr), and hydrogen bromide (HBr). Additionally, bromine occurs naturally in seawater as sodium bromide and in some minerals like bromargyrite.
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 ) ----------------------------------------------------
The melting point of NaBr is 747 oC.
Sodium and bromine are the elements in sodium bromide (NaBr) compound.
To prepare a 0.01N KBr solution, dissolve 0.74g of KBr in 1 liter of water. This will give you a solution with a molarity of 0.01N for KBr.
The chemical symbol for sodium bromide is NaBr.