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.
Bromine has an atomic mass of 79.904 g/mol, while the molar mass of NaBr is 102.89 g/mol. Therefore, the percent composition of bromine in NaBr is (79.904 g/mol / 102.89 g/mol) x 100 = 77.68%.
Sodium and bromine are the elements in sodium bromide (NaBr) compound.
NaBr is ionic. There is no compound by the formula NaBr3.
NaBr contains an ionic bond, where sodium (Na) donates an electron to bromine (Br) to form Na+ and Br-. This results in the attraction between the positively charged sodium ion and the negatively charged bromine ion, forming the ionic compound NaBr.
The ionic compound formed from sodium (Na) and bromine (Br) is sodium bromide, with the chemical formula NaBr.
Bromine has an atomic mass of 79.904 g/mol, while the molar mass of NaBr is 102.89 g/mol. Therefore, the percent composition of bromine in NaBr is (79.904 g/mol / 102.89 g/mol) x 100 = 77.68%.
Sodium and bromine are the elements in sodium bromide (NaBr) compound.
Sodium Bromide - ionic compound - NaBr.
There are two elements that make up the compound NaBr, or sodium bromide. These two elements are sodium and bromine.
NaBr is ionic. There is no compound by the formula NaBr3.
NaBr contains an ionic bond, where sodium (Na) donates an electron to bromine (Br) to form Na+ and Br-. This results in the attraction between the positively charged sodium ion and the negatively charged bromine ion, forming the ionic compound NaBr.
The ionic compound formed from sodium (Na) and bromine (Br) is sodium bromide, with the chemical formula NaBr.
The name of the compound with the symbol NaBr is sodium bromide. Sodium bromide is a sedative and hypnotic when used over a long period of time. It can be used to control refractory seizures and status epilepticus if no other medication will work.
When sodium and bromine combine, they form sodium bromide, which is an ionic compound. The reaction between sodium and bromine is a redox reaction, where sodium loses an electron to form a sodium ion (Na+) and bromine gains an electron to form a bromide ion (Br-). The resulting compound, sodium bromide (NaBr), is a white crystalline solid with a high melting point.
Sodium bromide has the chemical formula NaBr. Its molar mass is approximately 102.89 g/mol (22.99 g/mol for Na and 79.90 g/mol for Br). To find the percent composition, divide the molar mass of each element by the molar mass of the compound and multiply by 100. For sodium: (22.99 g/mol / 102.89 g/mol) * 100 ≈ 22.36%. For bromine: (79.90 g/mol / 102.89 g/mol) * 100 ≈ 77.64%. Therefore, the percent composition of sodium bromide is approximately 22.36% sodium and 77.64% bromine.
Bromine is an element and can't be "made" from any other element (except by a nuclear reaction). However, since the question asks for a sodium compound, one possibility is sodium bromide, which can be melted and electrolyzed to form bromine at the anode.
NaBr