Na- 1 x 23.0=23.0g
Cl- 1 x 35.5=35.5g
23.0g + 35.5g = 58.5 (58.5 is total)
(23.0g/58.5g) x 100 = 39.3% Na
(35.5g/58.5g) x 100 = 60.7% Cl
To make a 1% HCl solution from a 35% HCl solution, you would need to dilute the concentrated solution with water. The ratio of concentrated HCl to water would be approximately 1:34. So, to make 1% HCl, you would mix 1 part of the 35% HCl solution with 34 parts of water.
The percent composition of lithium in lithium bromide (LiBr) is approximately 7.7%. The percent composition of bromine in lithium bromide is approximately 92.3%.
It would be 0.1M HCl solution. In 1 liter there would be 3.65 g of HCl and the balance would be water.
The percent composition of C5H12 is: carbon (C) = 83.78%, hydrogen (H) = 16.22%.
To find the percent composition of each element in the compound, you first calculate the molar mass of Be (9.01 g/mol) and I (126.90 g/mol). Then, calculate the percent composition of each element by dividing the mass of the element by the total molar mass of the compound and multiplying by 100. The percent composition of Be is 5.14% and the percent composition of I is 94.86%.
Type your answer heZinc chloride (ZnCl2) and hydrogen (H2) react to form zinc (Zn) and hydrochloric acid (HCl) in the equation ZnCl2 + H2 ---> Zn + 2HCl. The percent composition of ZnCl2 is 52% Cl and 48% Zn; the percent composition of H2 is 100% H.re...
To make a 1% HCl solution from a 35% HCl solution, you would need to dilute the concentrated solution with water. The ratio of concentrated HCl to water would be approximately 1:34. So, to make 1% HCl, you would mix 1 part of the 35% HCl solution with 34 parts of water.
i dont knw
0.084%
The percent composition of lithium in lithium bromide (LiBr) is approximately 7.7%. The percent composition of bromine in lithium bromide is approximately 92.3%.
HCL is short for hydrochloric acid, or hydrochloride referring to the form/composition of the particular drug. <<adr>>
It would be 0.1M HCl solution. In 1 liter there would be 3.65 g of HCl and the balance would be water.
The percent composition of C5H12 is: carbon (C) = 83.78%, hydrogen (H) = 16.22%.
To find the percent composition of each element in the compound, you first calculate the molar mass of Be (9.01 g/mol) and I (126.90 g/mol). Then, calculate the percent composition of each element by dividing the mass of the element by the total molar mass of the compound and multiplying by 100. The percent composition of Be is 5.14% and the percent composition of I is 94.86%.
To find the percent composition, first calculate the total mass by adding the masses of Argon and Sulfur. Then, determine the percent composition of each element by dividing the mass of that element by the total mass and multiplying by 100. In this case, the percent composition of Argon is about 87.1%, and the percent composition of Sulfur is about 12.9%.
The percent composition of calcium fluoride is approximately 51.1% calcium and 48.9% fluorine.
Sodium bromide has a percent composition of approximately 22.8% sodium and 77.2% bromine.