No. There is no platinum ratio.
The ratio is 1:2The ratio is 1:2The ratio is 1:2The ratio is 1:2
The ratio of C12H22O11 to WHAT!
The ratio of volumes is directly proportional to the cube of the ratio of their sides. And, incidentally, all cubes are similar.
The ratio is 1:25 4 percent as a ratio is 0.04 : 1
The molar mass of sodium chloride is 58,44 g. The molar mass of sodium bicarbonate is 84,007 g. The ratio is 1,437.
When 2.10 g of NaHCO3 decomposes, it produces CO2, H2O, and Na2CO3. The molar ratio between NaHCO3 and H2O is 1:1. This means that 2.10 g of NaHCO3 will produce 18 g of water.
To find the mass of C6H8O7 needed to consume 0.42g of NaHCO3, you first need to balance the chemical equation for the reaction between them. Then, calculate the molar ratio between NaHCO3 and C6H8O7 to determine the mass needed. This involves stoichiometry and molar mass calculations to find the answer.
No, NaHCO3 is not an element, it is a compound. It is the chemical formula for sodium bicarbonate, which is commonly known as baking soda. It is composed of sodium (Na), hydrogen (H), carbon (C), and oxygen (O) atoms bonded together in a specific ratio.
The chemical formula for sodium hydrogen carbonate is NaHCO3.
To neutralize the HCl spill, we need to determine the amount of moles of HCl present first. The moles of HCl can be calculated as concentration (mol/L) x volume (L). Then, since NaHCO3 reacts in a 1:1 ratio with HCl, you would need the same number of moles of NaHCO3 as moles of HCl to neutralize the spill. From there, you can calculate the mass of NaHCO3 needed using its molar mass.
Divide 6.10 (g NaHCO3) by 84.007 (g.mol−1 NaHCO3) to get 0.0726 mol NaHCO3
To neutralize 150 mL of 0.44 M HCl, you would need an equal number of moles of NaHCO3. First, calculate the number of moles of HCl (mol = Molarity x Volume) used. Then, since 1 mol of NaHCO3 neutralizes 1 mol of HCl, you would need the same number of moles of NaHCO3. From that, you can calculate the mass of NaHCO3 needed using its molar mass.
Sodium Bicarbonate come from absorption of CO2 into NaOH Total reaction is 2NaOH + CO2 -> NaHCO3 + H2O Each mole of NaHCO3 come from absorption of 1 mole CO2 Molecular weight of NaHCO3 is 84 g/mol and CO2 is 44 g/mol It is that 84 g of NaHCO3 had 44 g of CO2 By weight ratio it is 52% CO2 in Sodium Bicarbonate.
To determine the mass of NaCH3CO2 obtained, first calculate the amount of CH3CO2H in moles using its molarity and volume. Then use the stoichiometry of the reaction to find the moles of NaHCO3 reacting. Finally, use the mole ratio between NaHCO3 and NaCH3CO2 to find the mass of NaCH3CO2 that can be obtained.
NaHCO3 is commonly called bicarbonate of soda or baking soda.
For this you need the atomic (molecular) mass of NaHCO3. Take the number of grams and divide it by the atomic mass. Multiply by one mole for units to cancel. NaHCO3=84.0 grams110 grams NaHCO3 / (84.0 grams) = 1.31 moles NaHCO3