Na3 3+, PO4 3-. Na is actually +, so Na3 is 3+.
This will give a precipitate (s) of the insoluble copper phosphate2Na3PO4 + 3CuSO4 --> Cu3(PO4)2(s) + 3Na2SO4or (better without the soluble and not reacting ions (tribuned ions):2PO43- + 3SO42- --> Cu3(PO4)2(s)
simply use the formula DeltaTf=iKfm i= van't hoff number; Na3PO4 (sodium phosphate) van't hoff number is 4 m= molality moles of solute/kg of solvent in this case H2O Kf= the given number for Na3PO4 (sodium phosphate) is is ???*C/m just multiply those three things and that gives you the freezing point depression so we take the normal freezing point of water 0*C and we subtract the number we just calculated and that will give us the freezing point
To find the number of sodium atoms in 0.2310 g of sodium, you need to convert the mass to moles using the molar mass of sodium (22.99 g/mol). Then, use Avogadro's number (6.022 x 10^23 atoms/mol) to convert moles to atoms. This calculation will give you the number of sodium atoms present in 0.2310 g of sodium.
Na-1 S-2
sodium phosphate dibasic means that there are two sodium molecules in the formula (contributing to two bases). the formula is Na2HPO4. the first step is to figure out the molecular weight of the compount with the help of a periodic table. This comes out to 142g/mole. To make 1M solution (M= moles/liter so I am assuming 1 liter) you need 142 grams of the compound in 1L of dH2o. The definition of molarity (which is moles over liters) makes the calculation easy for making 1L of the solution. However if you want to make less or more of the solution, you will have to calculate accordingly. For example if you wanted to make 50 ml (0.05L) of the solution, you would multiply 142 g/liter (g/mol x moles/liter will give g/liter) X 0.05 liter, which would give you 7.1 g. 7.1 g of Na2HPO4 is needed to make 50 ml of 1 M solution. 142 g is needed to make 1000ml (1 L) of 1M solution.
Formula: Na2C2O4
sodium
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This will give a precipitate (s) of the insoluble copper phosphate2Na3PO4 + 3CuSO4 --> Cu3(PO4)2(s) + 3Na2SO4or (better without the soluble and not reacting ions (tribuned ions):2PO43- + 3SO42- --> Cu3(PO4)2(s)
simply use the formula DeltaTf=iKfm i= van't hoff number; Na3PO4 (sodium phosphate) van't hoff number is 4 m= molality moles of solute/kg of solvent in this case H2O Kf= the given number for Na3PO4 (sodium phosphate) is is ???*C/m just multiply those three things and that gives you the freezing point depression so we take the normal freezing point of water 0*C and we subtract the number we just calculated and that will give us the freezing point
The equation for the reaction: Na3PO4 + 3KCl ---> 3NaCl + K3PO4 Shows that 1 mol of sodium phoshpate will yield 3 mol of sodium chloride. The number of moles of Na3PO4 present in the solution in this reaction = concentration x volume => 0.0250 mol.dm-3 x 0.03 dm3 = 0.00075 mol. When reacted with excess KCl, this will give 0.00075 mol x 3 = 0.00225 mol of NaCl. The mass of NaCl formed = mol x RFM => 0.00225 x 58.5 = 0.132 g
The unit formula for sodium bromide is NaBr. There is one sodium ion and one bromide ion. Add the atomic weights from the periodic table for sodium and bromine. This will give you the formula weight. Atomic weight Na = 22.989770 Atomic weight Br = 79.904 Formula weight NaBr= 22.989770 + 79.904 = 102.894 Save
The simplest reaction is:NaOH + CO2 = Na2CO3
NaCl is the chemical formula for sodium chloride, often known as "table salt". "Na+" is the formula for a sodium ion; "Cl-" is the formula for a chloride ion. The charges balance to give a ratio of 1 sodium ion : 1 chloride ion per molecule of NaCl.
The formula NaCl represents sodium chloride, which is commonly known as table salt. It is an ionic compound composed of sodium cations (Na+) and chloride anions (Cl-), in a 1:1 ratio.
This question does not give enough information and cannot be answered in its present form.
To find the number of sodium atoms in 0.2310 g of sodium, you need to convert the mass to moles using the molar mass of sodium (22.99 g/mol). Then, use Avogadro's number (6.022 x 10^23 atoms/mol) to convert moles to atoms. This calculation will give you the number of sodium atoms present in 0.2310 g of sodium.