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How many grams of ammonium carbonate would you need to make 2.10 liters of a 0.619 M solution?
Wiki User
∙ 6y ago
124,9 g grams of ammonium carbonate are needed.
Wiki User
∙ 6y ago
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How may grams of calcium carbonate will be needed to form 4.29 liters of carbon dioxide?
The answer is 19,288 g cacium carbonate.
How do you determine pH of sodium carbonate?
Having the Molarity, concentration, helps. Molarity = moles of solute ( gotten from the grams put into solution ) divided by Liters of solution. Then. - log( Molarity of compound ) = pH ====( if basic subtract from 14 )
How many moles of ammonium ions are in 8.778 of ammonium carbonate?
assuming that 8.778 is in grams then there are 0.1069 moles in 8.778 grams of ammonium carbonate here is the math:(NH4)2CO3 N2=14.01g H8=8.08g C=12.01g O3=48.00 14.01+8.08+12.01+48.00=82.10g/mole 8.778g X 1 mole/82.10g=0.1069moles
If there are 20 grams of NH3 dissolved in 100 grams of water at 30 degrees C. What type of solution will you have?
This is a homogeneous solution of ammonium hydroxide in water.
What is the molarity of a solution that is made by dissolving 75.0 grams of MgCl2 in 500.0 milliliters of solution?
Need moles MgCl2 75.0 grams MgCl2 (1 mole MgCl2/95.21 grams) = 0.7877 mole MgCl2 ================now, Molarity = moles of solute/Liters of solution ( 500.0 milliliters = 0.5 Liters ) Molarity = 0.7877 moles MgCl2/0.5 Liters = 1.58 M MgCl2 solution --------------------------------
How may grams of calcium carbonate will be needed to form 4.29 liters of carbon dioxide?
The answer is 19,288 g cacium carbonate.
How many grams per liter ammonium sulfate are in a saturated solution of ammonium sulfate?
744 g/L of ammonium sulphate, at 20 0C
How do you determine pH of sodium carbonate?
Having the Molarity, concentration, helps. Molarity = moles of solute ( gotten from the grams put into solution ) divided by Liters of solution. Then. - log( Molarity of compound ) = pH ====( if basic subtract from 14 )
How many moles of ammonium ions are in 8.778 of ammonium carbonate?
assuming that 8.778 is in grams then there are 0.1069 moles in 8.778 grams of ammonium carbonate here is the math:(NH4)2CO3 N2=14.01g H8=8.08g C=12.01g O3=48.00 14.01+8.08+12.01+48.00=82.10g/mole 8.778g X 1 mole/82.10g=0.1069moles
How many moles of ammonium ions are in 8.754g of ammonium carbonate?
Question : How many moles of ammonium ions are in 8.754g of ammonium carbonate? Let's use the criss-cross method: (NH4+, CO32-)→(NH4)2CO3 That menas in each mole of ammonium carbonate((NH4)2CO3 )there are 2moles of ammonium (NH4). Let's find the quantity of mole(N) in 8.75g of ammonium carbonate((NH4)2CO3 ). N=mass/Molar mass (#) Molar mass((NH4)2CO3 )=2(14.0067+41.0079)+12.011+315.999)g/mol=96.09g/mol. (#) Gives us : N=(8.754g)/(96.09g/mol)=0.0911mole Let's K be the quantity of ammonium mole included in 8.75g of ammonium carbonate. As I stated before each mole of ammonium carbonate contains 2moles of ammonium, therefore : K=2*0.0911mole=1.1822mole Best regards, BILL JESY FOREVER 7171.
If there are 20 grams of NH3 dissolved in 100 grams of water at 30 degrees C. What type of solution will you have?
This is a homogeneous solution of ammonium hydroxide in water.
How many grams of hcl are contained in 4 liters of a 12m solution?
438 grams.
What is the concentration of a solution with a volume of 3.3 mL that contains 12 grams of ammonium sulfite?
31M
How many grams of ammonium sulfate are needed to make a 0.25M solution at a concentration of 6M?
5.50
How many grams of a drug must be used to prepare 3 liters of a 2 percent solution?
1 percent = 10 grams 2 % = 20 grams x 3 liters = 60 grams
What is the molarity of a solution that contains 1.1 of lithium in 0.5 liters of solution?
If 1,1 is grams the molarity is 0,317.
What is the molarity of a solution that is made by dissolving 75.0 grams of MgCl2 in 500.0 milliliters of solution?
Need moles MgCl2 75.0 grams MgCl2 (1 mole MgCl2/95.21 grams) = 0.7877 mole MgCl2 ================now, Molarity = moles of solute/Liters of solution ( 500.0 milliliters = 0.5 Liters ) Molarity = 0.7877 moles MgCl2/0.5 Liters = 1.58 M MgCl2 solution --------------------------------
