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124,9 g grams of ammonium carbonate are needed.

Q: How many grams of ammonium carbonate would you need to make 2.10 liters of a 0.619 M solution?

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The answer is 19,288 g cacium 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 )

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

This is a homogeneous solution of ammonium hydroxide in water.

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 --------------------------------

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The answer is 19,288 g cacium carbonate.

744 g/L of ammonium sulphate, at 20 0C

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 )

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.

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

This is a homogeneous solution of ammonium hydroxide in water.

438 grams.

31M

5.50

1 percent = 10 grams 2 % = 20 grams x 3 liters = 60 grams

If 1,1 is grams the molarity is 0,317.

One liter of Benedict's solution contains 173 grams sodium citrate, 100 grams sodium carbonate, and 17.3 grams cupric sulfate pentahydrate.