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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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8.754 grams of ammonium carbonate equals 0.1822 moles of ammonium ions.

96.03 grams

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

The answer is 19,288 g cacium carbonate.

4.00 grams of NaOH have been added to 2.00 liters of water. The molarity of this solution is?

This is a homogeneous solution of ammonium hydroxide in water.

438 grams.

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

31M

5.50

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

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

no ; false

Molarity = moles of solute/Liters of solution need to find moles NH3 16.7 grams NH3 (1 mole NH3/17.034 grams) = 0.9804 moles NH3 --------------------------------now Molarity = 0.9804 moles NH3/1.50 Liters = 0.654 M -------------

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

Molarity = moles of solute/Liters of solution ( 100 mL = 0.1 Liters)Moles of solute (K2SO4) = Liters of solution * MolarityMoles K2SO4 = 0.1 Liters * 0.1 M= 0.01 moles K2SO4 (174.27 grams/1 mole K2SO4)= 1.7 grams potassium sulfate=======================Add that many grams potassium sulfate to your 100 mL.

Molarity = moles of solute/Liters of solution Get moles of Fe2O3 160.0 grams Fe2O3 (1 mole Fe2O3/159.7 grams) = 1.0 mole Molarity = 1.0 mole/1.0 liters = 1.0 M Fe2O3

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

31 grams HCl (1mol/36.458g ) = 0.850 moles 500cm^3 = 0.50 Liters Molarity = moles of solute/liters of solution Molarity = 0.850mol/0.50L = 1.7 Molarity

1.5 moles

3.33

0.33 M NaOH solution

Need to find moles NaCl. 526 grams NaCl (1 mole NaCl/58.44 grams) = 9.0 moles NaCl --------------------------------now, Molarity = moles of solute/Liters of solution or, for our purposes Liters of solution = moles of solute/Molarity Liters of solution = 9.0 moles NaCl/3.0 M = 3.0 liters in volume ---------------------------------