Chemistry

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

124,9 g grams of ammonium carbonate are needed.

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## Related Questions

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

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

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

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 )

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

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

Need moles HCl first.9.63 grams HCl (1 mole HCl/36.458 grams) = 0.2641 moles HClMolarity = moles of solute/Liters of solutionMolarity = 0.2641 moles HCl/1.5 Liters= 0.176 M HCl solution=================

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

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.

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

Get moles silver nitrate. 255 grams AgNO3 (1 mole AgNO3/169.91 grams) = 1.5008 moles AgCO3 --------------------------------Now; Molarity = moles of solute/Liters of solution ( 1500 ml = 1.5 Liters ) Molarity = 1.5008 moles AgNO3/1.5 Liters = 1.00 M AgNO3 ---------------------

The formula unit for the usual form of solid ammonium carbonate is (NH4)2CO3.H2O. This formula shows that each formula unit contains two atoms of nitrogen. Because nitrogen forms diatomic molecules at standard temperature and pressure, the number of moles of nitrogen is therefore the same as the number of formula units of ammonium carbonate, stated to be 650. The gram formula unit mass of this solid ammonium carbonate is 114.10. Therefore, 114.10(650) or 7.42 X 103 grams of the solid, to the justified number of significant digits, will be required.

Use liters to match other units. So. 100 ml = 0.1 liters Molarity = moles of solute/ volume ( liters in this case ) of solution 1 M NaCl = moles NaCl/0.1 liters = 0.1 moles NaCl (58.44 grams/1 mole NaCl) = 5.844 grams of NaCl needed to make solution.

Molarity = moles of solute/Liters of solution 0.343 M NaOH = moles NaOH.2.5 Liters = 0.8575 moles NaOH (39.998 grams/1 mole NaOH) = 34 grams of NaOH ----------------------------

There are 23.5 grams of solute contained for every 1,000,000 liters of solution that contains 21.7 ppm of SnC12. Therefore, there is 0.00008365 grams of solute in 3.5 liters of solution.&Ecirc;

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

Need moles aluminum oxide first. 51 grams Al2O3 (1 mole Al2O3/101.96 grams) = 0.5002 moles Al2O3 ======================Now, Molarity = moles of solute/Liters of solution (500 ml = 0.500 Liters ) Molarity =0.5002 moles Al2O3/0.500 Liters = 1.0 M Al2O3 solution ----------------------------

Molarity = moles of solute/Liters of solution ( 75.0 ml = 0.075 Liters ) Get moles NaBr 1.5 M NaBr = moles NaBr/0.075 Liters = 0.1125 moles NaBr (102.89 grams/1 mole NaBr) = 11.575 grams NaBr ( call it 12 grams ) ----------------------------------------------------

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