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What mass of nitrogen in grams is present in 148 grams of ammonium sulfate?
Ammonium sulfate contains 21% nitrogen by mass. To find the mass of nitrogen in 148 grams of ammonium sulfate, you would first calculate 21% of 148 grams, which equals 31.08 grams of nitrogen.
How many moles of ammomium hydroxide are present in 0.475 g?
To find the number of moles of ammonium hydroxide, you need to know its molar mass. The molar mass of NH4OH is approximately 35.05 g/mol. By dividing the given mass by the molar mass, you can calculate that there are 0.0136 moles of ammonium hydroxide present in 0.475 grams.
How many grams of aluminum hydroxide are obtained from 17.2 grams of aluminum sulfide?
To determine the grams of aluminum hydroxide obtained from 17.2 grams of aluminum sulfide, we need to consider the stoichiometry of the reaction between aluminum sulfide and water to form aluminum hydroxide. Given the balanced chemical equation, we can calculate the molar mass of aluminum hydroxide and use it to convert the mass of aluminum sulfide to grams of aluminum hydroxide formed.
What is 1 mole of ammonium nitrate?
One mole of ammonium nitrate is equal to its molar mass, which is approximately 80.04 grams. This quantity represents Avogadro's number of individual ammonium nitrate molecules.
How many grams of sodium hydroxide are needed to make 250. ml of a 7.80 M solution?
To find the grams of sodium hydroxide needed, you first calculate the moles required using the molarity equation (moles = Molarity x Volume). Then, convert moles to grams using the molar mass of sodium hydroxide, which is 40 g/mol. Finally, the calculation would be: (7.80 mol/L) x 0.250 L x 40 g/mol = 78 grams of sodium hydroxide.
How many grams of ammonium carbonate would you need to make 2.10 liters of a 0.619 M solution?
To find the amount of ammonium carbonate needed, use the formula: grams = moles * molar mass. First, calculate moles using the given volume and molarity: moles = volume (L) * molarity (mol/L). Then, multiply the moles by the molar mass of ammonium carbonate (96.09 g/mol) to find the grams needed.
How many grams of water are needed to produce one hundred fifty grams of magnesium hydroxide?
A lot
400.0 grams of ammonia react with 56.0 moles of sulfuric acid to produce ammonium sulfate?
The reaction between 400.0 grams of ammonia and 56.0 moles of sulfuric acid will produce ammonium sulfate. To calculate the amount of product formed, you would need to determine the limiting reactant, which in this case would be sulfuric acid as it is in lower amount compared to ammonia. The balanced chemical equation for the reaction would be 2NH3 + H2SO4 → (NH4)2SO4.
A volume of 40.0g of aqueous potassium hydroxide was titrated against a standard solution of sulfuric acid. What was the molarity of the solution if 25.7g of 1.50M was needed?
The molarity of the potassium hydroxide solution is 3.30M. This can be calculated by determining the number of moles of sulfuric acid used (0.0171 mol), then equating this to twice the number of moles of potassium hydroxide used (0.00855 mol), and finally dividing this by the volume of the potassium hydroxide solution (0.00258 L).
What is the mass of ammonium in 2.171021 molecules of ammonium hydroxide?
A mole of substance contains 6.02 x 10^23 of particles. Hence, number of moles = number of particles / 6.02 x 10^23 Number of moles = 2.17 x 10^21 / 6.02 x 10^23 = 0.003604mol Each Ammonium Hydroxide(NH4OH) molecule has one Ammonium ion(NH4+) Therefore, there is also 0.003604mol of Ammonium ion. Ar of Nitrogen(N) = 14g/mol Ar of Hydrogen(H) = 1g/mol Mr of NH4+ = 14+4(1) = 18g/mol Using the formula : mass = number of moles x Mr mass = 0.003604mol x 18g/mol = 0.06488g
What is the percent sulfuric acid contained in concentrated sulfuric acid?
Concentrated sulfuric acid typically contains around 98% sulfuric acid by weight. This means that for every 100 grams of concentrated sulfuric acid, 98 grams are sulfuric acid and the remaining 2 grams are water.
What is the answer and how did you get it using the balanced equation from problem 4 Determine the limiting reagent if you start with 200 grams of sodium hydroxide and 100 grams of sulfuric acid?
The chemical equation is:2 NaOH + H2SO4 = Na2SO4 + 2 H2OMolar mass of sodium hydroxide is 39,9971 g; molar mass of sulfuric acid is 98,079 g.2 . 39,9971 g NaOH----------------------98,079 g H2SO4200 g NaOH------------------------xx = (200 x 98,079)/2 . 39,9971 = 245 g H2SO4So sulfuric acid is the limiting reagent.
The formula for sulfuric acid is HËÃSOËÅ A mole of sulfuric acid would have a mass of grams?
The formula for sulfuric acid is H₂SO₄. A mole of sulfuric acid would have a mass of approximately 98 grams.
What is the amount of nitrogen in ammonium chloride?
about 24.99% nitrogen is present in ammonium chloride........
How many grams of boric acid would be needed to neutralize 35 grams of calcium hydroxide?
To neutralize calcium hydroxide, the molar ratio is 2:1 (2 moles of boric acid for every 1 mole of calcium hydroxide). Calculate the molar mass of boric acid (H3BO3) and calcium hydroxide (Ca(OH)2), then use these values to convert the mass of calcium hydroxide to moles and then to grams of boric acid.
How many grams of sulfuric acid will neutralize 10.0 g of sodium hydroxide?
For every mole of sodium hydroxide, you need 1 mole of sulfuric acid for neutralization. The molar mass of sodium hydroxide (NaOH) is 40.0 g/mol and sulfuric acid (H2SO4) is 98.1 g/mol. So, to neutralize 40 g of NaOH (1 mole), you would need 98.1 g of H2SO4 (1 mole). Therefore, to neutralize 10.0 g of NaOH, you would need 24.53 g of H2SO4.
What mass of nitrogen in grams is present in 148 grams of ammonium sulfate?
Ammonium sulfate contains 21% nitrogen by mass. To find the mass of nitrogen in 148 grams of ammonium sulfate, you would first calculate 21% of 148 grams, which equals 31.08 grams of nitrogen.
