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How many moles of NaHCO3 are present in 3.00 g sample?
To find the moles of NaHCO3 in a 3.00 g sample, first calculate the molar mass of NaHCO3 (84.01 g/mol). Then, divide the mass of the sample by the molar mass to obtain the moles of NaHCO3. For this sample, 3.00 g / 84.01 g/mol ≈ 0.036 moles of NaHCO3 are present.
What is 2.00 grams of baking soda in moles?
For this you need the atomic (molecular) mass of NaHCO3. Take the number of grams and divide it by the atomic mass. Multiply by one mole for units to cancel. NaHCO3=84.0 grams2.00 grams NaHCO3 / (84.0 grams) = .0238 moles NaHCO3
What mass of NaHCO3 would be needed to neutralize a spill consisting of 12.5 mL of 5.0 M HCl?
12.5 mL * 5.0 (m)mol/(m)L HCl = 62.5 mmol spilled HClneeds62.5 mmol NaHCO3 = 62.5 mmol * 84.01 (m)g/(m)mol NaHCO3 = 5250 mg NaHCO3 = 5.25 g pure NaHCO3
How many moles of hcl are needed to react with 0.35 g of NAHCO3?
The answer to the conversion is that 35.0 grams of hydrochloride (HCL) equals 0.76 moles. The conversion rate is 35.0 grams divided by 46 gram per mole. A mole is the molecular weight of a substance.
How many moles are contained in 86.6 grams in sodium bicarbonate?
Sodium Bicarbonate = NaHCO3 Molecular weight = Summation of individual molecular weights of elements =[(1)(22.99 g/mol) ] + [(1)(1.01 g/mol)] + [(1)(12.01 g/mol)] + [(3)(15.99 g/mol)] =84.01 g/molSo,(86.6 g) / (84.01 g/mol) = 1.031 mol
How many moles of NaHCO3 are present in 3.00 g sample?
To find the moles of NaHCO3 in a 3.00 g sample, first calculate the molar mass of NaHCO3 (84.01 g/mol). Then, divide the mass of the sample by the molar mass to obtain the moles of NaHCO3. For this sample, 3.00 g / 84.01 g/mol ≈ 0.036 moles of NaHCO3 are present.
How many moles are in 2.10 kg of NaHCO3?
For this you need the atomic (molecular) mass of NaHCO3. Take the number of grams and divide it by the atomic mass. Multiply by one mole for units to cancel. NaHCO3=84.0 grams110 grams NaHCO3 / (84.0 grams) = 1.31 moles NaHCO3
What is the solute mass of NaHCO3 if mol solute is 0.025 volume solution is 0.25 liters and molarity 0.100M?
All you need is the moles and the mass of NaHCO3 and you have the moles 0.025 moles NaHCO3 (84.008 grams/1 mole NaHCO3) = 2.1 grams of sodium bicarbonate ---------------------------------------------
What mass of NaHCO3 is needed to neutralize 150ml of 044M HCl?
Several part problem. Get molarity of NaHCO3. (150 ml)( M NaHCO3) = (150 ml)(0.44 M HCl) = 0.44 M NaHCO3 --------------------------- get moles NaHCO3 ( 150 ml = 0.150 Liters ) 0.44 M NaHCO3 = moles NaHCO3/0.150 Liters = 0.066 moles NaHCO3 ---------------------------------------get grams 0.066 moles NaHCO3 (84.008 grams/1 mole NaHCO3) = 5.54 grams NaHCO3 needed ---------------------------------------------answer
What is 2.00 grams of baking soda in moles?
For this you need the atomic (molecular) mass of NaHCO3. Take the number of grams and divide it by the atomic mass. Multiply by one mole for units to cancel. NaHCO3=84.0 grams2.00 grams NaHCO3 / (84.0 grams) = .0238 moles NaHCO3
What mass of NaHCO3 would be needed to neutralize a spill consisting of 12.5 mL of 5.0 M HCl?
12.5 mL * 5.0 (m)mol/(m)L HCl = 62.5 mmol spilled HClneeds62.5 mmol NaHCO3 = 62.5 mmol * 84.01 (m)g/(m)mol NaHCO3 = 5250 mg NaHCO3 = 5.25 g pure NaHCO3
If 2.10 g of NaHCO3 are decomposed then how many grams of water are produced?
NaHCO3 ----> H2O Mass 2.10g 0.045g RAM 84 g/moles 18 g/moles number of moles 0.025moles 0.025moles
How many moles of NaCl would form from 3.25 moles of NaHCO3?
To find the moles of NaCl formed from NaHCO3, we need to consider the stoichiometry of the reaction. The balanced equation is: 2 NaHCO3 -> Na2CO3 + H2O + CO2 Therefore, for every 2 moles of NaHCO3, we get 1 mole of NaCl. Therefore, 3.25 moles of NaHCO3 would produce 1.625 moles of NaCl.
How much carbon dioxide is produced when 336 g of sodium hydrogencarbonate breaks down completely?
honestly i am using this to pass my exams so plz send accurate awns ..... (⓿_⓿):P
How many moles of hcl are needed to react with 0.35 g of NAHCO3?
The answer to the conversion is that 35.0 grams of hydrochloride (HCL) equals 0.76 moles. The conversion rate is 35.0 grams divided by 46 gram per mole. A mole is the molecular weight of a substance.
How many moles are contained in 86.6 grams in sodium bicarbonate?
Sodium Bicarbonate = NaHCO3 Molecular weight = Summation of individual molecular weights of elements =[(1)(22.99 g/mol) ] + [(1)(1.01 g/mol)] + [(1)(12.01 g/mol)] + [(3)(15.99 g/mol)] =84.01 g/molSo,(86.6 g) / (84.01 g/mol) = 1.031 mol
Nahco3 ch3co2 na ch3co2 CO2 h20 what mass of nach3co2 can be obtained from mixing 15.0 grams of nahco3 with 125ml of 15 m ch3co2h?
To determine the mass of NaCH3CO2 obtained, first calculate the amount of CH3CO2H in moles using its molarity and volume. Then use the stoichiometry of the reaction to find the moles of NaHCO3 reacting. Finally, use the mole ratio between NaHCO3 and NaCH3CO2 to find the mass of NaCH3CO2 that can be obtained.
