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How many grams of HNO3 are in 2.6 mols of the compound?
[H=1,N=14,O=16] Gram molar weigt of HNO3=1+14+3x6 =15+48 =63g 1 mole ----------> 63 g 2.6 mole --------> 63x2.6 =163.8 g thus, 163.8g of HNO3 is present in 2.6 moles of compound
How many moles of HNO3 are present in 450 g of HNO3?
To determine the number of moles of HNO3 present in 450 g, we first need to calculate the molar mass of HNO3, which is approximately 63.01 g/mol. Next, we use the formula moles = mass/molar mass to find the number of moles. Therefore, 450 g of HNO3 is equal to 7.14 moles of HNO3.
How many grams of Cu would be needed to react with 2.0 mol HNO3?
The balanced chemical equation for the reaction between copper (Cu) and nitric acid (HNO3) is Cu + 4HNO3 -> Cu(NO3)2 + 2NO2 + 2H2O. From the equation, it can be seen that 1 mol of Cu reacts with 4 mol of HNO3. Therefore, to react with 2.0 mol of HNO3, 0.5 mol of Cu is needed. The molar mass of Cu is approximately 63.5 g/mol, so 0.5 mol of Cu would be equivalent to 31.75 grams.
What acid form RbOH plus HNO3?
The reaction of rubidium hydroxide (RbOH) with nitric acid (HNO3) would produce rubidium nitrate (RbNO3) and water (H2O).
How many grams are in 1 mole of nitric acid?
The molar mass of nitric acid (HNO3) is approximately 63 grams per mole.
What is the molarity of a solution dissolve 0.31 grams of HNO3 in 300ml of water?
Molarity = moles of solute/Liters of solution ( get moles of HNO3 and 300 ml = 0.300 Liters ) 0.31 grams Nitric acid (1 mole HNO3/63.018 grams) = 0.004919 moles HNO3 Molarity = 0.004919 moles HNO3/0.300 Liters = 0.0164 M HNO3
How many grams of HNO3 are in 2.6 mols of the compound?
[H=1,N=14,O=16] Gram molar weigt of HNO3=1+14+3x6 =15+48 =63g 1 mole ----------> 63 g 2.6 mole --------> 63x2.6 =163.8 g thus, 163.8g of HNO3 is present in 2.6 moles of compound
How many moles of HNO3 are present in 450 g of HNO3?
To determine the number of moles of HNO3 present in 450 g, we first need to calculate the molar mass of HNO3, which is approximately 63.01 g/mol. Next, we use the formula moles = mass/molar mass to find the number of moles. Therefore, 450 g of HNO3 is equal to 7.14 moles of HNO3.
How many grams of Cu would be needed to react with 2.0 mol HNO3?
The balanced chemical equation for the reaction between copper (Cu) and nitric acid (HNO3) is Cu + 4HNO3 -> Cu(NO3)2 + 2NO2 + 2H2O. From the equation, it can be seen that 1 mol of Cu reacts with 4 mol of HNO3. Therefore, to react with 2.0 mol of HNO3, 0.5 mol of Cu is needed. The molar mass of Cu is approximately 63.5 g/mol, so 0.5 mol of Cu would be equivalent to 31.75 grams.
What acid form RbOH plus HNO3?
The reaction of rubidium hydroxide (RbOH) with nitric acid (HNO3) would produce rubidium nitrate (RbNO3) and water (H2O).
Calculate the number of moles in 0.135 grams of HNO3. Choose the correct answer.?
2.14 x 10^(-3)
How many grams are in 1 mole of nitric acid?
The molar mass of nitric acid (HNO3) is approximately 63 grams per mole.
A nitric acid soulution is neutralized using sodium hydroxide how many grams of sodium hydroxide are needed to neutralize 2.50 L of 0.800 M Nitriric acid soulution?
Balanced equation first. HNO3 + NaOH >> NaNO3 + H2O Now, Molarity = moles of solute/volume of solution ( find moles HNO3 ) 0.800 M HNO3 = moles/2.50 Liters = 2 moles of HNO3 ( these reactants are one to one, so we can proceed to grams NaOH ) 2 moles NaOH (39.998 grams NaOH/1mole NaOH) = 79.996 grams of NaOH need to neutralize the acid. ( you do significant figures )
How many grams would be in four 4 moles of nitric acid HNO3?
Use the Equation Moles = mass(g) / Mr (Relative molecular mass). Algebraically rearrange mass(g) = Moles X Mr We have 4 moles Mr(HNO3) is From the atomic masses in the Periodi Table 1 x H = 1 x 1 = 1 1 x N = 1 x 14 = 14 3 x O = 3 x 16 = 48 1 + 14 + 48 = 63 Substitute in mass(g) = 4 x 63 mass(HNO3) = 252 grams. NB Bearing in mind that nitric acid is usually dissolved in water , the mass of 4 moles will be greater than 252 grams, in order to account for the mass of solvent(water).
What is the pH of a solution that contains 1.32 grams of nitric acid dissolved in 750 milliters of water?
Two steps. Find molarity of nitric acid and need moles HNO3.Then find pH. 1.32 grams HNO3 (1 mole HNO3/63.018 grams) = 0.020946 moles nitric acid ------------------------------------- Molarity = moles of solute/Liters of solution ( 750 milliliters = 0.750 Liters ) Molarity = 0.020946 moles HNO3/0.750 Liters = 0.027928 M HNO3 ----------------------------------finally, - log(0.027928 M HNO3) = 1.55 pH ==========( could call it 1.6 pH )
What happens when CaCO3 is added to HNO3?
When CaCO3 is added to HNO3, a chemical reaction occurs where CaCO3 reacts with HNO3 to produce Ca(NO3)2, CO2, and H2O. This reaction is a double displacement reaction where the calcium ions in CaCO3 switch places with the nitrate ions in HNO3.
How many moles of sodium hydroxide are required to neutralize 0.20 mol of nitric acid?
Balanced Equation. NaOH + HNO3 >> NaNO3 + H2O Now, Molarity = moles of solute/liters of solution 0.800M HNO3 + mol/2.50L mol of HNO3 = 2 2mol HNO3 (1mol NaOH/1molHNO3 )(39.998g NaOH/1mol NaOH ) = 79.996 grams
