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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 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
What is the oxidation numbers of HNO3?
The oxidation numbers of each element in HNO3 are: H is +1, N is +5, and O is -2. The sum of the oxidation numbers in a neutral compound like HNO3 must equal zero.
How many atoms of zinc react with 1.49 g HNO3?
To find out how many atoms of zinc react with 1.49 g of HNO3, you would first need to balance the chemical equation for the reaction. Then, you can convert the mass of HNO3 to moles using its molar mass. Finally, use the coefficients in the balanced equation to determine the mole ratio between zinc and HNO3 to find the number of atoms of zinc involved in the reaction.
How do you prepare 0.01 M from 65 percent HNO3?
The molar mass of HNO3 is 63 gram/mol and of water is 18 gram/mol. Lets assume that the density of the HNO3 + water mixture is equal to the density of water at room temperature is 1000 gram/L. 0.01 M is equal to 0.01 mol HNO3 per liter HNO3 + water. Multiplying the molar mass of HNO3 by the molarity shows this is equal to 0.63 gram HNO3 per liter HNO3 + water. Using the density of water shows that this is equal to 0.000063 gram HNO3 per gram water. This is therefore 0.0065 percent by weight. (The assumption that the density of the mixture is approximately equal to that of pure water seems justified because the amount of HNO3 in the mixture is very low) Assuming we start with an initial 65 percent by weight of HNO3 (10.3 M using the same type of calculation), water will need to be added to dilute to 0.0065 percent (0.01M). To find the required amount of water that is added a total and component mass balance can be used. The total mass balance is given by: Mt = M0 + Ma Here, M0 is the initial amount of water + HNO3, Ma is the added amount of (pure) water and Mt is the total mass of the water + HNO3 after adding more water. The component mass balance over HNO3 is given by: xtMt = x0M0 + xaMa= x0M0 Where xt is the final weight fraction of HNO3 in water, x0 is the initial weight fraction in water and xa is the weight fraction of HNO3 in pure water (which is logically equal to zero). Lets say we have a starting mass M0 = 100 gram HNO3 + water with the initial weight fraction x0 = 0.65 gram HNO3 per gram and we want a final weight fraction xt = 0.000065, rearranging for the total weight after having added water to reach this dilution gives: Mt = x0M0/xt = 0.65x100/0.000065 = 1000000 gram HNO3 + water The amount of water which needs to be added then simply follows from: Ma = Mt - M0 = 1000000 - 100 = 999900 gram water
How do you find the molar mass of HNO3?
To find the molar mass of HNO3, you would add up the atomic masses of hydrogen (H), nitrogen (N), and three oxygen atoms (O) in one molecule of HNO3. The molar mass of HNO3 is approximately 63.01 grams per mole.
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 moles of molecules are there in 250g of hydrogen nitrate?
To find the number of moles in 250g of hydrogen nitrate (HNO3), we first need to determine the molar mass of HNO3. The molar mass of HNO3 is 63.01 g/mol. Then, we can calculate the number of moles by dividing the given mass by the molar mass: 250g / 63.01 g/mol = approximately 3.97 moles of HNO3 molecules.
What is the gram molecular mass of HNO3 is?
63.01 g mol-1
What is the molar mass of HNO3?
The molar mass of Ag NO3 is 169.8731 g/mol
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
What is the oxidation numbers of HNO3?
The oxidation numbers of each element in HNO3 are: H is +1, N is +5, and O is -2. The sum of the oxidation numbers in a neutral compound like HNO3 must equal zero.
An oxoacid of nitrogen with N in the OS plus 3?
Nitric acid = HNO3
How many atoms of zinc react with 1.49 g HNO3?
To find out how many atoms of zinc react with 1.49 g of HNO3, you would first need to balance the chemical equation for the reaction. Then, you can convert the mass of HNO3 to moles using its molar mass. Finally, use the coefficients in the balanced equation to determine the mole ratio between zinc and HNO3 to find the number of atoms of zinc involved in the reaction.
How do you prepare 0.01 M from 65 percent HNO3?
The molar mass of HNO3 is 63 gram/mol and of water is 18 gram/mol. Lets assume that the density of the HNO3 + water mixture is equal to the density of water at room temperature is 1000 gram/L. 0.01 M is equal to 0.01 mol HNO3 per liter HNO3 + water. Multiplying the molar mass of HNO3 by the molarity shows this is equal to 0.63 gram HNO3 per liter HNO3 + water. Using the density of water shows that this is equal to 0.000063 gram HNO3 per gram water. This is therefore 0.0065 percent by weight. (The assumption that the density of the mixture is approximately equal to that of pure water seems justified because the amount of HNO3 in the mixture is very low) Assuming we start with an initial 65 percent by weight of HNO3 (10.3 M using the same type of calculation), water will need to be added to dilute to 0.0065 percent (0.01M). To find the required amount of water that is added a total and component mass balance can be used. The total mass balance is given by: Mt = M0 + Ma Here, M0 is the initial amount of water + HNO3, Ma is the added amount of (pure) water and Mt is the total mass of the water + HNO3 after adding more water. The component mass balance over HNO3 is given by: xtMt = x0M0 + xaMa= x0M0 Where xt is the final weight fraction of HNO3 in water, x0 is the initial weight fraction in water and xa is the weight fraction of HNO3 in pure water (which is logically equal to zero). Lets say we have a starting mass M0 = 100 gram HNO3 + water with the initial weight fraction x0 = 0.65 gram HNO3 per gram and we want a final weight fraction xt = 0.000065, rearranging for the total weight after having added water to reach this dilution gives: Mt = x0M0/xt = 0.65x100/0.000065 = 1000000 gram HNO3 + water The amount of water which needs to be added then simply follows from: Ma = Mt - M0 = 1000000 - 100 = 999900 gram water
How many total atoms are required to prepare 60g of hno3?
To calculate the total number of atoms in 60g of HNO3, you need to convert the mass of HNO3 to moles using its molar mass. Then, you can use Avogadro's number (6.022 x 10^23 atoms/mol) to find the total number of atoms.
How many grams would be in four 4 moles of nitric acid HNO3?
The molar mass of nitric acid (HNO3) is 63.01 g/mol. To find the total grams in 4 moles, you would multiply the molar mass by the number of moles: 63.01 g/mol x 4 mol = 252.04 grams. So, there would be 252.04 grams in four moles of nitric acid (HNO3).
