HNO3 does not react with water.It become diluted with water.
It is a neutralization.KNO3 formed with water. KOH+HNO3 -->KNO3+H2O
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
Nitric acid: HNO3 (acid) Sodium hydroxide: NaOH (base) This is therefore an acid-base reaction. Acid + Base --> Salt + Water Therefore: HNO3 + NaOH --> NaNO3 + H20 Or: Nitric acid + Sodium hydroxide --> Sodium Nitrate + Water
In water HNO3 forms hydronium ions (H3O+) and nitrate ions (NO3-)
Yes. HNO3 is an electrolyte. In water, it will dissolve into H+ ions and NO3- ions.
It is a neutralization.KNO3 formed with water. KOH+HNO3 -->KNO3+H2O
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
Nitric acid: HNO3 (acid) Sodium hydroxide: NaOH (base) This is therefore an acid-base reaction. Acid + Base --> Salt + Water Therefore: HNO3 + NaOH --> NaNO3 + H20 Or: Nitric acid + Sodium hydroxide --> Sodium Nitrate + Water
HNO3
In water HNO3 forms hydronium ions (H3O+) and nitrate ions (NO3-)
Yes. HNO3 is an electrolyte. In water, it will dissolve into H+ ions and NO3- ions.
HNO3 is an acid.Its property remains same in water.
HNO3 + H2O -> H3O + NO3 is very acidic. This is because HNO3 is a strong acid and almost completely dissociates in 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
If its soluble in water it is polar so yes!
NaNO3 and water
It produces Nitric acid(HNO3) and Nitrous acid(HNO2), when reacted with water