Nitric acid is not a very strong acid. Adding distilled water will cause it to be more diluted. Its acidity does not change. However, to neutralize it, you do not need more alkali as compared to before adding the distilled water.
To solve this problem, we basically have 2 equations and 2 unknowns. The unknowns are the (volume of water) & the (volume of 70 wt%) nitric acid to add. * This problem will assume that you are interested in making 1 L (or 1000 mL) of 5 wt% nitric acid solution. Equation 1: (volume of water) + (volume of 70 wt% nitric acid) = 1000 mL Equation 2: mass of nitric acid / [mass of water + mass of 70 wt% nitric acid solution] = 0.05 (0.05 is 5 wt%) * Remember that mass = density * volume * Remember that 70 wt% nitric acid solution mean that for 100 grams (gm) of this acid, then there's 70 grams of HNO3 * Remember that density of 70 wt% nitric acid solution is 1.413 gm/cm^3 * Remember that density of water is 1 gm/cm^3 Equation 2 is now re-written as: [(density of 70 wt% nitric acid soln)*(volume of 70 wt% nitric acid)*0.70] / [(volume water)*(1gm*cm^3) + (volume of 70 wt% nitric acid)*(1.413gm/cm^3)] = 0.05 Solving for the 2 equations gives answer to the 2 unknowns: Answer: To make 1000 mL of 5 wt% nitric acid solution, add 1) 51.63 mL of 70 wt% nitric acid solution 2) 948.37 mL of water
Mixing equal parts of 40 volume and 10 volume developers will result in 30 volume.
Equal parts of 30V + 40V= 35V.
You would dissolve 1 part HNO3 into 99 parts of your solvent.
Equal parts of 10V. Developer + 20V. Developer= 15 Volume Developer
It must be distilled water but yes. You can combine equal parts distilled water and 40 volume to get 20 volume developer.
Equal parts distilled water & 40 vol peroxide.
Equal parts distilled water & 40 vol peroxide.
Yes: 3 parts hydrochloric acid (HCl) and 1 part nitric acid (HNO3)
If they are put together in equal parts, then they will equalize the PH level.
The trust gets divided into 100 equal parts and given to the beneficiaries.
To solve this problem, we basically have 2 equations and 2 unknowns. The unknowns are the (volume of water) & the (volume of 70 wt%) nitric acid to add. * This problem will assume that you are interested in making 1 L (or 1000 mL) of 5 wt% nitric acid solution. Equation 1: (volume of water) + (volume of 70 wt% nitric acid) = 1000 mL Equation 2: mass of nitric acid / [mass of water + mass of 70 wt% nitric acid solution] = 0.05 (0.05 is 5 wt%) * Remember that mass = density * volume * Remember that 70 wt% nitric acid solution mean that for 100 grams (gm) of this acid, then there's 70 grams of HNO3 * Remember that density of 70 wt% nitric acid solution is 1.413 gm/cm^3 * Remember that density of water is 1 gm/cm^3 Equation 2 is now re-written as: [(density of 70 wt% nitric acid soln)*(volume of 70 wt% nitric acid)*0.70] / [(volume water)*(1gm*cm^3) + (volume of 70 wt% nitric acid)*(1.413gm/cm^3)] = 0.05 Solving for the 2 equations gives answer to the 2 unknowns: Answer: To make 1000 mL of 5 wt% nitric acid solution, add 1) 51.63 mL of 70 wt% nitric acid solution 2) 948.37 mL of water
There are an infinite number of equal parts in the whole. If you divide something into four equal parts, there will be four parts in the whole. If you divide something into 100,000 equal parts, there will be 100,00 parts in the whole.
Aqua regia
Equal parts of 30V + 40V= 35V.
Mixing equal parts of 40 volume and 10 volume developers will result in 30 volume.
bisect is to divive into equal parts.