If the ratio of carbon isn't correctly formed it will either burn too quickly or too slowly and if the correct ratio of sulphur (or sulfur for america) isn't represented it will burn too hot or cold and wont be as effective..
A precipitate
75 parts Potassium Nitrate ( KNO3 or saltpeter/saltpetre, or nitre/niter). 10 parts Sulphur/sulphur (S) 15 parts Charcoal (C)
The chemical reaction is:KCl + AgNO3 = AgCl(s) + KNO3
1:2
The ratio between the speed of light in vacuum, c, and the speed at which light travels in a material, v, is called the refractive index of the material.The refractive index of air for visible light is 1.000293, so the speed of light in air is c / 1.000293 ≈ 299,705,000 m/s. The refractive index of glass, depending on the type of glass, for visible light is around 1.5, so the light in glass travels at c / 1.5 ≈ 200,000,000 m/s.
=Black Powder:==75 parts Potassium Nitrate ( KNO3 or saltpeter/saltpetre, or nitre/niter). 10 parts Sulphur/sulphur (S) 15 parts Charcoal (C)==Generally, around the world in the 1800s, Black Powder came to be standardized at 15/2/3 or 15/3/2 of KNO/S/C, both to the same effect.=
The balanced equation is as follows: KIO3 + AgNO3 --> KNO3 + AgIO3
KI(aq) + AgNO3(aq) --> AgI(s) + KNO3(aq) is a double replacement reaction.
C=blog(1+s/n)
It is my understanding that gunpowder is a mixture of Sulfur, Charcoal, and Saltpeter or aka Potassium Nitrate. The unbalanced equation would look like this-- KNO3(s)+C(s)+S(s)---->N2(g)+CO2(g)+K2S(s). So the answer is yes, gunpowder does produce CO2.
KNO3(s) -----> K+(aq) + NO3-(aq)
The reaction is the following:AgNO3 + KI = KNO3 = AgI(s)
A precipitate
The first part of this answer is algebraic; see the example below for how it works out in practice. Suppose you have a total quantity N that you want to divide in the ratio a:b:c. Add a, b and c that is, a+b+c = s. Calulate N/s, the value of each unit in the ratio. then a*(N/s), b*(N/s) and c*(N/s) are the required amounts. Example: Divide 60 sweets in the ratio 2:3:5 N = 60 s = 2+3+5 = 10 therefore N/s = 60/10 = 6. [Therefore, each 1 in the ratio is worth 6 sweets]. The required division of the sweets is 2*6, 3*6 and 5*6 = 12, 18 and 30. This process can be extended to dividing quantities into ratios comprising four or more numbers in an analogous fashion.
Instead of waiting for the answer, I ended up solving it. lol 300grams of H20 X 110grams of KNO3/100 grams of H20 = 330 grams of KNO3 The grams of H20 both cancel out and leave you with 330 grams of KNO3 Containing the solubility of KNO3. At 60 degrees celsius the KNO3 grams were 110, which is (over) / 100grams of H20. Hope this helps with this workbook problem :)
75 parts Potassium Nitrate ( KNO3 or saltpeter/saltpetre, or nitre/niter). 10 parts Sulphur/sulphur (S) 15 parts Charcoal (C)
AgNO3+K2CrO4->KNO3+AgCrO4will be2AgNO3+K2CrO4=>2KNO3+Ag2CrO4