The source of alpha particles in a smoke detector is the Americium.
Can range anywhere from 1 AU to 10 AU
1) Calculate the number of moles in 1.15 liter of 0.100 M HNO3 solution.
M = n / V => n = M*V = 0.100M * 1.15 l = 0.115 moles
2) Calculate the mass of 0.115 moles of HNO3
mass = number of moles * molar mass
molar mass of HNO3 = 1.00 g/mol + 14.0 g/mol + 3*16.0g/mol = 63.0 g/mol
mass = 0.115 mol * 63.0g/mol = 7.245 g
3) Calculate the mass of 70.3% HNO3 solution that contains 7.245 grams of HNO3
% = (mass of solute / mass of solution) * 100
=> mass of solution = mass of solute * 100 / % = 7.245 g * 100 / 70.3%
mass of soltuion = 10.3 g.
4) Convert 10.3 grams of HNO3 solution into volume, using density, D
D =mass / Volume => Volume = mass / D
=. Volume = 10.3 g / 1.41 g/cm^3 = 7.30 cm^3
Answer: 7.30 cm^3
The simplest ratios compare only two values, but ratios comparing three or more values are also possible. In any situations in which two or more distinct numbers or quantities are being compared, ratios are applicable. By describing quantities in relation to each other, they explain how chemical formulas can be duplicated or recipes in the kitchen expanded.
Please let me know if I’m wrong
Answer:
identical conditions, separate samples of O2 and an unknown gas were allowed to effuse through identical membranes simultaneously. After a certain amount of time, it was found that 6.23 mL of O2 had passed through the membrane, but only 3.85 mL of of the unknown gas had passed through. What is the molar mass of the unknown gas
identical conditions, separate samples of O2 and an unknown gas were allowed to effuse through identical membranes simultaneously. After a certain amount of time, it was found that 6.23 mL of O2 had passed through the membrane, but only 3.85 mL of of the unknown gas had passed through. What is the molar mass of the unknown gas
Explanation:
identical conditions, separate samples of O2 and an unknown gas were allowed to effuse through identical membranes simultaneously. After a certain amount of time, it was found that 6.23 mL of O2 had passed through the membrane, but only 3.85 mL of of the unknown gas had passed through. What is the molar mass of the unknown gas