The decomposition of so3 to so2 gas and o2 gas can be described in the balanced chemical equation:
2so3(g) = 2 so2 (g) + 02(g).
so assuming a complete reaction, the ratio of so2 gas to total products is 2/3 while that of 02 is 1/3.
Subtracting water's water vapor pressure, 760-40 mm hG = 720 mm Hg.
then the products partial pressures are
so2 = 2/3 * (720) = 480 mm Hg.
o2 = 720-480 = 240 mm Hg.
<span>Mr = 13 g / mol
mass = 5 g
Mol = 5/13 mol :)</span>
8 H2= 0.90 g , hope that helps
you with your work
Answer:
350 g dye
0.705 mol
2.9 × 10⁴ L
Explanation:
The lethal dose 50 (LD50) for the dye is 5000 mg dye/ 1 kg body weight. The amount of dye that would be needed to reach the LD50 of a 70 kg person is:
70 kg body weight × (5000 mg dye/ 1 kg body weight) = 3.5 × 10⁵ mg dye = 350 g dye
The molar mass of the dye is 496.42 g/mol. The moles represented by 350 g are:
350 g × (1 mol / 496.42 g) = 0.705 mol
The concentration of Red #40 dye in a sports drink is around 12 mg/L. The volume of drink required to achieve this mass of the dye is:
3.5 × 10⁵ mg × (1 L / 12 mg) = 2.9 × 10⁴ L
Answer:
1.88 × 10²² Molecules of CO
Explanation:
At STP for an ideal gas,
Volume = Mole × 22.4 L/mol
Or,
Mole = Volume / 22.4 L/mol
Mole = 0.7 L / 22.4 L/mol
Mole = 0.03125 moles
Now,
No. of Molecules = Moles × 6.022 × 10²³ Molecules/mol
No. of Molecules = 0.03125 × 6.022 × 10²³ Molecules/mol
No. of Molecules = 1.88 × 10²² Molecules of CO
