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:
2.8 L
Explanation:
From the question given above, the following data were obtained:
Number of mole (n) = 0.109 mole
Pressure (P) = 0.98 atm
Temperature (T) = 307 K
Gas constant (R) = 0.0821 atm.L/Kmol
Volume (V) =?
The volume of the helium gas can be obtained by using the ideal gas equation as follow:
PV = nRT
0.98 × V = 0.109 × 0.0821 × 307
0.98 × V = 2.7473123
Divide both side by 0.98
V = 2.7473123 / 0.98
V = 2.8 L
Thus, the volume of the helium gas is 2.8 L.
Answer:
im gunna save my answer so no one can take it:)
Explanation:
I tried my best
The balanced chemical equation is:
2H2 + O2 ---> 2H2O
We are given the amount of the product produced from the reaction. This will be the starting point for the calculations.
355 g H2O ( 1 mol H2O/ 18.02 g H2O) ( 1 mol O2 / 2 mol H2O ) ( 32 g O2 / 1 mol O2 ) = 315.205 g O2
