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:
A)16.3% (Hexane's formula is C6H14)
Explanation:
first of all open the menu
To find the net ionic equation we must first write the balanced equation for the reaction. We must bear in mind that the reagents Ca(NO3)2 and Na2S are in the aqueous state and as product we will have CaS in the solid state, since it is not soluble in water and NaNO3 in the aqueous state.
The balanced equation of the reaction will be:
Ca(NO3)2(aq) + → Ca(aq) + 2Na(s)NO3Now, c(aq)ompounds in the aqueous state can be written in their ionic form, so the reaction will transform into:Na2S +
So, the answer will be option A
Answer:
0.35 milli moles of ethanol can be theoretically be produced under these conditions.
Explanation:
Moles of glucose = milli mole
Moles of ADP = 0.35 milli mole
Moles of Pi = 0.35 milli mole
Moles of ATP = 0.70 milli mole
As we can see that ADP and Pi are in limiting amount which means tat they are limiting reagent. So, the moles of ethanol produced will depend upon the moles of ADP and Pi.
According to reaction, 2 moles of ADP gives 2 moles of glucose.
Then 0.35 milli moles of ADp will give :
of ethanol
0.35 milli moles of ethanol can be theoretically be produced under these conditions.