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Cloud [144]
3 years ago
9

Red #40 has an acute oral LD50 of roughly 5000 mg dye/1 kg body weight. This means if you had a mass of 1 kg, ingesting 5000 mg

of Red #40 will have a 50% of killing you. Determine the amount of dye a 70 kg person would need to eat to reach his or her LD50. How many moles of dye would that be? The molar mass of Red #40 is 496.42 g/mol. Using the calculated concentration of Red #40 dye in the sports drink, how many of liters of sports drink would it take to reach this LD50?
Chemistry
1 answer:
FrozenT [24]3 years ago
8 0

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

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love history [14]

Answer:

density = 8.9 g/ml

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V = ... ?

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3 0
3 years ago
Can anyone please help me!! What is d. Beat the eggs?? A physical or a chemical change?
Stells [14]

Answer:

physical change

Explanation:

im not 100% but thats what i would pick because all you are doing is changing the state they are in. the eggs arnt chemically changing at all. i think ur overthinking it

6 0
3 years ago
Consider 0.01 m aqueous solutions of each of the following. a) NaI; b) CaCl2; c) K3PO4; and d) C6H12O6 (glucose) Arrange the sol
stealth61 [152]

Answer:

The solutions are ordered by this way (from lowest to highest freezing point):  K₃PO₄ < CaCl₂ < NaI < glucose

Option d, b, a and c

Explanation:

Colligative property: Freezing point depression

The formula is: ΔT = Kf . m . i

ΔT = Freezing T° of pure solvent - Freezing T° of solution

We need to determine the i, which is the numbers of ions dissolved. It is also called the Van't Hoff factor.

Option d, which is glucose is non electrolyte so the i = 1

a. NaI →  Na⁺  +  I⁻        i =2

b. CaCl₂ →  Ca²⁺  +  2Cl⁻      i =3

c. K₃PO₄ → 3K⁺ + PO₄⁻³     i=4

Potassium phosphate will have the lowest freezing point, then we have the calcium chloride, the sodium iodide and at the end, glucose.

7 0
3 years ago
An experimental spacecraft consumes a special fuel at a rate of 372 L/min. The density of the fuel is 0.730 g/mL and the standar
Karolina [17]

Explanation:

First, we will calculate fuel consumption is as follows.

         372 L/min \times 1000 ml/L \times 0.730 g/ml \times \frac{1}{60} min/s

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Now, we will calculate the power as follows.

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Thus, we can conclude that maximum power (in units of kilowatts) that can be produced by this spacecraft is 1.19 \times 10^{5} kW.

5 0
3 years ago
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user100 [1]

Answer:

The correct option is: When the amount of acid and base are equal

Explanation:

Titration is an analytic method that is used to determine the concentration of an<em> unknown solution</em>, called <em>titrand</em>.

In this method, standard solution of known concentration, called <em>titrant</em>, is taken in the burette and added drop-wise to the titrand solution in the flask, until the endpoint is reached.

In case of an acid-base titration, a <em>pH indicator</em> is used, which changes the color of the solution when the endpoint is reached.

<u>The </u><u>endpoint</u><u> indicates the </u><u>equivalence point</u><u> of an acid-base titration, where the </u><em><u>concentration of the acid and base is equal</u></em><u>. </u>

<u>Therefore, the </u><u>correct option</u><u> is: </u><u>When the amount of acid and base are equal</u>

3 0
3 years ago
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