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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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algol13

Answer:

P_2=135.73psi

Explanation:

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In this case we use the Boyle's law which allows us to understand the volume-pressure behavior as an inversely proportional relationship:

V_1P_1=V_2P_2

Whereas we solve for P_2 as the required final pressure:

P_2=\frac{V_1P_1}{V_2} =\frac{27.0L*14.7psi}{3.00L}\\ \\P_2=135.73psi

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7 0
3 years ago
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Would it be C???...idk<br> HELP
givi [52]

Answer:

C.  Butanal , is the aldehyde

Explanation:

A . It is carboxylic acid : ---COOH group

B. It is Ester : ----COOR group , Here R = CH3

C. It is Aldehyde : -----CHO group

D. It is ketone : ----C=O group

See image :

8 0
3 years ago
What superhero gained his powers by becoming a transgenic organism?
Karolina [17]

Answer:

B.spiderman

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7 0
2 years ago
How many moles of each reactant are needed to produce 3.60 x 10^2g ch3oh
zysi [14]
Methanol is prepared by reacting Carbon monoxide and Hydrogen gas,

                                   CO  +  2 H₂    →    CH₃OH

Calculating Moles of CO:
                                       According to equation,

             32 g (1 mole) of CH₃OH is produced by  =  1 Mole of CO
So,
             3.60 × 10² g of CH₃OH is produced by  =  X Moles of CO

Solving for X,
                       X  =  (3.60 × 10² g × 1 Mole) ÷ 32 g

                       X  =  11.25 Moles of CO

Calculating Moles of H₂:
                                       According to equation,

             32 g (1 mole) of CH₃OH is produced by  =  2 Mole of H₂
So,
             3.60 × 10² g of CH₃OH is produced by  =  X Moles of H₂

Solving for X,
                       X  =  (3.60 × 10² g × 2 Mole) ÷ 32 g

                       X  =  22.5 Moles of H₂

Result:
            3.60 × 10² g of CH₃OH
is produced by reacting 11.25 Moles of CO and 22.5 Moles of H₂.
3 0
3 years ago
2 C4H10 + 13 O2--&gt; 8 CO2 + 10 H2O
AleksAgata [21]

Answer:

4.14 x 10²⁴ molecules CO₂

Explanation:

2 C₄H₁₀ + 13 O₂ --> 8 CO₂ + 10 H₂O

To find the number of CO₂ molecules, you need to start with 100 grams of butane (C₄H₁₀), convert to moles (using the molar mass), convert to moles of CO₂ (using coefficients from equation), then convert to molecules (using Avagadro's number). The molar mass of C₄H₁₀ is calculated using the quantity of each element (subscript) multiplied by the number on the periodic table. The ratios should be arranged in a way that allows for units to be cancelled.

4(12.011g/mol) + 10(1.008 g/mol) = 58.124 g/mol C₄H₁₀

100 grams C₄H₁₀          1 mol C₄H₁₀             8 mol CO₂          
--------------------------  x  ----------------------  x  ---------------------  
                                        58.124 g              2 mol C₄H₁₀          

    6.022 x 10²³ molecules
x  ------------------------------------  =  4.14 x 10²⁴ molecules CO₂
              1 mol CO₂

7 0
2 years ago
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