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Rom4ik [11]
3 years ago
7

If alcohol is ingested rapidly, the _____ will rise sharply within twenty minutes. blood pressure ABC LD50 BAC

Chemistry
1 answer:
GaryK [48]3 years ago
5 0
The Blood Alcohol Level (BAC) will rise. Because it usually peaks about 30 minutes after drinking, a sharp spike at 20 minutes makes sense. 

BAC.
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Using the formula M1V1 = M2V2 , how many milliliters of a 2.50 M hydrochloric acid solution is required to make 100.0 mL of a 0.
olga nikolaevna [1]

30ml of a 2.50 M hydrochloric acid solution is required to make 100.0 mL of a 0.750 M solution.

<h3>What is HCl?</h3>

HCl is an acid which is made up of hydrogen and chlorine gas.

By the formula of dilution

M1V1 = M2V2

where V1 = 2.50 m

M2 = 100.0

V2 = 0.750 m

M1 × 2.50 m = 100.0 × 0.750 m

M1 = 30 ml

Thus, the M1 is 30 ml option a is correct.

Learn more about HCl

brainly.com/question/3637432

8 0
2 years ago
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How are t cells and b cells same
tigry1 [53]
T cells and B cells are similar because they both deal with fighting of viruses.
5 0
3 years ago
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Question 4 (1 point)<br> If you have exactly one million atoms how many moles do you have?
rjkz [21]

Answer:

                       1.66 × 10⁻¹⁸  Moles

Explanation:

                     As we know one mole of any substance contains 6.022 × 10²³ particles (atoms, ions, molecules or formula units). This number is also called as Avogadro's Number.

The relation between Moles, Number of Atoms and Avogadro's Number is given as,

             Number of Moles  =  Number of Atoms ÷ 6.022 × 10²³ Atoms/mol

Putting values,

              Number of Moles  =  1.0 × 10⁶ Atoms ÷ 6.022 × 10²³ Atoms/mol

              Number of Moles  =  1.66 × 10⁻¹⁸  Moles

7 0
3 years ago
Part III.The two reactions involved in quantitatively determining the amount of iodate in solution are: IO3-(aq) 5 I-(aq) 6 H (a
slega [8]

Answer:

\large \boxed{\math{\dfrac{\text{6 mol thiosulfate}}{\text{1 mol iodate}}}}

Explanation:

The I₂ is the common substance in the two equations.

(1) IO₃⁻ + 5I⁻ + 6H⁺ ⟶ 3I₂ + 3H₂O

{2) I₂ + 2S₂O₃²⁻ ⟶ 2I⁻ + S₄O₆²⁻

From Equation (1), the molar ratio of iodate to iodine is

\dfrac{\text{I}_{2}}{\text{IO}_{3}^{-}} = \dfrac{3}{1}

From Equation (2), the molar ratio of iodine to thiosulfate is

\dfrac{\text{S$_{2}$O}_{3}^{2-}}{\text{I}_{2}} = \dfrac{2}{1}

Combining the two ratios, we get

\text{Stoichiometric factor} = \dfrac{\text{S$_{2}$O}_{3}^{2-}}{\text{IO}_{3}^{-}} = \dfrac{\text{S$_{2}$O}_{3}^{2-}}{\text{I}_{2}} \times \dfrac{\text{I}_{2}}{\text{IO}_{3}^{-}} = \dfrac{2}{1} \times \dfrac{3}{1} = \mathbf{\dfrac{6}{1}}\\\\\text{The stoichiometric factor is $\large \boxed{\mathbf{\dfrac{\text{6 mol thiosulfate}}{\text{1 mol iodate}}}}$}

7 0
3 years ago
Please help. im freaking out rn. i have like 40 missing assignments please
Katarina [22]

Answer:

I'm pretty sure its the one that says very little at the beginning but if I get it wrong I'm sorry

5 0
3 years ago
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