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d1i1m1o1n [39]
4 years ago
15

3% Hydrogen Peroxide has an oral LD50 of 900 mg/kg. Acetic Acid has an oral LD50 of 3310 mg/kg. Which one is more hazardous to c

onsume?
A.They are both toxic
B. Hydrogen Peroxide
Chemistry
1 answer:
Lina20 [59]4 years ago
4 0
Answer is: hydrogen peroxide <span>is more hazardous to consume.
</span>Median lethal dose<span>, </span><span>LD50</span><span> (</span>lethal dose, 50%) <span>is a measure of the </span>lethal dose<span> of a </span>toxin. LD50<span> used as a general indicator of a substance's </span>acute toxicity, lower LD50<span> is indicative of increased toxicity (in this case </span>900 mg/kg is lower than <span>3310 mg/kg).</span>
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Licemer1 [7]

Answer:

an element is not in an atom

Explanation:

because the only things in a atom are electrons neutrons and protons

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A 0.75M solution of CH3OH is prepared in 0.500 kg of water. How many moles of CH3OH are needed?
4vir4ik [10]

Answer:

We need 0.375 mol of CH3OH to prepare the solution

Explanation:

For the problem they give us the following data:

Solution concentration 0,75 M

Mass of Solvent is 0,5Kg

knowing that the density of water is 1g / mL,  we find the volume of water:

                           d = \frac{g}{mL} \\\\ V= \frac{g}{d}  = \frac{500g}{1 \frac{g}{mL} } = 500mL = 0,5 L

Now, find moles of CH_{3} OH are needed using the molarity equation:

                           M = \frac{ moles }{ V (L)} \\\\\\molesCH_{3}OH  = M . V(L) = 0,75 M . 0,5 L\\\\molesCH_{3}OH = 0,375 mol

therefore the solution is prepared using 0.5 L of H2O and 0.375 moles of CH3OH,  resulting in a concentration of 0,75M

5 0
3 years ago
Dinitrogen tetraoxide, N2O4, decomposes to nitrogen dioxide, NO2, in a first-order process. If k = 1.5 x 103 s-1 at 5 ºC and k =
Arada [10]

Answer:

The activation energy for the decomposition = 33813.28 J/mol

Explanation:

Using the expression,

\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )

Wherem  

k_1\ is\ the\ rate\ constant\ at\ T_1

k_2\ is\ the\ rate\ constant\ at\ T_2

E_a is the activation energy

R is Gas constant having value = 8.314 J / K mol  

Thus, given that, E_a = ?

k_2=4.0\times 10^3s^{-1}

k_1=1.5\times 10^3s^{-1}  

T_1=5\ ^0C  

T_2=25\ ^0C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (5 + 273.15) K = 278.15 K  

T = (25 + 273.15) K = 298.15 K  

T_1=278.15\ K

T_2=298.15\ K

So,

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E_a=-\ln \frac{1.5\times \:10^3}{4.0\times \:10^3}\:\times \frac{8.314}{\left(\frac{1}{278.15}-\frac{1}{298.15}\right)}

E_a=-\frac{8.314\ln \left(\frac{1.5\times \:10^3}{4\times \:10^3}\right)}{\frac{1}{278.15}-\frac{1}{298.15}}

E_a=-\frac{689483.53266 \ln \left(\frac{1.5}{4}\right)}{20}

E_a=33813.28\ J/mol

<u>The activation energy for the decomposition = 33813.28 J/mol</u>

8 0
3 years ago
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Found the choices. Pls see attachment. 

The statements that explains this phenomenon are:
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