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mamaluj [8]
3 years ago
10

Why are infants and children often more susceptible to the effects of toxic substances than adult? (Explain)

Chemistry
1 answer:
puteri [66]3 years ago
6 0

From what I know personally from learning about this type of subject in the past is that children are more susceptible to the effects of toxic substances more than adults are, certain scientific data specifically indicate that children actually are for all intents and purposes more basically susceptible to toxic exposure than adults because they for the most part have proportionally definitely more intake of food contaminants, basically active developmental processes, basically multiple exposure pathways and really susceptible kind of socio-behavioral activities in a particularly major way.

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What is the definition of the molar mass of an element or compound?
VashaNatasha [74]

Answer:

The molar mass is the mass of a given chemical element or chemical compound (g) divided by the amount of substance(mol) .

Explanation:

hope it may help you

mark as brainlist please

8 0
3 years ago
How many hours will it take for the concentration of methyl isonitrile to drop to 15.0 % of its initial value?
SpyIntel [72]

Answer:

The concentration  of methyl isonitrile will become 15% of the initial value after 10.31 hrs.

Explanation:

As the data the rate constant is not given in this description, However from observing the complete question  the rate constant is given as a rate constant of 5.11x10-5s-1 at 472k .

Now the ratio of two concentrations is given as

ln (\frac{C}{C_0})=-kt

Here C/C_0 is the ratio of concentration which is given as 15% or 0.15.

k is the rate constant which is given as 5.11 \times 10^{-5} \, s^{-1}

So time t is given as

ln (\frac{C}{C_0})=-kt\\ln(0.15)=-5.11 \times 10^{-5} \times t\\t=\frac{ln(0.15)}{-5.11 \times 10^{-5} }\\t=37125.6 s\\t=37125.6/3600 \\t= 10.31 \, hrs

So the concentration will become 15% of the initial value after 10.31 hrs.

7 0
3 years ago
Read 2 more answers
A 10.0 mL sample of 0.25 M NaOH(aq) is titrated with 0.10 M HCl(aq) (adding HCl to NaOH). Determine which region on the titratio
Anna11 [10]

Answer:

1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.

2) The pH of the solution after adding HCl is 12.6

Explanation:

10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.

nNaOH=\frac{0.25mol}{L} .10.0 \times 10^{-3} L=2.5 \times 10^{-3}mol

nHCl=\frac{0.10mol}{L} \times 15.0 \times 10^{-3} L=1.5 \times 10^{-3}mol

There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.

                    NaOH       +       HCl       ⇒       NaCl      +         H₂O

Initial          2.5 × 10⁻³         1.5 × 10⁻³               0                      0

Reaction    -1.5 × 10⁻³        -1.5 × 10⁻³          1.5 × 10⁻³          1.5 × 10⁻³

Final            1.0 × 10⁻³               0                 1.5 × 10⁻³          1.5 × 10⁻³

The concentration of NaOH is:

[NaOH]=\frac{1.0 \times 10^{-3} mol }{25.0 \times 10^{-3} L} =0.040M

NaOH is a strong base so [OH⁻] = [NaOH].

Finally, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log 0.040 = 1.4

pH = 14 - pOH = 14 - 1.4 = 12.6

5 0
3 years ago
What is the total probability of finding a particle in a one-dimensional box in level n = 4 between x = 0 and x = L/8?
Lubov Fominskaja [6]

Answer:

P = 1/8

Explanation:

The wave function of a particle in a one-dimensional box is given by:

\psi = \sqrt \frac{2}{L} sin(\frac{n \pi x}{L})

Hence, the probability of finding the particle in the  one-dimensional box is:

P = \int_{x_{1}}^{x_{2}} \psi^{2} dx

P = \int_{x_{1}}^{x_{2}} (\sqrt \frac{2}{L} sin(\frac{n \pi x}{L}))^{2} dx

P = \frac{2}{L} \int_{x_{1}}^{x_{2}} (sin^{2}(\frac{n \pi x}{L}) dx

Evaluating the above integral from x₁ = 0 to x₂ = L/8 and solving it, we have:

P = \frac{2}{L} [\frac{L}{16} (1 - 4\frac{sin(\frac{n \pi}{4})}{n \pi})]

P = \frac{1}{8} (1 - 4\frac{sin(\frac{n \pi}{4})}{n \pi})    

Solving for n=4:

P = \frac{1}{8} (1 - 4\frac{sin(\frac{4 \pi}{4})}{4 \pi})    

P = \frac{1}{8} (1 - \frac{sin (\pi)}{\pi})    

P = \frac{1}{8}

I hope it helps you!

7 0
3 years ago
Hi, can someone help me with this.
AleksAgata [21]
It is 0.5474 but you can put 0.5 hope this helps
4 0
3 years ago
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