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2 years ago

7

11mg of cyanide per kilogram of body weight is lethal for 50% of domestic chickens. If a chicken weighs 3kg, how many grams of c

yanide would it need to ingest to kill 50% of domestic chickens?

1 answer:

yanalaym [24]2 years ago

7 0

Answer:

$0.033g$

Explanation:

Hello,

In this case, since 11 mg per kilogram of body weight has the given lethality, the mg that turn out lethal for a chicken weighting 3 kg is computed by using a rule of three:

$11mg\longrightarrow 1kg\\\\?\ \ \ \ \ \ \longrightarrow 3kg$

Thus, we obtain:

$?=\frac{3kg*11mg}{1kg}\\ \\?=33mg$

That in grams is:

$=33mg*\frac{1g}{1000mg} \\\\=0.033g$

Regards.

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A sample of H2SO4 contains 2.02 g of hydrogen, 32.07 g of sulfur, and 64.00 g of oxygen. How many grams of sulfur and grams of o

Gemiola [76]

From the chemical formula of sulfuric acid, we can see the molar ratio:

H : S : O
2 : 1 : 4

Now, we convert the mass of hydrogen given into the moles of hydrogen. This is done using

Moles = mass / Mr
Moles = 7.27 / 1
Moles = 7.27

Therefore, the moles will be:

S = 7.27 / 2 = 3.64 moles
O = 7.27 * 2 = 14.54 moles

Now, the respective masses are:
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2 years ago

A 0.2-mm-thick wafer of silicon is treated so that a uniform concentration gradient of antimony is produced. One surface contain

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Answer:

- 0.0249% Sb/cm

$-1.2465 * 10^9 \frac{atoms}{cm^3.cm}$

Explanation:

Given that:

One surface contains 1 Sb atom per 10⁸ Si atoms and the other surface contains 500 Sb atoms per 10⁸ Si atoms.

The concentration gradient in atomic percent (%) Sb per cm can be calculated as follows:

The difference in concentration = $\delta_c$

The distance $\delta_x$ = 0.2-mm = 0.02 cm

Now, the concentration of silicon at one surface containing 1 Sb atom per 10⁸ silicon atoms and at the outer surface that has 500 Sb atom per 10⁸ silicon atoms can be calculated as follows:

$\frac{\delta_c}{\delta_c} = \frac{(1/10^8 -500/10^8)}{0.02cm} *100%$

= - 0.0249% Sb/cm

b) The concentration $(c_1)$ of Sb in atom/cm³ for the surface of 1 Sb atoms can be calculated by using the formula:

$c_1 = \frac{(8 si atoms/unit cells)(1/10^3)}{(lattice parameter)^3/unit cell}$

Lattice parameter = 5.4307 Å; To cm ; we have

= $5.4307A^0* \frac{10^{-8}cm}{ A^0}$

$c_1 = \frac{(8 si atoms/unit cells)(1/10^8)}{(5.4307*10^{-8}cm)^3/unit cell}$

= $0.00499*10^{17}atoms/cm^3$

The concentration $(c_2)$ of Sb in atom/cm³ for the surface of 500 Sb can be calculated as follows:

$c_1 = \frac{(8 si atoms/unit cells)(500/10^8)}{(5.4307*10^{-8}cm)^3/unit cell}$

= $\frac{4*10^{-3}}{1.601*10^{-22}}$

= $2.4938*10^{17}atoms/cm^3$

Finally, to calculate the concentration gradient

$(\frac{\delta _c}{\delta_ x}) = \frac{c_1-c_2}{\delta_x}$

$(\frac{\delta _c}{\delta_ x}) = \frac{0.00499*10^{17}-2.493*10^{17}}{0.02}$

$= -1.2465 * 10^9 \frac{atoms}{cm^3.cm}$

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2 years ago

An electron is on a -2.5 eV energy level. The electron is struck by a 2.5 eV photon. What will most likely happen?

kati45 [8]

The correct answer would be C

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2 years ago

In redox half-reactions, a more positive standard reduction potential means I. the oxidized form has a higher affinity for elect

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Answer:

The 1st and 4th options are correct

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Explanation:

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Standard reduction potential which is also referred to as standard cell potential can be described as the potential difference that exist between cathode and anode of the cell. In the standard reduction potential most times the species will be reduced which is usually analysed in a reduction half reaction.

(Standard Hydrogen Electrode) is utilized when determining the Standard reduction or potentials of a chemical specie. this is because of Hydrogen having zero reduction and oxidation potentials, as a result of this a measured potential of any species is compared with that of Hydrogen, the difference helps to know the potential reduction of that particular specie.

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2 years ago

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STatiana [176]

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According to photoelectric equation:

$E=h\nu=\frac{hc}{\Lambda }$

E = Energy of the photon of light used to produce free chlorine atoms

$\nu$ = frequency of the light used to produce free chlorine atoms

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$40,318.83\times 10^{-23}J=\frac{hc}{\Lambda }=\frac{6.626\times 10^{-34} J.s\times 3\times 10^8 m/s}{\lambda }$

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The longest wavelength of light that will produce free chlorine atoms in solution is 493 nm.

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2 years ago