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irakobra [83]
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?
Chemistry
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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Calculate the theoretical density of iron, and then determine the number of vacancies per cm3 needed for a BCC iron crystal to h
Ann [662]

Answer:

Therefore the theoretical density of iron is 7.877 g/cm³ .

Therefore the number of vacancy per cm³ is 3.27 \times 10^{20}

Explanation:

BCC structure contains 2 atoms per cell.

BCC : Body centered cubic.

Atomic mass of iron = 55.845 gram/mole.

Atomic mass of a chemical element is the mass of 1 mole of the chemical element.

Density: Density of a matter is the ratio of mass of the matter and volume of the matter.

Avogrdro Number is number of atoms per 1 mole.

Avogrdro Number= 6.023×10²³

Lattice parameter of iron = 2.866×10⁻⁸ cm

The theoretical density:

\rho=\frac{\textrm{(x atom/cell)} \times \textrm{(atomic mass)}}{\textrm{avogadro's number}\times \textrm{(lattice parameter)}^3}

 =\frac{2\times 55.845 }{(6.023\times 10^{23})\times (2.866\times 10^{-8})^3}   g/cm³      [x= 2,Since it is BCC structure]

=7.877 g/cm³

Therefore the theoretical density of iron is 7.877 g/cm³ .

Now we have to find out the number of unit cell of iron crystal having density  7.874 g/cm³.

\rho=\frac{\textrm{(x atom/cell)} \times \textrm{(atomic mass)}}{\textrm{avogadro's number}\times \textrm{(lattice parameter)}^3}

\Rightarrow 7.874 =\frac{x\times 55.845}{6.023\times 10^{23}\times (2.866\times 10^{-8})^3}

\Rightarrow x= \frac{7.874\times 6.023\times 10^{23}\times (2.866\times 10^{-8})^3}{55.845}

⇒ x =1.9923 atom/cell

Therefore the vacancy of atom per cell = (2- 1.9923)=0.0077

Vacancy\  per\ cm^3=\frac{\textrm{Vacancy per cm} ^3  }{\textrm{( lattice parameter)} ^3}

                          =\frac{0.0077}{(2.866\times 10^-8)^3}

                           =3.27 \times 10^{20}

Therefore the number of vacancy per cm³ is 3.27 \times 10^{20}

   

   

5 0
1 year ago
A sample of hydrate sodium carbonate has a mass of 71.5 g. after heating, the anhydrous sodium carbonate has a mass of 26.5 gram
drek231 [11]

Answer:

62.9%

Explanation:

Step 1: Given data

  • Mass of hydrate sodium carbonate (mNa₂CO₃.xH₂O): 71.5 g
  • Mass of anhydrous sodium carbonate (mNa₂CO₃): 26.5 g

Step 2: Calculate the mass lost of water

We will use the following expression.

mH₂O = mNa₂CO₃.xH₂O - mNa₂CO₃

mH₂O = 71.5 g - 26.5 g = 45.0 g

Step 3: Calculate the percent of water in the hydrate sodium carbonate

We will use the following expression.

%H₂O = mH₂O / mNa₂CO₃.xH₂O × 100%

%H₂O = 45.0 g / 71.5 g × 100%

%H₂O = 62.9%

3 0
2 years ago
The question is on the pic, thanks :)
Inessa05 [86]
It’s the BOA not the dog or kangaroo
8 0
1 year ago
Which compound could serve as a reactant in a neutralization reaction ?
Veseljchak [2.6K]
 The   compound that  could serve  as  a reactant   in the neutralization reaction is    H2SO4
     
  Explanation
   Neutralization reaction occur  between an acids  and a base. H2SO4             (  sulfuric acid)  is  a strong  acid.  It can be neutralized by  strong base such as NaOH ( sodium  hydroxide)

  Example of neutralization reaction is

2NaOH + H2SO4 → Na2SO4  + 2H2O
7 0
1 year ago
A bottle of antiseptic hydrogen peroxide (H2O2) is labeled 3.0% (v/v). How many mL H2O2 are in a 400.0 mL bottle of this solutio
PolarNik [594]

Answer:

= 12 mL H202

Explanation:

Given that, the concentration of H2O2 is given antiseptic = 3.0 % v/v

It implies that, 3ml  H2O2 is present in 100 ml of solution.

Therefore, to calculate the amount of H202 in 400.0 mL bottle of solution;

we have;

 (3.0 mL/ 100 mL) × 400 mL

= 12 mL H202

6 0
2 years ago
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