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GrogVix
5 days 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:
lions [2.7K]5 days ago
7 0

Answer:

0.033g

Explanation:

Hello,

In this scenario, since 11 mg of cyanide per kilogram of body weight is lethal for 50% of domestic chickens, we can determine the lethal amount for a chicken weighing 3 kg using a proportion:

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

Thus, we find:

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

Which equates to grams as follows:

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

Regards.

You might be interested in
Compounds A and B are colorless gases obtained by combining sulfur with oxygen. Compound A results from combining 6.00 g of sulf
lorasvet [2668]

Answer:

1.5

Explanation:

It is given that:

Compound A and B originate from Sulfur + Oxygen.

Compound A:

6g sulfur + 5.99g Oxygen

Compound B:

8.6g sulfur + 12.88g oxygen

By comparing the ratios:

Compound A:

S: O = 6.00: 5.99

S/0 = 6.0g S / 5.99g O

Compound B:

S: O = 8.60: 12.88

S / O = 8.60g S / 12.88g O

The mass ratio of A and that of B

(6.0g S / 5.99g O) ÷ (8.60g S / 12.88g O)

(6.0 g S / 5.99g O) × (12.88g O / 8.60g S)

(6 × 12.88) / (5.99 × 8.60)

= 77.28 / 51.514

= 1.50017

= 1.5

4 0
13 days ago
An experimenter studying the oxidation of fatty acids in extracts of liver found that when palmitate (16:0) was provided as subs
eduard [2645]

Answer:

Fatty acids with an even number of carbons, like palmitate, undergo complete β-oxidation in the liver mitochondria, resulting in CO₂, as acetyl-CoA, their end product, can enter the TCA cycle.

On the other hand, odd-number fatty acids such as undecanoic acid generate acetyl-CoA and propionyl-CoA during their final pass. To allow entry into the TCA cycle, propionyl-CoA must go through additional processes, including carboxylation.

The conversion of CO2 and propionyl-CoA into methylmalonyl-CoA is facilitated by propionyl-CoA carboxylase, a biotin-dependent enzyme that is inhibited by avidin. In contrast, the oxidation of palmitate does not require carboxylation.

Explanation:

Fatty acids with an even number of carbons, such as palmitate, are completely oxidized to CO₂ in the liver mitochondria due to the ability of their oxidation product, acetyl-CoA, to enter the TCA cycle where it is further oxidized to CO₂.

Undecanoic acid is classified as an odd-number fatty acid, consisting of 11 carbon atoms. The last stage of β-oxidation for odd-number fatty acids, like undecanoic acid, produces a five-carbon fatty acyl substrate that is oxidized and split into acetyl-CoA and propionyl-CoA. To enter the TCA cycle, propionyl-CoA needs additional reactions such as carboxylation. Since the oxidation occurs using a liver extract, CO₂ must be supplied externally for propionyl-CoA carboxylation, enabling the complete oxidation of undecanoic acid.

The conversion of CO2 and propionyl-CoA into methylmalonyl-CoA is catalyzed by propionyl-CoA carboxylase, which contains biotin. The function of biotin is to activate CO₂ before it is transferred to the propionate group. The addition of avidin obstructs the complete oxidation of undecanoic acid as it binds very tightly to biotin, thereby hindering the activation and transfer of CO₂ to propionate.

In contrast, palmitate oxidation does not require carboxylation, meaning that the presence of avidin doesn't influence its oxidation.

6 0
5 days ago
How many moles of nitrogen gas are there in 6.8 liters at room temperature and pressure (293 K and 100 kPa)?
Alekssandra [2891]
Utilize the ideal gas law:

n = PV / RT

P = 100kPa = 100 x 1000 x (9.8 x 10^{-6}) = 0.98 atm
Convert kPa to atm, where 1 Pa = 9.8 x 10^{-6} atm.
T = 293 K
V = 6.8 L
R = 1/12
Substituting all values leads to:
n = 0.272
4 0
1 month ago
A 25.0-g sample of ice at -6.5oC is removed from the freezer and allowed to warm until it melts. Given the data below, select al
KiRa [2857]

Answer:

B, D

Explanation:

We need to recognize that the ice will rise in temperature from -6.5 ºC to 0 ºC for it to change into water.

Let's define q₁ as the heat needed to warm the ice to 0ºC, and q₂ as the heat for the transition from solid to liquid.

The calculation for q₁ is as follows:

q₁ = s x m x ΔT, where s represents the specific heat of ice (2.09 J/gºC), m is the mass, and ΔT is the temperature difference.

For q₂, the enthalpy of fusion is computed as:

q₂ = C x ΔT

with C indicating the specific heat for the phase transition, denoted as AH in kJ/mol.

All necessary data for computing q₁, q₂, and the total heat change (q₁ + q₂) is provided.

q₁ = 25.0 g x (2.09 J/gºC) x (0 - (-6.5 ºC))

q₁ = 339.6 J = 0.339 kJ

q₂ = (25 g/18 g/mol) x 6.02 kJ/mol = 1.39 x 6.02 kJ = 8.36 kJ

Combining these values gives us qtotal = 0.339 kJ + 8.36 kJ = 8.70 kJ.

Now we can answer the question:

(a) False, AH refers to the heat capacity during melting.

(b) True, as we concluded earlier.

(c) False, there’s only one phase transition from solid (ice) to liquid.

(d) True based on our calculations above.

(e) False, according to our findings.

7 0
1 month ago
5. Rubbing alcohol is a commonly used disinfectant and has a cooling effect when applied to the skin. The active ingredient in r
VMariaS [2860]

Answer:

The designation of 70% (vol/vol) indicates

that it contains 70% (vol/vol), meaning 70 ml of isopropanol is included in 100 ml of rubbing alcohol solution.

If it were 200 ml, then naturally, it would contain 70*2 = 140 ml of isopropanol required.

8 0
16 days ago
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