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icang [17]
3 years ago
12

Guys please help me...

Chemistry
1 answer:
valentina_108 [34]3 years ago
7 0

Answer:

On the farm there are 70 chickens.

Explanation:

We can directly <u>work out the number of ducks from the given number of pigs</u>, with the <em>ratio of ducks to pigs</em>:

  • 36 pigs * \frac{5Ducks}{9Pigs} = 20 ducks

Then we<u> work out the number of chickens</u>, using the ration of chickens to ducks:

  • 20 ducks * \frac{7Chickens}{2Ducks} = 70 chickens

If on the farm there are 36 pigs, there are 70 chickens as well.

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C because it is maximumly contracted
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Answer:

21091mg of aspirin the person need to consume

Explanation:

To solve this question we must find the mass of the person in kg. Knowing the lethal dose for aspirin is 400mg/kg of person, we can find the amount of aspirin that the person need to consume to get a lethal dose:

<em>Mass person:</em>

116lb * (1kg / 2.2lb) = 52.7kg

<em>Lethal dose:</em>

52.7kg * (400mg / kg) =

<h3>21091mg of aspirin the person need to consume</h3>
3 0
3 years ago
A 3.4 g sample of an unknown monoprotic organic acid composed of C,H, and O is burned in air to produce 8.58 grams of carbon dio
Pavlova-9 [17]

Answer:

C_7H_6O_2

Explanation:

Hello there!

In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.

(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:

n_C=8.58gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.195molC

And the moles of hydrogen due to the produced 1.50 grams of water:

n_H=1.50gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O}  =0.166molH

Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:

m_O=3.4g-0.195molC*\frac{12.01gC}{1molC}-0.166molH*\frac{1.01gH}{1molH} =0.89gO\\\\n_O=0.89gO*\frac{1molO}{16.0gO}=0.0556molO

Thus, the subscripts in the empirical formula are:

C=\frac{0.195}{0.0556}=3.5 \\\\H=\frac{0.166}{0.0556}=3\\\\O=\frac{0.0556}{0.0556}=1\\\\C_7H_6O_2

As they cannot be fractions.

(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:

n_{KOH}=0.279L*0.1mol/L\\\\n_{KOH}=0.0279mol

Which are equal to the moles of the acid:

n_{acid}=0.0279mol

And the molar mass:

MM_{acid}=\frac{3.4g}{0.0279mol} =121.86g/mol

(3) Finally, since the molar mass of the empirical formula is:

7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol

Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:

C_7H_6O_2

Best regards!

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