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zlopas [31]
3 years ago
14

The average adult male has a total blood volume of 5.0 L. After drinking a few beers, he has a BAC of 0.10. What mass of alcohol

is circulating in his blood?
Chemistry
2 answers:
Amiraneli [1.4K]3 years ago
8 0

Answer:5 grams  of alcohol is circulating in his blood

Explanation:

BAC (Blood alcohol content) means that level of alcohol in the blood stream.

Given BAC of an adult male = 0.10 = 0.10 g/100 ml

Total blood volume = 5 L = 5000 ml

MAss of alcohol in 5000 ml of blood:

\frac{0.10 g}{100 ml}\times 5000 ml=5 g

5 grams  of alcohol is circulating in his blood

stellarik [79]3 years ago
7 0
  B A C ( Blood Alcohol Content ) of 0.10 means that there are 0.10 g of alcohol for every  dl  of blood.
  5 L = 50 dl
  50 * 0.10 g = 5 g
  In his blood is circulating 5 grams of alcohol. 
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<u>Explanation:</u>

To calculate the concentration of base, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is KOH

We are given:

n_1=2\\M_1=123\times 10^{-4}M=0.0123M\\V_1=50mL\\n_2=1\\M_2=0.00945M\\V_2=?mL

Putting values in above equation, we get:

2\times 0.0123\times 50=1\times 0.00945\times V_2\\\\V_2=130.16mL

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Why is a complete atom electrically neutral?
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The air bags in cars are inflated when a collision triggers the explosive, highly exothermic decomposition of sodium azide (NaN3
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Answer : The mass of NaN_3 required is, 166.4 grams.

Explanation :

First we have to calculate the moles of nitrogen gas.

Using ideal gas equation:

PV=nRT

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P = Pressure of N_2 gas = 1.00 atm

V = Volume of N_2 gas = 113 L

n = number of moles N_2 = ?

R = Gas constant = 0.0821L.atm/mol.K

T = Temperature of N_2 gas = 85^oC=273+85=358K

Putting values in above equation, we get:

1.00atm\times 113L=n\times (0.0821L.atm/mol.K)\times 358K

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Now we have to calculate the moles of sodium azide.

The balanced chemical reaction is,

2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)

From the balanced reaction we conclude that

As, 3 mole of N_2 produced from 2 mole of NaN_3

So, 3.84 moles of N_2 produced from \frac{2}{3}\times 3.84=2.56 moles of NaN_3

Now we have to calculate the mass of NaN_3

\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3

Molar mass of NaN_3 = 65 g/mole

\text{ Mass of }NaN_3=(2.56moles)\times (65g/mole)=166.4g

Therefore, the mass of NaN_3 required is, 166.4 grams.

3 0
3 years ago
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