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3 years ago

ASAP worth 15 points also

Chemistry

1 answer:

Taya2010 [7]3 years ago

7 0

The person above me is correct I took a test on this so it’s the right answer

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Blood alcohol concentration (BAC) is a measure of alcohol in your body, expressed as grams of alcohol per 100ml of blood. For a

ella [17]

Answer:

A 71 kg person will get a BAC of 0.05 when drinking 0.3442 L of beer

Explanation:

The mass of blood in a human body is approximately 8%, so if a person weighs 71 kg, the mass of blood would be:

71 kg * 8/100 = 5.68 kg of blood.

Using blood density, we can calculate the volume that 5.68 kg of blood occupies:

5.68 kg * $\frac{1m^{3}}{1025kg}$ = 0.0055415 m³

We convert m³ into mL, keeping the unit that we want to convert to in the numerator; and the unit that we want to convert in the denominator:

$0.0055415m^{3}*\frac{1000L}{1m^{3}} *\frac{1000mL}{1L}=5541.5mL$

Now we calculate the amount of alcohol that would be needed in the bloodstream to get a BAC of 0.05:

$\frac{5541.5mL}{100mL}*0.05g$ = 2.77 g of alcohol are needed in the bloodstream in order to have a BAC of 0.05

The amount of alcohol that needs to be ingested is higher than 2.77 g, due to the fact that only 17% of the alcohol goes into the bloodstream, so:

2.77 g $*\frac{100}{17} =$ 16.29 g of alcohol need to be ingested

Then we use the alcohol concentration of beer to calculate the volume of beer needed, using the alcohol density. First we convert the alcohol density to g/L, making sure the units that we want to convert cancel each other:

$789\frac{kg}{m^{3}}*\frac{1000g}{1kg} *\frac{1m^{3}}{1000L} =789g/L$

Now we use the density to calculate the litres of alcohol needed, keeping in mind that 16.29 g of alcohol are needed:

$16.29g*\frac{1L}{789g}=$ 0.02065 L of alcohol are needed.

Finally we calculate the litres of beer needed, keeping in mind the concentration of alcohol in beer:

$0.02065L_{alcohol}*\frac{100L_{beer}}{6L_{alcohol}} =$ 0.3442 L of beer are needed.

4 0

3 years ago

Most rocks contain more than 1 type of blank

mestny [16]

Answer:

hhhhhhhhhhh

Explanation:

hhhhhhhhhhhhh

4 0

2 years ago

Read 2 more answers

How many grams of N2 are in 44.8L at STP?

lutik1710 [3]

Answer: 10 i think

Explanation:

3 0

3 years ago

1. The statement that the first ionization energy for an oxygen atom is lower than the first ionization energy for a nitrogen at

Alex73 [517]

Answer:

too much questions

Explanation:

sry I can't answer u

8 0

3 years ago

Read 2 more answers

Calculate the mass of vanadium(V) oxide (V2O5) that contains a million (1.0 *10^6) vanadium atoms. Be sure your answer has a uni

sweet [91]

Answer : The mass of vanadium(V) oxide will be $1.51\times 10^{-16}g$

Explanation : Given,

Number of atoms of $V_2O_5$ = $1.0\times 10^{6}$

Molar mass of $V_2O_5$ = 181.88 g/mole

In $V_2O_5$ , there are 2 atoms of vanadium and 5 atoms of oxygen.

First we have to determine the moles of $V_2O_5$ .

As, $2\times 6.022\times 10^{23}$ number of vanadium atom present in 1 moles of $V_2O_5$

So, $1.0\times 10^{6}$ number of vanadium atom present in $\frac{1.0\times 10^{6}}{2\times 6.022\times 10^{23}}=8.3\times 10^{-19}$ moles of $V_2O_5$

Now we have to determine the mass of $V_2O_5$ .

$\text{Mass of }V_2O_5=\text{Moles of }V_2O_5\times \text{Molar mass of }V_2O_5$

$\text{Mass of }V_2O_5=(8.3\times 10^{-19}mole)\times (181.88g/mole)=1.51\times 10^{-16}g$

Therefore, the mass of vanadium(V) oxide will be $1.51\times 10^{-16}g$

4 0

4 years ago

Read 2 more answers