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Readme [11.4K]
3 years ago
7

"how many molecules of carbon dioxide exit your lungs when you exhale 5.00 × 10−2 mol of carbon dioxide, co2?"

Chemistry
2 answers:
IrinaVladis [17]3 years ago
5 0
We are going to use Avogadro's constant to calculate how many molecules of

carbons dioxide exist in lungs:

when 1 mole of CO2 has 6.02 x 10^23 molecules, so how many molecules in

CO2 when the number of moles is 5 x 10^-2

number of molecules = moles of CO2 * Avogadro's number

                                     = 5 x 10^-2  * 6.02 x 10^23

                                     = 3 x 10^22 molecules 

∴ There are 3 x 10^22 molecules in CO2 exist in lungs 
SpyIntel [72]3 years ago
3 0
The number of CO2  molecules  that   exit  is calculated   as   follows

by use  of Avogadro   law  constant

1   mole  = 6.02 x10^23  molecules
5.00 x10^-2 mol   =?   molecules

{( 6.02  x10^23)x (5.00  x10^-2) / 1}  = 3.01 x10^22  molecules

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Answer : The volume of hydrogen gas at STP is 4550 L.

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 100.0 atm

P_2 = final pressure of gas at STP = 1 atm

V_1 = initial volume of gas = 50.0 L

V_2 = final volume of gas at STP = ?

T_1 = initial temperature of gas = 27.0^oC=273+27.0=300K

T_2 = final temperature of gas at STP = 0^oC=273+0=273K

Now put all the given values in the above equation, we get:

\frac{100.0atm\times 50.0L}{300K}=\frac{1atm\times V_2}{273K}

V_2=4550L

Therefore, the volume of hydrogen gas at STP is 4550 L.

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3 years ago
Write the formula for the polyatomic ion in KOH. Express as an ion..
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KOH is a compound containing two ions, K+ and OH-. 
<span>The polyatomic ion present is OH- which is called hydroxide. </span>
<span>The compound is named potassium hydroxide.</span>
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3 years ago
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Mannoheptulose is a sugar found in avocados. If each C-C bond contains 76 kcal of energy and each C-H bond contains 91 kcal, how
nadya68 [22]

Answer:

Total Kcal energy produced in the catabolism of mannoheptulose = 1184 Kcal

Explanation:

The molecular formula of mannoheptulose is C₇H₁₄O₇.

The structure is as shown in the attachment below.

Number of C-C bonds present in mannoheptulose = 6

Number of C-H bonds present in mannoheptulose = 8

Since the each C-C bond contains 76 Kcal of energy,

Amount of energy present in six C-C bonds = 6 * 76 = 456 Kcal

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amount of energy present in eight C-H bonds = 8 * 91 = 728 Kcal

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3 years ago
A sample of oxalic acid (a diprotic acid of the formula H2C2O4) is dissolved in enough water to make 1.00 L of solution. A 100.0
OleMash [197]

<u>Answer:</u> The mass of original oxalic acid sample is 6.75 grams

<u>Explanation:</u>

To calculate the concentration of acid, 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_2C_2O_4

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

We are given:

n_1=2\\M_1=?M\\V_1=100.0mL\\n_2=1\\M_2=0.750M\\V_2=20.0mL

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2\times M_1\times 100.0=1\times 0.750\times 20.0\\\\M_1=\frac{1\times 0.750\times 20.0}{2\times 100.0}=0.075M

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Volume of solution = 1.00 L

Putting values in above equation, we get:

0.075M=\frac{\text{Mass of oxalic acid}}{90g/mol\times 1L}\\\\\text{Mass of oxalic acid}=(0.075\times 90\times 1)=6.75g

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