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Tanzania [10]
3 years ago
6

The air all around us is a mixture of gasses containing nitrogen, oxygen, carbon dioxide, argon

Chemistry
1 answer:
KiRa [710]3 years ago
4 0

Answer:

Total pressure = 4.57 atm

Explanation:

Given data:

Partial pressure of nitrogen = 1.3 atm

Partial pressure of oxygen = 1824 mmHg

Partial pressure of carbon dioxide = 247 torr

Partial pressure of argon = 0.015 atm

Partial pressure of water vapor = 53.69 kpa

Total pressure = ?

Solution:

First of all we convert the units other into atm.

Partial pressure of oxygen = 1824 mmHg / 760 = 2.4 atm

Partial pressure of carbon dioxide = 247 torr / 760 = 0.325 atm

Partial pressure of water vapor = 53.69 kpa / 101 = 0.53 atm

Total pressure = Partial pressure of N +  Partial pressure of O +  Partial pressure of CO₂ +  Partial pressure of Ar +  Partial pressure of water vapor

Total pressure = 1.3 atm + 2.4 atm + 0.325 atm + 0.015 atm + 0.53 atm

Total pressure = 4.57 atm

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Answer:

B is the answer

Explanation:

Because it a molecular mass of one

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72.40% iron,27.60% Oxygen Empirical formula
VARVARA [1.3K]

Explanation:

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72.40/ 56 27.60/16

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3 years ago
The distance between the centers of the two oxygen atoms in an oxygen molecule is 1.21 x 10-8 cm. What is this distance in inche
Sergio [31]

Answer:

d=4.75\times 10^{-9}\ \text{inches}

Explanation:

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We know that,

1 cm = 0.393 inches

We can solve it as follows :

1.21\times 10^{-8}\ cm=0.393\times 1.21\times 10^{-8}\\\\=4.75\times 10^{-9}\ \text{inches}

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5 0
2 years ago
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 64.0 mL of 0.0600 M EDTA . Titration of the excess unreacted EDTA req
tigry1 [53]

Answer:

the concentration of Cd^{2+}  in the original solution= 0.0088 M

the concentration of Mn^{2+} in the original solution = 0.058 M

Explanation:

Given that:

The volume of the sample  containing Cd2+ and Mn2+ =  50.0 mL; &

was treated with 64.0 mL of 0.0600 M EDTA

Titration of the excess unreacted EDTA required 16.1 mL of 0.0310 M Ca2+

i.e the strength of the Ca2+ = 0.0310 M

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Volume of newly freed EDTA = \frac{Volume\ of \ Ca^{2+}* Sample \ of \ strength }{Strength \ of EDTA}

= \frac{14.2*0.0310}{0.0600}

= 7.3367 mL

concentration of  Cd^{2+} = \frac{volume \ of \  newly  \ freed \ EDTA * strength \ of \ EDTA }{volume \ of \ sample}

= \frac{7.3367*0.0600}{50}

= 0.0088 M

Thus the concentration of Cd^{2+} in the original solution = 0.0088 M

Volume of excess unreacted EDTA = \frac{volume \ of \ Ca^{2+} \ * strength \ of Ca^{2+} }{Strength \ of \ EDTA}

= \frac{16.1*0.0310}{0.0600}

= 8.318 mL

Volume of EDTA required for sample containing Cd^{2+}   and  Mn^{2+}  = (64.0 - 8.318) mL

= 55.682 mL

Volume of EDTA required for Mn^{2+}  = Volume of EDTA required for

                                                                sample containing  Cd^{2+}   and  

                                                             Mn^{2+} --  Volume of newly freed EDTA

Volume of EDTA required for Mn^{2+}  = 55.682 - 7.3367

= 48.3453 mL

Concentration  of Mn^{2+} = \frac{Volume \ of EDTA \ required \ for Mn^{2+} * strength \ of \ EDTA}{volume \ of \ sample}

Concentration  of Mn^{2+} =  \frac{48.3453*0.0600}{50}

Concentration  of Mn^{2+}  in the original solution=   0.058 M

Thus the concentration of Mn^{2+} = 0.058 M

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Answer is: Peer review.

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4 0
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