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son4ous [18]
3 years ago
10

Intravenous saline solutions are often administered to patients in the hospital. Normal saline solution containers has a concent

ration of 0.20 M and a volume of 0.10L of solution. How many moles of sodium chloride were dissolved? *
2 points
0.02 mol
2.0 mol
0.5 mol
5 mol
Chemistry
1 answer:
zalisa [80]3 years ago
8 0

0.02 mol NaCl

Explanation:

(0.2M NaCl)×(0.10 L) = 0.02 mol NaCl

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ASAP MULTIPLE CHOICE
OleMash [197]
The mass of electrons , i just did the same test and got it right
7 0
3 years ago
A mixture of 75 mole% methane and 25 mole% hydrogen is burned with 25% excess air. Fractional conversions of 90% of the methane
son4ous [18]

Solution :

Consider a mixture of methane and hydrogen.

Take the basis as 100 moles of the mixture.

The mixture contains 75% of methane and 25% of hydrogen by mole and it is burned with 25% in excess air.

Moles of methane = 0.75 x 100

Moles of hydrogen = 0.25 x 100

The chemical reactions involved during the reaction are :

$CH_4+2O_2 \rightarrow CO_2 + 2H_2O$

$CH_4+1.5O_2 \rightarrow CO+2H_2O$

$H_2+0.5O_2 \rightarrow H_2O$

The fractional conversion of methane is 90%

Number of moles of methane burned during the reaction is = 0.9 x 75

                                                                                                   = 67.5

Moles of methane leaving = initial moles of methane - moles of methane burned

                                           = 75 - 67.5

                                           = 7.5 moles

Fractional conversion of hydrogen is 85%

The number of moles of hydrogen burned during the reaction is = 0.85 x 25

                                                                                                   = 21.25

Moles of hydrogen leaving = initial moles of hydrogen - moles of hydrogen burned

                                           = 25 - 21.25

                                           = 3.75 moles

Methane undergoing complete combustion is 95%.

$CO_2$ formed is = 0.95 x 67.5

                       = 64.125 moles

$CO$ formed is = 0.05 x 67.5

                       = 3.375 moles

Oxygen required for the reaction is as follows :

From reaction 1, 1 mole of the methane requires 2 moles of oxygen for the complete combustion.

Hence, oxygen required is = 2 x 75

                                            = 150 moles

From reaction 3, 1 mole of the hydrogen requires 0.5 moles of oxygen for the complete combustion.

Hence, oxygen required is = 0.5 x 25

                                            = 12.5 moles

Therefore, total oxygen is = 150 + 12.5 = 162.5 moles

Air is 25% excess.

SO, total oxygen supply = 162.5 x 1.25 = 203.125 moles

Amount of nitrogen = $203.125 \times \frac{0.79}{0.21} $

                                = 764.136 moles

Total oxygen consumed = oxygen consumed in reaction 1 + oxygen consumed in reaction 2 + oxygen consumed in reaction 3

Oxygen consumed in reaction 1 :

1 mole of methane requires 2 moles of oxygen for complete combustion

 = 2 x 64.125

 = 128.25 moles

1 mole of methane requires 1.5 moles of oxygen for partial combustion

= 1.5 x 3.375

= 5.0625 moles

From reaction 3, 1 mole of hydrogen requires 0.5 moles of oxygen

= 0.5 x 21.25

= 10.625 moles.

Total oxygen consumed = 128.25 + 5.0625 + 10.625

                                        = 143.9375 moles

Total amount of steam = amount of steam in reaction 1 + amount of steam in reaction 2 + amount of steam in reaction 3

Amount of steam in reaction 1 = 2 x 64.125 = 128.25 moles

Amount of steam in reaction 2 = 2 x 3.375 = 6.75 moles

Amount of steam in reaction 3  = 21.25 moles

Total amount of steam = 128.25 + 6.75 + 21.25

                                     = 156.25 moles

The composition of stack gases are as follows :

Number of moles of carbon dioxide = 64.125 moles

Number of moles of carbon dioxide = 3.375 moles

Number of moles of methane = 7.5 moles

Number of moles of steam = 156.25 moles

Number of moles of nitrogen = 764.136 moles

Number of moles of unused oxygen = 59.1875 moles

Number of moles of unused hydrogen = 3.75 moles

Total number of moles of stack  gas

= 64.125+3.375+7.5+156.25+764.136+59.1875+3.75

= 1058.32 moles

Concentration of carbon monoxide in the stack gases is

$=\frac{3.375}{1058.32} \times 10^6$

= 3189 ppm

b).  The amount of carbon monoxide in the stack gas can be decreased by increasing the amount of the excess air. As the amount of the excess air increases, the amount of the unused oxygen and nitrogen in the stack gases will increase and the concentration of CO will decrease in the stack gas.  

6 0
3 years ago
Balance the following chemical equation <br> __Mg+__O2–&gt;__MgO
mario62 [17]

Answer:

2Mg^+ +O2 right arrow 2MgO

Explanation:

6 0
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Which is not a characteristic of a mineral?
Eduardwww [97]

Answer:

i also had this question:P

Explanation:

8 0
2 years ago
which pair shares the same empirical formula. c2h2 and c6h6, c2h2 and c2h4, ch2 and c6h6, ch and c2h4
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Answer:

Answer: CH₃ and C₂H₆ have same empirical formula.

Explanation:

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