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likoan [24]
3 years ago
10

Which produces the greatest number of ions when one mole dissolves in water?

Chemistry
1 answer:
Arada [10]3 years ago
7 0
~Hello there!

Your question: Which produces the greatest number of ions when one mole dissolves in water?

Your answer: Na2SO4<span> can dissociate into three ions, whereas the other choices produce only two ions (or in the case of sucrose, none) when dissolved in water.

Any queries ^?

Happy Studying!
</span>
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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
Un globo lleno de helio tenia un volumen de 8.5 L en el suelo a 20°C y a una presión de 750 torr. Cuando se le soltó, el globo s
Ray Of Light [21]

Answer:

El volumen del gas era 12.95 L

Explanation:

Se relaciona la presión y el volumen mediante la ley de Boyle, que dice:

“El volumen ocupado por una determinada masa gaseosa a temperatura constante, es inversamente proporcional a la presión”

La ley de Boyle se expresa matemáticamente como:  P*V=k

Por otro lado, la Ley de Charles consiste en la relación que existe entre el volumen y la temperatura absoluta de una cierta cantidad de gas ideal, el cual se mantiene a una presión constante. Esta ley dice que cuando la cantidad de gas y de presión se mantienen constantes, el cociente que existe entre el volumen y la temperatura siempre tendrán el mismo valor:  

\frac{V}{T}=k

Por último, la Ley de Gay Lussac dice que la temperatura absoluta y la presión son directamente proporcionales. Es decir, cuando se mantiene todo lo demás constante, mientras suba la temperatura de un gas subirá también su presión. Y mientras la temperatura del gas baje, lo mismo ocurrirá con la presión:

\frac{P}{T}=k

Combinado las mencionadas tres leyes se obtiene:

\frac{P*V}{T} =k

Cuando se desean estudiar dos diferentes estados, uno inicial y una final de un gas, se puede aplicar:

\frac{P1*V1}{T1} =\frac{P2*V2}{T2}

Recordando que la temperatura debe usarse en grados Kelvin, conoces los siguientes datos:

  • P1: 750 torr
  • V1: 8.5 L
  • T1: 20°C= 293°K (siendo 0°C=273°K)
  • P2: 425 torr
  • V2: ?
  • T2: -20°C= 253 °K

Reemplazando:

\frac{750 torr*8.5 L}{293K} =\frac{425 torr*V2}{253 K}

Resolviendo:

V2=\frac{750 torr*8.5 L}{293K} *\frac{253 K}{425 torr}

V2= 12.95 L

<u><em>El volumen del gas era 12.95 L</em></u>

<u><em></em></u>

5 0
2 years ago
7. Which of the following would lower the pitch of a string instrument's sound?
vfiekz [6]

The correct option is C.

The pitch of a string refers to the quality of sound that is produced by the string when it vibrates. There are three basic factors that affect the quality of pitch of a string, these are: the tension, the thickness of the string and the length of the string. The higher the thickness of the string, the lower the pitch of the string, thus, increasing the thickness of the string will lower the pitch.

3 0
3 years ago
Read 2 more answers
What is not within the scope of chemistry
ankoles [38]

hello!

your answer will be

The origin of the matter does not usually fall into chemistry

have a goed day

5 0
3 years ago
a mixture consisting of only of lithium chloride, lithium carbonate, and lithium nitrate was analyzed
Dafna11 [192]

A mixture consisting of only of lithium chloride, lithium carbonate, and lithium nitrate was analyzed by the help of concentric hydrochloric acid.

<h3>What is concentric hydrochloric acid?</h3>

The concentric hydrochloric acid is used for the analysis of organic and inorganic mixture like  lithium chloride, lithium carbonate, and lithium nitrate it releases fumes and cannot be touched with normal hands.

Always use a dropper to use it and for the chemical analysis.

Therefore, mixture consisting of only of lithium chloride, lithium carbonate, and lithium nitrate was analyzed by the help of concentric hydrochloric acid.

Learn more about concentric hydrochloric acid, here,

brainly.com/question/24586675

#SPJ1

6 0
1 year ago
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