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allochka39001 [22]
2 years ago
11

A solution that contains many dissolved molecules in a fixed amount of solution is called

Chemistry
1 answer:
vfiekz [6]2 years ago
3 0

Answer:

  • <em>A solution that contains many dissolved molecules in a fixed amount of solution is called</em> <u>concentrated</u> (first choice)

Explanation:

Concentration is the term used to express the amount of solute in a solution. The concentration is a measure of how much solute is dissolved.

The term <em>concentrated </em>is a qualitative form to describe that a solution has a high concentration; this is, it <em>contains many dissolved molecules</em> in certain volume of solution).

The opposite to concentrated is <em>diluted </em>(choice 3). This is, a diluted solution contains a <em>small amount of solute</em> (molecules dissolved) in a fixed amount of solution.

The other two choices (2 and 4) use the terms <em>strong</em> and <em>weak</em>. Those terms refere to a special kind of solutions, acids and bases, and designate how much they ionize (dissociate) in the solution: a strong acid or base dissociates in a high percent, while a weak acid or base dissociates poorly.

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Gnoma [55]

Answer:

Option 2, 3, and also 6 btw

Explanation:

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2 years ago
Do you believe the research subjects in the original study were appropriate for this experiment
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4 0
3 years ago
Read 2 more answers
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
Match the prefixes. 1. 1 hex- 2. 2 eth- 3. 3 prop- 4. 4 hept- 5. 5 non- 6. 6 dec- 7. 7 pent- 8. 8 but- 9. 9 meth- 10. 10 oct-
erma4kov [3.2K]

Hi!

All these prefixes are commonly employed in the nomenclature in organic chemistry. These are associated with organic compounds, such as alkanes, alkenes, alcohols and carboxylic acids -and these prefixes are known as word root.

<h3>The answers would be:</h3><h3>a) 1 - meth</h3>

Meth is associated with the number one, in that it signifies the presence of one carbon in the compound in concern. For instance, a one carbon alkane would be known as methane -with <em>ane </em>being indicative of the compound being an alkane, and meth denoting the presence of one carbon

<h3>b) 2 - eth</h3>

Eth is the associated prefix with the number 2 as it denotes the presence of 2 carbon atoms in the compound. For instance, the compound ethene gets it name for having 2 carbon atoms (denoted by eth). The eth acts as a prefix to the characteristic <em>ene</em> of alkenes.

<h3>c) 3 - prop</h3>

Prop is the prefix that signifies the presence of 3 carbons in the compound in concern. For instance, propanol is a three carbon alcohol, which we can tell by the name. Prop indicating three carbons in the compound, and <em>anol</em> being indicative of the compound being an alcohol.

<h3>d) 4 - but</h3>

But is associated with the number one, in that it denotes the presence of 4 carbons in the compound. For instance, a 4 carbon carboxylic acid would be known as butanoic acid -with anoic<em> </em>being indicative of the compound being a carboxylic acid, and but signifying the presence of four carbons.

<h3>e) 5 - pent</h3>

Pent is the associated prefix with the number five as it denotes the presence of five carbon atoms in the compound. For instance, the compound pentene gets it name for having five carbon atoms (denoted by pent). The pent acts as a prefix to the characteristic <em>ene</em> of alkenes.

<h3>f) 6 - hex</h3>

Hex is the prefix that signifies the presence of six carbons in the compound in question. For instance, a six carbon alkane would be known as hexane -with <em>ane </em>being indicative of the compound being an alkane, and hex denoting the presence of one carbon

<h3>g) 7 - hept</h3>

Hept is the associated prefix with the number seven as it denotes the presence of seven carbon atoms in the compound. For instance, the compound heptene gets it name for having seven carbon atoms (denoted by hept). The hept acts as a prefix to the characteristic <em>ene</em> of alkenes.

<h3>h) 8 - oct</h3>

Oct is the prefix that is associated with the number 8. This is because it denotes the presence of eight carbon atoms in the compound. For instance, the compound octane gets it name for having eight carbon atoms (denoted by oct). The oct acts as a prefix to the characteristic <em>ane</em> of alkanes.

<h3>i) 9 - non</h3>

Non is the prefix that is associated with the number nine, and thus denotes the presence of nine carbon atoms in the compound. For instance, for alkanes, a nonane would be a nine carbon atom alkane, with non denoting the presence of nine carbon atoms, and <em>ane</em> being indicative that the compound it an alkane.

<h3>j) 10- dec</h3>

Dec is the prefix that is associated with the number nine, and thus denotes the presence of nine carbon atoms in the compound. For instance, for alkanes, a decane would be an alkane with ten carbon atoms, with dec denoting the presence of ten carbon atoms, and <em>ane</em> being indicative that the compound it an alkane.

<h3>Hope this helps!</h3>
4 0
3 years ago
100 points to the person who can answer these 4 questions
Vinil7 [7]

Answer:

....,................................

Explanation:

1= A

2=D

3=C

4=C

8 0
2 years ago
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