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adell [148]
3 years ago
11

What is the definition of to squeeze a gas into a smaller space

Chemistry
1 answer:
ollegr [7]3 years ago
6 0
I think the best word here would be compress. It is to compress that is defined as to squeeze a gas into a smaller space. Compression is the reduction of volume which cause an increase in pressure of the gas. Hope this answers the question. Have a nice day.
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5.25 ml of substance A has a mass of 3.9 g and 6.24 ml of substance B has a mass of 4.4 g. Which liquid is more dense?
Viktor [21]
Substance A because it weighs less
5 0
2 years ago
In a chemical reaction, water decomposes to form hydrogen and oxygen. What is the correct term for the water?
gulaghasi [49]
The answer is: [D]: a reactant.
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7 0
3 years ago
Read 2 more answers
Draw the lewis structure of the hypochlorite ion clo- . include lone pairs
Dmitry_Shevchenko [17]

Solution:- Hypochlorite ion ClO^- has one Cl and one O atom. Cl has 7 valence electrons and O has 6 valence electrons. Since there is one negative charge on the ion,

total valence electrons = 7 + 6 +1 = 14

(note:- if there is negative charge then it is added  and if there is positive charge then it is subtracted while calculating the valence electrons)

Both Cl and O atoms wants to complete their octet and so for this we put a single bond between them. Single bond means two electrons, so the remaining electrons would be 14 - 2 = 12

It means 12 electrons will be placed as lone pair of electrons. To complete the octet, we put 6 dots around each of the atom. Oxygen is more electron negative than Cl, so we show the -1 charge for oxygen.


6 0
3 years ago
The normal freezing point of water (H2O) is 0.00 oC and its Kf value is 1.86 oC/m. A nonvolatile, nonelectrolyte that dissolves
adelina 88 [10]

Answer:

3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C

Explanation:

One colligative property is the freezing point depression due the addition of a solute. The equation is:

ΔT=Kf*m*i

<em>Where ΔT is change in temperature = 0.400°C</em>

<em>Kf is freezing point constant of the solvent = 1.86°C/m</em>

<em>m is molality of the solution (Moles of solute / kg of solvent)</em>

<em>And i is Van't Hoff constant (1 for a nonelectrolyte)</em>

Replacing:

0.400°C =1.86°C/m*m*1

0.400°C / 1.86°C/m*1 = 0.215m

As mass of solvent is 280.0g = 0.2800kg, the moles of the solute are:

0.2800kg * (0.215moles / 1kg) = 0.0602 moles of solute must be added.

The mass of ethylene glycol must be added is:

0.0602 moles * (62.10g / mol) =

3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C

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6 0
3 years ago
In an organic structure, you can classify each of the carbons as follows: Primary carbon (1o) = carbon bonded to just 1 other ca
Alla [95]

The question is incomplete, complete question is :

In an organic structure, you can classify each of the carbons as follows: Primary carbon (1°) = carbon bonded to just 1 other carbon group Secondary carbon (2°) = carbon bonded to 2 other carbon groups Tertiary carbon (3°) = carbon bonded to 3 other carbon groups Quaternary carbon (4°) = carbon bonded to 4 other carbon groups How many carbons of each classification are in the structure below? How many total carbons are in the structure? How many primary carbons are in the structure? How many secondary carbons are in the structure? How many tertiary carbons are in the structure? How many quaternary carbons are in the structure?

Structure is given in an image?

Answer:

There are 10 carbon atoms in the given structures out of which 6 are 1° , 1 is 2° , 2 are 3° and 1 is 4°.

Explanation:

Total numbers of carbon = 10

Number of primary carbons that is carbon joined to just single carbon atom = 6

Number of secondary carbons that is carbon joined to two carbon atoms = 1

Number of tertiary carbons that is carbon joined to three carbon atoms = 2

Number of quartenary carbons that is carbon joined to four carbon atoms = 1

So, there are 10 carbon atoms in the given structures out of which 6 are 1° , 1 is 2° , 2 are 3° and 1 is 4°.

3 0
3 years ago
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