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Anna71 [15]
3 years ago
14

ILL GIVE POINTS TO WHOEVER HELPS!!!

Chemistry
1 answer:
kirza4 [7]3 years ago
4 0
Yeah no one is gonna read all that babe.
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If you are given an ideal gas with pressure (P) = 259,392.00 Pa and temperature (T) = 2.00 oC of 1 mole Argon gas in a volume of
3241004551 [841]

Answer:

 R = 0.064 dm³ atm K⁻¹ mol⁻¹

Explanation:

Answer:

Explanation:

Data Given:

volume of gas V = 8.8 dm³

no. of mole of gas (n) = 1 mole

Pressure P = 259,392.00 Pa  

Convert Pascal (Pa) to atm (atmospheric pressure)

As

101,325 Pascals = 1 atm

So,

259,392.00 Pa  = 2 atm

Then Pressure (P) = 2 atm

Temperature T = 2.00 °C  

change the temperature from °C to K

As  to convert °C to K the below formula used

                   0°C + 273.15 = 273.15K

So, for 2 °C

                    2°C  + 273.15 =  275.15 K

So,

Temperature T  = 275.15 K

ideal gas constant = ?

formula used for Ideal gases

                            PV = nRT

as we have to find R of the gas:

we will rearrange the ideal gas equation as below:

R = PV / nT ........................................... (1)

Put value in equation (1)

                 R = 2atm x 8.8 dm³ / 1 mole x 275.15 K

                  R = 17.6 atm. dm³ / 275.15 mol. K

                  R = 0.064 dm³ atm K⁻¹ mol⁻¹

So the value of R is 0.064 dm³ atm K⁻¹ mol⁻¹

and the unit of R (ideal gas constant) is dm³ atm K⁻¹ mol⁻¹

7 0
3 years ago
draw the lewis structure for CO2, H2CO3, HCO3-, and CO3 2-.Rank these in order of increasing attraction to water molecules. Expl
gavmur [86]

Answer:

The structures are attached in file.

Hydrogen bonding and intermolecular forces is the reason for ranks allotted.

Explanation:

In determining Lewis structure, we calculate the overall number of valence electrons available for bonding.  Making carbon (the least electronegative atom) the central atom in the structure, we allocate valence electrons until each atom has achieved stability.

In order of decreasing affinity to water molecules:

CO_{3}^{2-}  > HCO_{3} ^{2-} > H_{2} CO_{3}

This is due to the fact that the CO_{3}^{2-}will accept protons more readily than the bicarbonate ion, HCO_{3} ^{2-}. Carbonic acid, H_{2} CO_{3} will not accept any more protons, hence it is the least attractive to water molecule, even though soluble.

3 0
3 years ago
stbank, Question 075 Get help answering Molecular Drawing questions. Compound A, C6H12 reacts with HBr/ROOR to give compound B,
Law Incorporation [45]

Answer:

Explanation:

In this case we want to know the structures of A (C6H12), B (C6H13Br) and C (C6H14).

A and C reacts with two differents reagents and conditions, however both of them gives the same product.

Let's analyze each reaction.

First, C6H12 has the general formula of an alkene or cycloalkane. However, when we look at the reagents, which are HBr in ROOR, and the final product, we can see that this is an adition reaction where the H and Br were added to a molecule, therefore we can conclude that the initial reactant is an alkene. Now, what happens next? A is reacting with HBr. In general terms when we have an adition of a molecule to a reactant like HBr (Adding electrophyle and nucleophyle) this kind of reactions follows the markonikov's rule that states that the hydrogen will go to the carbon with more hydrogens, and the nucleophyle will go to the carbon with less hydrogen (Atom that can be stabilized with charge). But in this case, we have something else and is the use of the ROOR, this is a peroxide so, instead of follow the markonikov rule, it will do the opposite, the hydrogen to the more substituted carbon and the bromine to the carbon with more hydrogens. This is called the antimarkonikov rule. Picture attached show the possible structure for A. The alkene would have to be the 1-hexene.

Now in the second case we have C, reacting with bromine in light to give also B. C has the formula C6H14 which is the formula for an alkane and once again we are having an adition reaction. In this case, conditions are given to do an adition reaction in an alkane. bromine in presence of light promoves the adition of the bromine to the molecule of alkane. In this case it can go to the carbon with more hydrogen or less hydrogens, but it will prefer the carbon with more hydrogens. In this case would be the terminal hydrogens of the molecules. In this case, it will form product B again. the alkane here would be the hexane. See picture for structures.

8 0
3 years ago
Please give me the reasons of the solution!
irinina [24]

1. The answer is option E, that is None of the above is correct.

As a polymer becomes more crystalline,

its melting point doesn't decreases, its density doesn't decreases, its stiffness doesn't decreases and its yield stress doesn't decreases.

2. The answer is option B, that is the molecules are arranged in sheets, with their long axes parallel and their ends aligned as well.

In the smectic A liquid-crystalline phase, molecules are arranged in sheets, with their long axes parallel and their ends aligned as well.

3.  For a substitutional alloy to form, the two metals combined must have similar atomic radii and chemical bonding properties.

6 0
3 years ago
What is a warm front
Galina-37 [17]

Answer:the Boundary of an advancing mass of warm air, in particular the leading edge of the warm sector of a low-pressure system.

Explanation:

3 0
3 years ago
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