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4 years ago

In the reaction 2k + 2h2o → 2koh + h2, which compound has an element ratio of 2 : 1? h2o h2 k koh

1 answer:

4 0

The compound that has the element ratio 2:1 is 2H20.
In 2H2O, four atoms of hydrogen react with two hydrogen of oxygen to form molecules of water. Four divided by two will give two, that implies ratio 2;1. This implies that, two atoms of hydrogen will react with one atom of oxygen to give a molecule of water.<span />

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5.00 g of an unknown compound was dissolved in 250.0 mL of water. The unknown compound was found to be a non-electrolyte and the

Zielflug [23.3K]

Answer:

This unknown compound has a molar mass of 33.74g/mole

Explanation:

<u>Step 1:</u> Given data

The formula for the osmotic pressure is:

π = MRT = 14.5 atm

⇒ with M = Concentration of the solution

⇒ with R = gas constant = 0.08206 L atm / mol K

⇒ with T = absolute temperature (in Kelvin) = 298K

<u>Step 2:</u> Calculate concentration

M = π/ RT

M = 14.5 / (0.08206 * 298)

M = 0.593 M this is 0.593 moles per L

<u>Step 3:</u> Calculate number of moles

Since the volume is only 0.250 L, the amount of moles is:

0.593 * 0.250 L = 0.1482 moles

<u>Step 4:</u> Calculate molar mass

The molar mass is : mass / moles

Molar mass of this unknown compund = 5g / 0.1482 moles = 33.74 g/mole

This unknown compound has a molar mass of 33.74g/ mole

4 0

3 years ago

Which reactant will be the best reactant for a nucleophilic aromatic substitution?

tangare [24]

The reactant that will be the best reactant for a nucleophilic aromatic substitution is NO₂- NO₂. The correct option is b.

<h3>What is nucleophilic aromatic substitution?</h3>

Nucleophilic aromatic substitution is a substitution process of nucleophile substance is substituted by halides in an aromatic ring. Aromatic compounds contain this type of substitution.

In option b, the compound is the one nitroxide group substituted by halogen, that is fluorine. The fluorine group is substituted in these given aromatic compounds.

Thus, the correct option is b, NO₂- NO₂.

To learn more about nucleophilic aromatic substitution, refer to the link:

brainly.com/question/28265482

#SPJ4

The question is incomplete. Your most probably complete question is given below:

NO₂F

NO₂- NO₂-F

CH₃-O

CH₃-O-F

4 0

2 years ago

What’s the partial pressure of carbon dioxide in a container that holds 5 moles of carbon dioxide, 3 moles of nitrogen, and 1 mo

satela [25.4K]

Answer:

(

)

0.58

atm

Explanation:

(

)

Total pressure

⋅

Mole fraction of CO

1.05

⋅

1.05 ⋅ 5/9 = 0.58 atm

Explanation:

7 0

3 years ago

Please help me find the answer , thank you.

USPshnik [31]

Answer:

the answeer is definitely silicon.

4 0

3 years ago

Consider the following molecules and the description of the bonding present in each: CH3CH2CH2CH3CH3CH2CH2CH3 (C−CC−C and C−HC−H

denis23 [38]

CH3CH2CH2CH3 < CH3CH2CHO < CH3CHOHCH3

Explanation:

Boiling point trend of Butane, Propan-1-ol and Propanal.

Butane is a member of the CnH2n+2 homologous series is an alkane. Alkanes have C-H and C-C bonds which have Van der waals dispersion forces which are temporary dipole-dipole forces (forces caused by the electron movement in a corner of the atom). This bond is weak but increases as the carbon chain/molecule increases.

In Propan-1-ol(Primaryalcohol), there is a hydrogen bond present in the -OH group. Hydrogen bond is caused by the attraction of hydrogen to a highly electronegative element like Cl-, O- etc. This bond is stronger than dispersion forces because of the relative energy required to break the hydrogen bond. Alcohols (CnH2n+1OH) also experience van der waals dispersion forces on its C-C chain and C-H so as the Carbon chain increases the boiling point increases in the homologous series.

Propanal which is an Aldehyde (Alkanal) with the general formula CnH2n+1CHO. This molecule has a C-O, C-C and C-H bonds only. If you notice, the Oxygen is not bonded to the Hydrogen so there is no hydrogen bond but the C-O bond has a permanent dipole-dipole force caused by the electronegativity of oxygen which is bonded to carbon. It also has van der waals dispersion forces caused by the C-C and C-H as the carbon chain increases down the homologous series. The permanent dipole-dipole forces are not as easy to break as van der waals forces.

In conclusion, the hydrogen bonds present in alcohols are stronger than the permanent dipole-dipole bonds in the aldehyde and the van der waals forces in alkanes (irrespective of the carbon chain in Butane). So Butane < Propanal < Propan-1-ol

3 0

3 years ago