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Natalija [7]
3 years ago
8

Which molecule below has hydrogen bonding? Which molecule below has hydrogen bonding? H2 CH3CH2OH HCl CH4 all of the above

Chemistry
1 answer:
QveST [7]3 years ago
3 0

Answer:

CH_3CH_2OH

Explanation:

Hydrogen bonding:-

Hydrogen bonding is a special type of the dipole-dipole interaction and it occurs between hydrogen atom that is bonded to highly electronegative atom which is either fluorine, oxygen or nitrogen atom.

Partially positive end of the hydrogen atom is attracted to partially negative end of these atoms which is present in another molecule. It is strong force of attraction between the molecules.

Thus, hydrogen must be linked to electronegative atom which is oxygen, fluorine and nitrogen which is in CH_3CH_2OH and thus, it will shown hydrogen bonding.

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Name 3 different uses for nuclear technology
AlekseyPX

Agriculture, space exploration, and also for medical purposes.

Hope I helped :)

6 0
3 years ago
How many molecules of oxygen are needed to make 3.5 oxygens of water?​
max2010maxim [7]

Answer:

57.6g

Explanation:

So, if in one mole of water, 16 g of oxygen atom is present. Then, in 3.6 moles of water, the mass of oxygen present will be 3.6×16=57.6g. Therefore, the amount of oxygen present in 3.6 g water is option (B)- 57.6 g.

8 0
2 years ago
How many ml of 2.50 M NaOH solution are required to make a 525 mL of 0.150 M?
larisa [96]

Answer:

31.5 mL of a 2.50M NaOH solution

Explanation:

Molarity (M) is an unit of concentration defined as moles of solute (In this case, NaOH), per liter of solvent. That is:

Molarity = moles solute / Liter solvent

If you want to make 525mL (0.525L) of a 0.150M of NaOH, you need:

0.525L × (0.150mol / L) = <em>0.07875 moles of NaOH</em>

<em />

If you want to obtain these moles from a 2.50M NaOH solution:

0.07875mol NaOH × (1L / 2.50M) = 0.0315L = <em>31.5 mL of a 2.50M NaOH solution</em>

7 0
3 years ago
Read 2 more answers
Use the solubility generalizations on the information page to predict if one or more precipitates will form when aqueous solutio
neonofarm [45]

Answer:

3 (NH4)2SO4(aq) + 2 Al(NO3)3(aq) → 6 NH4NO3(aq) + Al2(SO4)3(aq)

Explanation:

In solubility rules, all ammonium and nitrates ions are solubles and all sulfates are soluble except the sulfates that are produced with Ca²⁺, Sr²⁺, Ba²⁺, Ag⁺ and Pb²⁺. That means the NH4NO3 and the Al2(SO4)3 produced are both <em>soluble and no precipitate is predicted. </em>

The reaction is:

<h3>3 (NH4)2SO4(aq) + 2 Al(NO3)3(aq) → 6 NH4NO3(aq) + Al2(SO4)3(aq)</h3>
6 0
3 years ago
What is the pressure in atmospheres of the gas remaining in the flask? Ignore the volume of solid NH4Cl produced by the reaction
Mashcka [7]

Answer:

a) HCl is the limiting reagent.

b) Mass of NH₄Cl formed = 6.68 g

c) Pressure of the gas remaining in the flask = 1.742 atm

Explanation:

The complete Question is presented in the attached image to this solution.

To solve this question, we first need to obtain the limiting regaent for this reaction.

The limiting reagent is the reagent that is in short supply in the reaction and is used up in the reaction. It determines the amount of products that will be formed and the amount of other reactants that will be required for the reaction.

NH₃ (g) + HCl (g) ⟶ NH₄Cl (s)

1 mole of NH₃ reacts with 1 mole of HCl

we first convert the masses of the gases available to number of moles.

Number of moles = (Mass/Molar Mass)

Molar mass of NH₃ = 17.031 g/mol, Molar mass of HCl = 36.46 g/mol

Number of moles of NH₃ = (4.55/17.031) = 0.2672 mole

Number of moles of HCl = (4.55/36.46) = 0.1248 mole

Since 1 mole of NH₃ reacts with 1 mole of HCl

It is evident that HCl is in short supply and is the limiting reagent.

NH₃ is in excess.

So, to calculate the amount of NH₄Cl formed,

1 mole of HCl gives 1 mole of NH₄Cl

0.1248 mole of HCl will also gove 0.1248 mole of NH₄Cl

Mass (Number of moles) × (Molar Mass)

Molar mass of NH₄Cl = 53.491 g/mol

Mass of NH₄Cl formed = 0.1248 × 53.491 = 6.68 g

c) The gas remaining in the flask is NH₃

0.1248 mole of NH₃ is used up for the reaction, but 0.2672 mole was initially available for reaction,

The amount of NH₃ left in the reacting flask is then

0.2672 - 0.1248 = 0.1424 mole.

Using the ideal gas Equation, PV = nRT

We can obtain the rrequired pressure of the remaining gas in the flask

P = Pressure = ?

V = Volume = 2.00 L

n = number of moles = 0.1424 mole

R = molar gas constant = 0.08205 L.atm/mol.K

T = absolute temperature in Kelvin = 25 + 273.15 = 298.15 K

P = (nRT/V)

P = (0.1424×0.08205×298.15/2) = 1.742 atm

Hope this Helps!!!

7 0
3 years ago
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