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Serjik [45]
3 years ago
12

Explain the sharing of electrons between a water molecule that forms four hydrogen bonds with the other four water molecules

Chemistry
1 answer:
Kazeer [188]3 years ago
8 0

Answer:

The sharing of electrons between a water molecule that forms four hydrogen bonds with the other four water molecules:

Explanation:

The hydrogen bond is a weak electrostatic force of attraction that exists between a covalently bonded H-atom and a highly electronegative atom like N,O or F.

In the case of the water molecule,

the highly electronegative atom is Oxygen and the intermolecular hydrogen bond in water is as shown below:

Thus H-bond is a weak electrostatic attraction formed between H-atom and O-atom in water.

You might be interested in
__ SnCl4 → __ Sn + __ Cl2
Kisachek [45]

Answer:

Sn (s) + 2 Cl2 (g) → SnCl4 (l)

This is an oxidation-reduction (redox) reaction:

4 Cl0 + 4 e- → 4 Cl-I

(reduction)

Sn0 - 4 e- → SnIV

(oxidation)

Cl2 is an oxidizing agent, Sn is a reducing agent.

Reactants:

Sn

Names: Tin source: wikidata, accessed: 2019-09-07source: ICSC, accessed: 2019-09-04source: NIOSH NPG, accessed: 2019-09-02, Sn source: wikidata, accessed: 2019-09-07, Element 50 source: wikidata, accessed: 2019-09-07

Appearance: White crystalline powder source: ICSC, accessed: 2019-09-04; Gray to almost silver-white, ductile, malleable, lustrous solid. source: NIOSH NPG, accessed: 2019-09-02

Cl2

Names: Chlorine source: ICSC, accessed: 2019-09-04source: NIOSH NPG, accessed: 2019-09-02, Molecular chlorine source: NIOSH NPG, accessed: 2019-09-02

Appearance: Greenish-yellow compressed liquefied gas with pungent odour source: ICSC, accessed: 2019-09-04; Greenish-yellow gas with a pungent, irritating odor. [Note: Shipped as a liquefied compressed gas.] source: NIOSH NPG, accessed: 2019-09-02

Products:

SnCl4 – Tetrachlorostannane source: wikipedia, accessed: 2019-09-28source: wikidata, accessed: 2019-09-02, Tin tetrachloride source: wikipedia, accessed: 2019-09-28source: wikidata, accessed: 2019-09-02source: ICSC, accessed: 2019-09-04, Tin(IV) chloride source: wikipedia, accessed: 2019-09-28

Other names: Stannic chloride source: wikipedia, accessed: 2019-09-28source: wikidata, accessed: 2019-09-02source: ICSC, accessed: 2019-09-04, Tin(iv) chloride (anhydrous) source: ICSC, accessed: 2019-09-04

Appearance: Colorless to slightly yellow fuming liquid source: wikipedia, accessed: 2019-09-28; Colourless or slightly yellow fuming liquid with pungent odour source: ICSC, accessed: 2019-09-04

6 0
2 years ago
Name the following aromatic hydrocarbon:
Tresset [83]

3-Methylpentane is the IUPAC name for the substance.

whether in a continuous chain or a ring, the longest chain of carbons joined by a single bond serves as the basis for IUPAC nomenclature. According to a precise set of priorities, all deviations—whether they involve numerous bonds or atoms other than carbon and hydrogen—are denoted by prefixes or suffixes.

+3-Methylpentane is the IUPAC name for the substance in question. It has a lengthy chain of 5 carbon atoms, which gives it the prefix pent-, and a single bond is what gives it the postfix -ane (alkane). Given that the methyl group is present at the third carbon, it is 3-methylpentane.

Learn more about IUPAC Nomenclature here-

brainly.com/question/14379357

#SPJ9

8 0
1 year ago
1) When 2.38g of magnesium is added to 25.0cm of 2.27 M hydrochloric acid, hydrogen gas is released.
Andrej [43]

Answer:

a. HCl.

b. 0.057 g.

c. 1.69 g.

d. 77 %.

Explanation:

Hello!

In this case, since the reaction between magnesium and hydrochloric acid is:

Mg+2HCl\rightarrow MgCl_2+H_2

Whereas there is 1:2 mole ratio between them.

a) Here, we can identify the limiting reactant as that yielded the fewest moles of hydrogen gas product via the 1:1 and 2:1 mole ratios:

n_{H_2}^{by\  HCl}=0.025L*2.27\frac{molHCl}{1L}*\frac{1molH_2}{2molHCl}  =0.0284molH_2\\\\n_{H_2}^{by\  Mg}=2.38gMg*\frac{1molMg}{24.3gMg}*\frac{1molH_2}{1molMg}=0.0979molH_2

Thus, since hydrochloric yields fewer moles of hydrogen than magnesium, we realize it is the limiting reactant.

b) Here, we use the molar mass of gaseous hydrogen (2.02 g/mol) to compute the mass:

m_{H_2}=0.0284molH_2*\frac{2.02gH_2}{1molH_2}=0.057gH_2

c) Here, we compute the mass of magnesium associated with the yielded 0.0248 moles of hydrogen:

m_{Mg}^{reacted}=0.0284molH_2*\frac{1molMg}{1molH_2}*\frac{24.3gMg}{1molMg}  =0.690gMg

Thus, the mass of excess magnesium turns out:

m_{Mg}^{excess}=2.38g-0.690g=1.69gMg

d) Finally, we compute the percent yield, considering 0.044 g is the actual yield and 0.057 g the theoretical yield:

Y=\frac{0.044g}{0.057g} *100\%\\\\Y=77\%

Best regards!

8 0
3 years ago
An 80.0-gram sample of a gas was heated from 25 °C to 225 °C. During this process, 346 J of work was done by the system and its
german
Hello!

First you need to calculate q 
<span>delta U is change in internal energy </span>

<span>delta U = q + w </span>
<span>q is heat and w work done </span>
<span>here work was done by the system means energy leaving the system so w is negative </span>

<span>delta U = q + w </span>

<span>q = delta U - w = 6865 J - (-346 J) = 7211 J = 7.211 KJ </span>

<span>q = m x c x delta T </span>

<span>7211 J = 80.0 g x c x (225-25) °C </span>

<span>c = 0.451 J /g °C
</span>
Hope this Helps! Have A Wonderful Day! :)
5 0
3 years ago
If you begin with 2.7 g Al and 4.05 g Cl2, what mass of AlCl3 can be produced?
Tresset [83]
<span>atomic weights: Al = 26.98, Cl = 35.45 In this reaction; 2Al = 53.96 and 3Cl2 = 212.7 Ratio of Al:Cl = 53.96/212.7 = 0.2537 that is approximately four times the mass Cl is needed. Step 2: (a) Ratio of Al:Cl = 2.70/4.05 = 0.6667 since the ratio is greater than 0.2537 the divisor which is Cl is not big enough to give a smaller ratio equal to 0.2537. so Cl is limiting (b)since Cl is the limiting reactant 4.05g will be used to determine the mass of AlCl3 that can be produced. From Step 1: 212.7g of Cl will produce 266.66g AlCl3 212.7g = 266.66g 4.05g = x x = 5.08g of AlCl3 can be produced (c) Al:Cl = 0.2537 Al:Cl = Al:4.05 = 0.2537 mass of Al used in reaction = 4.05 x 0.2537 = 1.027g Excess reactant = 2.70 - 1.027 = 1.67g King Leo · 9 years ago</span>
8 0
3 years ago
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