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2 years ago

Which of these statements about hydrogen bonds is not true?

Chemistry

2 answers:

madreJ [45]2 years ago

8 0

A is false Hydrogen bonds have nothing to do with boiling point

just olya [345]2 years ago

5 0

B as the bonds are a cause of the high boiling temp for water

i could be wrong

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How many kilojoules of heat are released when 32.0 g of NaOH are dissolved in water? (The molar heat of solution of NaOH is –445

Katen [24]

The energy release when dissolving 1 mol of NaOH in water is 445.1 kJ
the mass of NaOH to be dissolved is 32.0 g
The number of NaOH moles in 32.0 g - 32.0 g / 40 g/mol = 0.8 mol
the energy released whilst dissolving 1 mol of NaOH - 445.1 kJ
when dissolving 0.8 mol - the energy released is 445.1 kJ/mol x 0.8 mol
therefore heat released is - 356.08 kJ
answer is -356.08 kJ

6 0

3 years ago

Read 2 more answers

How many electrons would be transferred in either a voltaic or electrolytic cell that uses the following half reactions

omeli [17]

Answer:

Numbers of electrons transferred in the electrolytic or voltaic cell is 6 electrons.

Explanation:

$Fe^{3+} (aq) + 3 e^-\rightarrow Fe (s) ,E^o = -0.036 V$

$Mg^{2+}(aq) + 2 e^- \rightarrow Mg (s),E^o = -2.37 V$

The substance having highest positive reduction $E^o$ potential will always get reduced and will undergo reduction reaction.

Reduction : cathode

$Fe^{3+} (aq) + 3 e^-\rightarrow Fe (s) ,E^o = -0.036 V$ ..[1]

Oxidation: anode

$Mg(s)\rightarrow Mg^{2+}(aq) + 2 e^-,E^o = 2.37 V$ ..[2]

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{red,cathode}-E^o_{red,anode}$

$E^o_{cell}=-0.036V-(-2.37 V)=2.334 V$

The overall reaction will be:

2 × [1] + 3 × [2] :

$2Fe^{3+} (aq) + 3Mg(s)+6e^-\rightarrow 2Fe (s)+3Mg^{2+}(aq)+6e^-$

Electrons on both sides will get cancelled :

$2Fe^{3+} (aq) + 3Mg(s)\rightarrow 2Fe (s)+3Mg^{2+}(aq)$

Numbers of electrons transferred in the electrolytic or voltaic cell is 6 electrons.

5 0

3 years ago

Fan blades would be an analogy for which atomic model?

julia-pushkina [17]

I believe the correct answer from the choices listed above is option A. Fan blades would be an analogy for electron cloud model. Austrian physicist Erwin Schrödinger (1887-1961) developed an “Electron Cloud Model<span>” in 1926. It consisted of a dense nucleus surrounded by a cloud of electrons. Hope this helps.</span>

4 0

3 years ago

CaO + H2O -> Ca(OH)2

yawa3891 [41]

The % yield of Ca(OH)₂ : 62.98%

<h3>Further eplanation </h3>

Percent yield is the compare of the amount of product obtained from a reaction with the amount you calculated

General formula:

Percent yield = (Actual yield / theoretical yield )x 100%

An actual yield is the amount of product actually produced by the reaction. A theoretical yield is the amount of product that you calculate from the reaction equation according to the product and reactant coefficients

Reaction

CaO + H₂O ⇒ Ca(OH)₂

mass CaO= 4.2 g

mol CaO(MW=56,0774 g/mol) :

$\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{4.2}{56,0774 g/mol}\\\\mol=0.075$

mol Ca(OH)₂ based on mol CaO

mol ratio CaO : Ca(OH)₂,= 1 : 1, so mol Ca(OH)₂ = 0.075

mass Ca(OH)₂(MW=74,093 g/mol) ⇒ theoretical

$\tt mass=mol\times MW\\\\mass=0.075\times 74,093 g/mol\\\\mass=5.557~g$

% yield :

$\tt =\dfrac{actual}{theoretical}\times 100\%\\\\=\dfrac{3.5}{5.557}\times 100\%\\\\=62.98\%$

8 0

2 years ago

The isotope \left.\begin{array}{r}212 \\ 83\end{array}\right? Bi has a half-life of 1.01 yr. What mass (in mg) of a 2.00-mg samp

anyanavicka [17]

Half-life is the length of time it takes for half of the radioactive atoms of a specific radionuclide to decay. A good rule of thumb is that, after seven half-lives, you will have less than one percent of the original amount of radiation.

half-life, in radioactivity, the interval of time required for one-half of the atomic nuclei of a radioactive sample to decay (change spontaneously into other nuclear species by emitting particles and energy), or, equivalently, the time interval required for the number of disintegrations per second of a radioactive.

<h3>What affects the half-life of an isotope?</h3>

Since the chemical bonding between atoms involves the deformation of atomic electron wavefunctions, the radioactive half-life of an atom can depend on how it is bonded to other atoms. Simply by changing the neighboring atoms that are bonded to a radioactive isotope, we can change its half-life.

Learn more about half life of an isotope here:

<h3>brainly.com/question/13979590</h3><h3 /><h3>#SPJ4</h3>

5 0

1 year ago