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iogann1982 [59]
3 years ago
13

if carbon which has an electrongavity of 2.5 bonds with hydrogen, which has an electrongavity of 2.1, the bond between the two a

toms will be classified as a(n) covalent bond​
Chemistry
1 answer:
sukhopar [10]3 years ago
8 0

Answer:

If carbon which has an electrongavity of 2.5 bonds with hydrogen, which has an electrongavity of 2.1, the bond between the two atoms will be classified as a(n) covalent bond → Totally TRUE

Explanation:

We must follow the Pauling's rules to solve these question.

The electonegativity of an element is defined as the ability it has to attract electrons that link it to another element. It is a periodic property, which allows predicting the polarity of the bond formed between two atoms, as well as its covalent or ionic character.

The Pauling scale values ​​range from 0.7 for Francium (the least electronegative element) to 4 for Fluorine, the most electronegative. Using the electronegativities table we can predict the polarity of some compounds

ΔEn (C-H) = 2.5 - 2.1 = 0.4 → This is a covalent bond.

ΔEn ≤ 0.4 → Covalent non polar bond

0.5 < ΔEn < 1.7 → Covalent polar bond

ΔEn ≥ 1.7 → Ionic bond

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Problem PageQuestion Aqueous sulfuric acid will react with solid sodium hydroxide to produce aqueous sodium sulfate and liquid w
trasher [3.6K]

Answer:

maximum mass of sodium sulphate=132g

Explanation:

firstly balance the chemical reaction,

H_2 SO_4_aq+2NaOH_s=Na_2 SO_4_aq+2H_2 O_l

molecular weight of H2SO4=98g/mole;

Molecular weight of NaOH=40g/mole;

number of mole of H2SO4 given=91g/(98g/mole)=0.93mole;

number of mole of NaOH given=116g/(40g/mole)=2.9mole

moles are:

number of mole of H2SO4=0.93

number of mole of NaOH=2.9

From the balanced equation,

1mole of H2SO4 reacts with 2 moles of NaOH;

hence ;

0.93 mole of H2SO4 will react with 2*0.93 moles of NaOH means

moles of NaOH that reacts wih H2SO4=1.86mole but we have given 2.9 mole therefore

NaOH will be excess reagent and H2SO4 will be the limitting reagent

and mass of product depends on limitting reagent i.e mass of H2SO4.

1mole of H2SO4 produces 1 mole of Na2SO4;

0.93 mole produces 0.93 mole of Na2SO4;

maximum mass of Na2SO4 produces=mole*molecular weight;

Molecular weight of Na2SO4=142g/mole;

maximum mass of Na2SO4 produces=0.93*142=132g;

maximum mass ofNa2SO4 produces=132g;

6 0
3 years ago
A baseball is traveling (+20 m/s) and is hit by a bat. It leaves the bat traveling (-30 m/s). What is the change in the velocity
Tpy6a [65]
D.

A more accurate answer would be -50 m/s. 50 m/s is the change in speed.
4 0
3 years ago
Read 2 more answers
Name two elements that you would expect to have properties very much like those of calcium
neonofarm [45]
Atoms in the same group (column) of the periodic table have similar properties. This is because all elements in the same group have the same amount of valence (outer) electrons.

So, any element directly above or below Calcium on the periodic table would have similar properties.

These would be Magnesium, Barium, Strontium, Radium, and Beryllium.
3 0
3 years ago
A block of ice has little thermal energy. If you place it in a warm room, it melts into warm liquid water, which has much more e
Naily [24]

<u>Answer:</u>

<em>The energy to turn the ice into water:</em>

  • The energy that is required to change the state of ice into a liquid is obtained in the form of heat energy from the ambient temperature of the warm room.
  • Once this heat energy is absorbed, the individual molecules of ice gain kinetic energy and start vibrating faster.
  • Yet, the temperature of the ice remains constant until the ice reaches its melting point because this energy is first utilised to break all the bonds of the lattice structure of the ice.
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8 0
3 years ago
Use the de Broglie's Wave Equation to find the wavelength of an electron moving at 7.3 × 106 m/s. Please show your work. Note: h
ad-work [718]

Answer:

\lambda=9.96\times 10^{-11}\ m

Explanation:

The expression for the deBroglie wavelength is:

\lambda=\frac {h}{m\times v}

Where,  

\lambda is the deBroglie wavelength  

h is Planck's constant having value 6.62607\times 10^{-34}\ Js

m is the mass of electron having value 9.11\times 10^{-31}\ kg

v is the speed of electron.

Given that v = 7.3\times 10^6\ m/s

Applying in the equation as:

\lambda=\frac {h}{m\times v}

\lambda=\frac{6.62607\times 10^{-34}}{9.11\times 10^{-31}\times 7.3\times 10^6}\ m

\lambda=\frac{10^{-34}\times \:6.626}{10^{-25}\times \:66.503}\ m

\lambda=\frac{6.626}{10^9\times \:66.503}\ m

\lambda=9.96\times 10^{-11}\ m

6 0
3 years ago
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