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Setler79 [48]
3 years ago
5

The electronegativity of C = 2.5, O = 3.5. What type of bond is between C-O? ______ ______ bond

Chemistry
1 answer:
Schach [20]3 years ago
5 0

Answer:

Polar covalent bond

Explanation:

On the basis of electronegativity bond could be ionic bond, polar and non pole covalent bond.

Ionic bond:

It is the bond which is formed by the transfer of electron from one atom to the atom of another element.  

Both bonded atoms have very large electronegativity difference. The atom with large electronegativity value accept the electron from other with smaller value of electronegativity.  The electronegativity difference between bonded atoms is greater than 1.7.

For example:

Sodium chloride is ionic compound. The electronegativity of chlorine is 3.16 and for sodium is 0.93. There is large difference is present. That's why electron from sodium is transfer to the chlorine. Sodium becomes positive and chlorine becomes negative ion. Both atoms are bonded together electrostatic attraction occur between anion and cations.

Covalent bond:

It is formed by the sharing of electron pair between bonded atoms.  

The atom with larger electronegativity attract the electron pair more towards it self and becomes partial negative while the other atom becomes partial positive.  There are two type of covalent bond. Polar and non polar covalent bond. When electronegativity difference is 0.4 - 1.7 bond is polar covalent bond when it is less than 0.4 the bond is non polar covalent.

For example:

In CO the electronegativity of oxygen is 3.5 and carbon is 2.5. That's why electron pair attracted more towards oxygen, thus oxygen becomes partial negative and carbon becomes partial positive. and bond is polar covalent.

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In the winter, a heated home in the Northeast might be maintained at a temperature of 78°F. What is this temperature on the Cels
masya89 [10]

Answer:  25.6^0C,  298.6K

Explanation:

Temperature of the gas is defined as the degree of hotness or coldness of a body. It is expressed in units like ^0C and K  

These units of temperature are inter convertible.

We are given:

Temperature of the gas = 78^0F

Converting this unit of temperature into ^0C by using conversion factor:

^oC=\frac{5}{9}\times (^oF-32)

^0C=\frac{5}{9}\times (78^oF-32)

25.6^0C

Converting this unit of temperature into K by using conversion factor:

K=t^0C+273

K=25.6+273

298.6K

Thus the temperature on the Celsius and Kelvin scales are 25.6^0C  and 298.6K respectively.

6 0
3 years ago
PLEASE HELP ME!!!
kondor19780726 [428]
<h3>Answer:</h3>

0.424 J/g °C

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right  

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtraction Property of Equality<u> </u>

<u>Chemistry</u>

<u>Thermochemistry</u>

Specific Heat Formula: q = mcΔT

  • q is heat (in Joules)
  • m is mass (in grams)
  • c is specific heat (in J/g °C)
  • ΔT is change in temperature
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] m = 38.8 g

[Given] q = 181 J

[Given] ΔT = 36.0 °C - 25.0 °C = 11.0 °C

[Solve] c

<u>Step 2: Solve for Specific Heat</u>

  1. Substitute in variables [Specific Heat Formula]:                                             181 J = (38.8 g)c(11.0 °C)
  2. Multiply:                                                                                                             181 J = (426.8 g °C)c
  3. [Division Property of Equality] Isolate <em>c</em>:                                                         0.424086 J/g °C = c
  4. Rewrite:                                                                                                             c = 0.424086 J/g °C

<u>Step 3: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

0.424086 J/g °C ≈ 0.424 J/g °C

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