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Sauron [17]
4 years ago
14

What is the definition of a covalent bond?

Chemistry
2 answers:
valentinak56 [21]4 years ago
8 0
A covalent bond is the bond formed by the sharing of a pair of electrons by two atoms.

Your answer is D: A bond between two atoms.
koban [17]4 years ago
7 0
D. A bond between two atoms
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The density of benzene is 0.879g/mL. The volume (with 1 decimal) of 131.9g sample of benzene is how many mL?
timurjin [86]

Answer:

150.1 mL

Explanation:

Step 1: Given data

  • Density of benzene (ρ): 0.879 g/mL
  • Mass of the sample of benzene (m): 131.9 g
  • Volume of the sample of benzene (V): ?

Step 2: Calculate the volume of the sample of benzene

Density is an intrinsic property. It is equal to the quotient between the mass and the volume of the sample of benzene.

ρ = m/V

V = m/ρ

V = 131.9 g/(0.879 g/mL)

V = 150.1 mL

5 0
3 years ago
The normal boiling point of bromine is 58.8°C, and its enthalpy of vaporization is 30.91 kJ/mol. What is the approximate vapor p
saul85 [17]

Answer : The vapor pressure of bromine at 10.0^oC is 0.1448 atm.

Explanation :

The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of bromine at 10.0^oC = ?

P_2 = vapor pressure of propane at normal boiling point = 1 atm

T_1 = temperature of propane = 10.0^oC=273+10.0=283.0K

T_2 = normal boiling point of bromine = 58.8^oC=273+58.8=331.8K

\Delta H_{vap} = heat of vaporization = 30.91 kJ/mole = 30910 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

\ln (\frac{1atm}{P_1})=\frac{30910J/mole}{8.314J/K.mole}\times (\frac{1}{283.0K}-\frac{1}{331.8K})

P_1=0.1448atm

Hence, the vapor pressure of bromine at 10.0^oC is 0.1448 atm.

4 0
3 years ago
Sulfur has 16 electrons. When it combines with magnesium. It acquires A -2 charge. How many electrons does it have now?
Rainbow [258]
18 electrons. With a -2 charge, that means it gains two electrons.
8 0
3 years ago
Five drops of 1.0M NaOH increase the pH of a solution from 3.0 to 4.6. Five additional drops of 1.0M NaOH increase the pH to 4.7
Andrej [43]
The solution is buffered with a weak acid and it's conjugate base. Since NaOH is a base, you can use a weak acid as it's buffer.
5 0
3 years ago
how are the relationships of elements in a group different from the relationships of elements in a group
alex41 [277]
Same group = same properties, same # of valence electrons
Same period = same # of atomic orbitals
3 0
4 years ago
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