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3 years ago

How do the electrons in bonds (bonding domains) differ from lone pairs (non-bonding domains)?

1 answer:

max2010maxim [7]3 years ago

5 0

The electrons in bonds (bonding domains) differ from lone pairs (non-bonding domains) is because the bonding domains are bonded to the central atom vs the lone pairs are just stuck on as extra electrons. The difference of bonding domains from non-bonding domains is that the bonding domains are bonded to the central atom and the non-bonding domains are just stuck on as extra electrons.

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A measurement of 5.685 × 10-6 is equivalent to 0.000 000 568 5. True False

Alecsey [184]

The answer is false.

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3 years ago

Aqueous solutions of isopropyl alcohol are commonly sold as rubbing alcohol. The boiling point of isopropyl alcohol is 82.4 °C.

Monica [59]

Answer:

This is due to more hydrogen bonding in ethylene glycol than it is in isopropyl alcohol

Explanation:

The boiling point of isopropyl alcohol is 82.4 °C it contains only a single OH group, hence intermolecular hydrogen bonding is solely responsible for it's boiling point, whereas Ethylene glycol (CH2OHCH2OH) contains 2-OH group and both intermolecular and intramolecular hydrogen bonding are responsible for the higher boiling point of ethylene glycol at 198 °C.

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3 years ago

The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute th

MrRissso [65]

The question is incomplete, complete question is :

The frequency factors for these two reactions are very close to each other in value. Assuming that they are the same, compute the ratio of the reaction rate constants for these two reactions at 25°C.

$\frac{K_1}{K_2}=?$

Activation energy of the reaction 1 , $Ea_1$ = 14.0 kJ/mol

Activation energy of the reaction 2, $Ea_1$ = 11.9 kJ/mol

Answer:

0.4284 is the ratio of the rate constants.

Explanation:

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

The expression used with catalyst and without catalyst is,

$\frac{K_2}{K_1}=\frac{A\times e^{\frac{-Ea_2}{RT}}}{A\times e^{\frac{-Ea_1}{RT}}}$

$\frac{K_2}{K_1}=e^{\frac{Ea_1-Ea_2}{RT}}$

where,

$K_2$ = rate constant reaction -1

$K_1$ = rate constant reaction -2

Activation energy of the reaction 1 , $Ea_1$ = 14.0 kJ/mol = 14,000 J

Activation energy of the reaction 2, $Ea_1$ = 11.9 kJ/mol = 11,900 J

R = gas constant = 8.314 J/ mol K

T = temperature = $25^oC=273+25=298 K$

Now put all the given values in this formula, we get

$\frac{K_1}{K_2}=e^{\frac{11,900- 14,000Jl}{8.314 J/mol K\times 298 K}}=2.3340$

0.4284 is the ratio of the rate constants.

7 0

3 years ago

(0904) How many atoms are there in 56.2 grams of Krypton, Kr?

Luba_88 [7]

Answer:

56.2÷6.02×10^23

=9.34×10^23

Explanation:

Divide the given mass of the atom by the mass of an Atom (the avogadro's constant) to find the number of atoms in the given mass.

8 0

3 years ago

What is the molarity of 225 grams of Cu(NO2)2 in a total volume of 2.59 L?

Kay [80]

Answer:

The molarity is 0.56 $\frac{moles}{L}$

Explanation:

In a mixture, the chemical present in the greatest amount is called a solvent, while the other components are called solutes. Then, the molarity or molar concentration is the number of moles of solute per liter of solution.

In other words, molarity is the number of moles of solute that are dissolved in a given volume.

The Molarity of a solution is determined by:

$Molarity (M)=\frac{number of moles of solute}{Volume}$

Molarity is expressed in units ( $\frac{moles}{liter}$ ).

Then you must know the number of moles of Cu(NO₂)₂. For that it is necessary to know the molar mass. Being:

Cu: 63.54 g/mol
N: 14 g/mol
O: 16 g/mol

the molar mass of Cu(NO₂)₂ is:

Cu(NO₂)₂= 63.54 g/mol + 2*(14 g/mol + 2* 16 g/mol)= 155.54 g/mol

Now the following rule of three applies: if 155.54 g are in 1 mole of the compound, 225 g in how many moles are they?

$moles=\frac{225 g*1 mole}{155.54 g}$

moles= 1.45

So you know:

number of moles of solute= 1.45 moles
volume=2.59 L

Replacing in the definition of molarity:

$Molarity=\frac{1.45 moles}{2.59 L}$

Molarity= 0.56 $\frac{moles}{L}$

The molarity is 0.56 $\frac{moles}{L}$

5 0

3 years ago