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dsp73
3 years ago
15

The molecule NH3 contains all single bonds. true false

Chemistry
2 answers:
dsp733 years ago
8 0
<h3><u>Answer;</u></h3>

True

<h3><u>Explanation</u>;</h3>
  • The molecule NH3 contains all single bonds.
  • NH3 has a three single covalent  bond among its nitrogen and hydrogen atoms,because one valence electron of each of three atom of hydrogen is shared with three electron.
  • There are three covalent bonds are in NH3 . Each hydrogen make a single bond with nitrogen and there is also a pair of electron which is unpaired from nitrogen.
morpeh [17]3 years ago
7 0

Answer:

true

Explanation:

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A substance that is ______ will NOT dissolve in a solvent. A) freezing B) insoluble C) evaporating D) soluble
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insoluble

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The first-order rate constant for the reaction of methyl chloride (CH3Cl) with water to produce methanol (CH3OH) and hydrochlori
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Answer:

K(48.5°C) = 1.017 E-8 s-1

Explanation:

  • CH3Cl + H2O → CH3OH + HCl

at T1 = 25°C (298 K) ⇒ K1 = 3.32 E-10 s-1

at T2 = 48.5°C (321.5 K) ⇒ K2 = ?

Arrhenius eq:

  • K(T) = A e∧(-Ea/RT)
  • Ln K = Ln(A) - [(Ea/R)(1/T)]

∴ A: frecuency factor

∴ R = 8.314 E-3 KJ/K.mol

⇒ Ln K1 = Ln(A) - [Ea/R)*(1/T1)]..........(1)

⇒ Ln K2 = Ln(A) - [(Ea/R)*(1/T2)].............(2)

(1)/(2):

⇒ Ln (K1/K2) = (Ea/R)* (1/T2-1/T1)

⇒ Ln (K1/K2) = (116 KJ/mol/8.3134 E-3 KJ/K.mol)*(1/321.5 K - 1/298 K)

⇒ Ln (K1/K2) = (13952.37 K)*(- 2.453 E-4 K-1)

⇒ Ln (K1/K2) = - 3.422

⇒ K1/K2 = e∧(-3.422)

⇒ (3.32 E-10 s-1)/K2 = 0.0326

⇒ K2 = (3.32 E-10 s-1)/0.0326

⇒ K2 = 1.017 E-8 s-1

7 0
3 years ago
Calculate the standard enthalpy change for the reaction at 25 ∘ C. Standard enthalpy of formation values can be found in this li
meriva

Answer:

The standard enthalpy change for the reaction at 25^{0}\textrm{C} is -2043.999kJ

Explanation:

Standard enthalpy change (\Delta H_{rxn}^{0}) for the given reaction is expressed as:

\Delta H_{rxn}^{0}=[3mol\times \Delta H_{f}^{0}(CO_{2})_{g}]+[4mol\times \Delta H_{f}^{0}(H_{2}O)_{g}]-[1mol\times \Delta H_{f}^{0}(C_{3}H_{8})_{g}]-[5mol\times \Delta H_{f}^{0}(O_{2})_{g}]

Where \Delta H_{f}^{0} refers standard enthalpy of formation

Plug in all the given values from literature in the above equation:

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4 0
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Answer:

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Step-by-step explanation:

The <em>superscripts</em> in an electron configuration tell us how many electrons are in a subshell.

If the electron configuration is 1s¹ 2s¹2p², the total number of electrons is

1 + 1 + 2 = 4

The atom contains four electrons.

<em>Note</em>: this atom is in an <em>excited state</em>, because the 1s and 2s subshells can each hold one more electron.

7 0
3 years ago
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