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ollegr [7]
3 years ago
10

The train traveled 100 miles in 4 hours how fast was the train traveling

Chemistry
2 answers:
DanielleElmas [232]3 years ago
7 0

Answer: the train was going 25 mph

Explanation: 100 divided by 4 is 25

Triss [41]3 years ago
3 0

Answer:

the train traveled 25 Miles per hour

Explanation:

100 divided by 4 is 25

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What name would you give to an orange smelling ester formed by adding octanol to acetic acid
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The name of the ester is octyl acetate.

Explanation:

Octanol will react with the acetic acid to form octyl acetate. You may find the structure of the compounds in the attached picture.

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structure of organic compounds

brainly.com/question/14122960

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How many moles of lithium metal would need to react with excess oxygen gas to produce 4 moles of lithium oxide in the following
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7 0
3 years ago
Given the following thermodynamic data, calculate the lattice energy of LiCl:
tiny-mole [99]

Answer:

\boxed{\text{-862 kJ/mol}}

Explanation:

One way to calculate the lattice energy is to use Hess's Law.

The lattice energy U is the energy released when the gaseous ions combine to form a solid ionic crystal:

Li⁺(g) + Cl⁻(g) ⟶ LiCl(s); U = ?

We must generate this reaction rom the equations given.

(1)  Li(s) + ½Cl₂ (g) ⟶ LiCl(s);      ΔHf°     = -409 kJ·mol⁻¹

(2) Li(s) ⟶ Li(g);                          ΔHsub =    161 kJ·mol⁻¹

(3) Cl₂(g) ⟶ 2Cl(g)                     BE        =   243 kJ·mol⁻¹

(4) Li(g) ⟶Li⁺(g) +e⁻                   IE₁         =   520 kJ·mol⁻¹

(5) Cl(g) + e⁻ ⟶ Cl⁻(g)                EA₁       =  -349 kJ·mol⁻¹

Now, we put these equations together to get the lattice energy.

                                                <u>E/kJ </u> 

(5) Li⁺(g) +e⁻ ⟶ Li(g)                520

(6) Li(g) ⟶ Li(s)                         -161

(7) Li(s) + ½Cl₂(g) ⟶ LiCl(s)     -409

(8) Cl(g) ⟶ ½Cl₂(g)                   -121.5

(9) Cl⁻(g) ⟶ Cl(g) + e⁻               <u>+349</u>

      Li⁺(g) +  Cl⁻(g) ⟶ LiCl(s)     -862

The lattice energy of LiCl is \boxed{\textbf{-862 kJ/mol}}.

3 0
2 years ago
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