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uranmaximum [27]
3 years ago
9

A car driving east along the freeway at 60 m/s hits some traffic and after 7 seconds is

Chemistry
1 answer:
Vladimir79 [104]3 years ago
4 0
Do u know how to solve area
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2) Which one<br>more active in between<br>Li and Na and why ?<br>IS​
Blababa [14]

Explanation:

what's li and na stand for ?

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3 years ago
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AKALI METALS ARE HIGLY_______ SINCE THEY ______ easily to form ions. HELP ASAP
kvv77 [185]

Answer:

Reactive and lose 1 electron

Explanation:

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3 years ago
Is a Animal and Plant cell living
Mila [183]
Yes Animal and plant are cell living
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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
3 years ago
TRIPLE POINTS!!! I NEED HELP!!! PLEASE EXPLAIN TOO!!! What is the energy of a quantum of light with a frequency of 4.31 x 1014 1
Lelechka [254]

Answer:

The answer is

2.86 \times  {10}^{ - 19}  \: J

Explanation:

The energy of a quantum of light can be found by using the formula

<h3>E = hf</h3>

where

E is the energy

f is the frequency

h is the Planck's constant which is

6.626 × 10-³⁴ Js

From the question

f = 4.31 × 10¹⁴ Hz

We have

E = 4.31 × 10¹⁴ × 6.626 × 10-³⁴

We have the final answer as

2.86 \times  {10}^{ - 19}  \: J

Hope this helps you

3 0
3 years ago
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