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Alisiya [41]
3 years ago
8

The bond dissociation enthalpies of the H–H bond and the H–Cl bond are 435 kJ mol–1 and 431 kJ mol–1 , respectively. The ∆Hºf of

HCl(
g. is –92 kJ mol–1 . What is the bond dissociation enthalpy of the Cl–Cl bond?
Chemistry
1 answer:
Mrac [35]3 years ago
5 0
First, we write the reaction equation for the formation of HCl from hydrogen and chlorine:
H₂ + Cl₂ → 2HCl

Now, we substitute the values of the enthalpies, denoting that of the Cl-Cl bond by x:
435 + x = -92 + 2(431)
x = 335
The bond dissociation enthalpy for the Cl-Cl bond is 335 kJ/mol
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3 years ago
Are all elements abbreviated with one or two letters? If not, name some that aren't.
mixas84 [53]
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4 years ago
Read 2 more answers
Can some body please help me with this Stoichiometry stuff
andriy [413]

Answer:

See explanations

Explanation:

Stoichiometry is very easy to master if you understand the ‘mole concept’ and how it is used to define and describe chemical process mathematically. A ‘mole’ – in chemistry – is the mass of substance containing one Avogadro’s Number of particles. That is, N₀ = 6.023 x 10²³ particles / mole. When working with chemical reactions and equations data should be first converted to moles using the following conversations:

1 mole = 1 formula weight = 6.023 x 10²³ particles = 22.4 liters at STP(0⁰, 1atm).

In this problem you are given the equation Na + H₂O => NaOH + H₂. ‘Reading the equation’ there is 1 mole of Na, 1 mole of water, 1 mole of NaOH and 1 mole of H₂. In another example 3H₂ + N₂ => 2NH₃ there are 3 moles of H₂, 1 mole of N₂ and 2 moles of NH₃. The mole values can be multiples or fractions but if one mole value increases all the remaining mole values increase or decrease proportionally. For example:

Using the equation Na + H₂O => NaOH + H₂, one could apply a 2 before the Na but all the following formulas would need be increased by a factor of 2. If one applies ½ to the Na then all the following formulas would need be cut in half also and the reaction stoichiometry would still be valid. The fact that the equation is written with coefficients of 1 is that it is in the smallest whole number ratio of coefficients. This then implies the reaction formula is in ‘standard form’. This also implies the equation conditions are at 0⁰C & 1atm pressure and 1 mole of any gas phase substance occupies 22.4 Liters volume. Such is the significance of converting given data to moles as all other substance mass (in moles) are proportional.  

For your 1st problem, 1.76 x 10²⁴ formula units of Na will react with water (usually read as an excess) to produce (?) grams of H₂.

1st write the equation followed by listing the givens below the respective formulas… That is…

                         Na                      +            H₂O       => NaOH    +         H₂,

Given:      1.76 x 10²⁴ atoms                excess             ---------          ? grams

Convert atoms Na to moles = 1.76 x 10²⁴atoms/6.023 x 10²³atoms/mole

=2.922moles Na produces=>2.922moles H₂(because moles Na=moles H₂).

Convert moles to grams =>2.922moles H₂  x  2.000 grams H₂/mole H₂

=5.8443 grams H₂  

2nd problem, 3.5 moles Na will react with H₂O (in excess) to produce (?) moles of NaOH.

Again write equation and assign values to each formula unit in the equation.

                         Na                      +            H₂O        =>           NaOH    +    H₂,

Given:            3.5moles                       excess                      ? grams       ----

Since coefficients of balanced std equation are equal then moles Na equals moles of NaOH, that is, 3.5 moles Na produces => 3.5 moles NaOH

Convert moles NaOH to grams => 3.5 moles NaOH x 40 g NaOH/mole NaOH =  140 grams NaOH    

3rd problem, 2.75 x 10²⁵ molecules H₂O will react with (?) atoms of Na.

Same procedure, convert to moles, solve problem by ratios then convert to needed dimension at end of problem.

                         Na          +            H₂O                              =>       NaOH    +    H₂

Given:           ? atoms          2.75 x 10²⁵ molecules H₂O    =>     NaOH  + H₂  

Convert to moles =>  2.75 x 10²⁵ molecules H₂O / 6.023 x 10²³ molecules H₂O/mole H₂O = 45.658 moles H₂O =>  45.658 moles Na (equal coefficients)

Convert moles Na to atoms Na  =>   45.658 moles Na x 6.023 x 10²³atoms Na/mole Na = 2.75 x 10²⁵ atoms Na.

Note => Problem 3 could have been solved by inspection b/c coefficients are equal, however, always go through a process that you can justify and defend even if it does take longer. Never assume anything. Depend on what you know, not what you 'think' you know.  

Master the mole concept and you master a lot of chemistry! Good luck.

                             

5 0
3 years ago
A solution contains 0.021 M Cl? and 0.017 M I?. A solution containing copper (I) ions is added to selectively precipitate one of
slavikrds [6]

Answer:

A precipitate will begin to form at [Cu+] = 3.0 *10^-10 M

The precipitate formed is CuI

Explanation:

Step 1: Data given

The solution contains 0.021 M Cl- and 0.017 M I-.

Ksp(CuCl) = 1.0 × 10-6

Ksp(CuI) = 5.1 × 10-12.

Step 2:  Calculate [Cu+]

Ksp(CuCl) = [Cu+] [Cl-]  

1.0 * 10^-6  = [Cu+] [Cl-]  

1.0 * 10^-6  = [Cu+] [0.021]  

[Cu+] = 1.0 * 10^-6 / 0.021

[Cu+] = 4.76 *10^-5 M

Ksp(CuI) = [Cu] [I]  

5.1 * 10^-12  = [Cu+] [I-]  

5.1 * 10^-12 =[Cu+] [0.017]  

[Cu+] = 5.1 * 10^-12 / 0.017

[Cu+] = 3.0 *10^-10 M

[Cu+]from CuI hast the lowest concentration

A precipitate will begin to form at [Cu+] = 3.0 *10^-10 M

The precipitate formed is CuI

5 0
3 years ago
A piece of a metal having a mass of 150.0000 grams is heated to 99.3 9C. When the hot metal is submerged in 50.000 grams of wate
NNADVOKAT [17]

Answer:

Explanation:

a ) The temperature of water increases from 21.3°C to 48.4°C .

b ) Heat flow of water ( q) is positive because there is rise in temperature .

c ) The temperature of metal decreases from 99.3⁰C to 48.4⁰C .

d ) Heat flow of metal is negative because there is fall in temperature .

e ) heat loss of metal = heat gain by water

heat gain by water ( q )  = 50 x 4.2 x ( 48.4 - 21.3 )

= 5691 J

q = 5690 J . ( 3 significant figures )

heat loss of metal = mass x specific heat x fall in temp = 5691

150 x specific heat x ( 99.3 - 48.4 ) = 5691

specific heat = 5691 / (150 x 50.9 )

= .745 J / g C   ( three significant figures )

4 0
3 years ago
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