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GREYUIT [131]
2 years ago
8

What element with 26 electrons?

Chemistry
2 answers:
marysya [2.9K]2 years ago
4 0

Answer:

Our new chat

Explanation:

arsen [322]2 years ago
3 0

Answer:

Iron - Fe

Explanation:

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What is the notation for the enthalpy of solution?
Tomtit [17]
The enthalpy<span> of </span>solution<span>, </span>enthalpy<span> of dissolution, or heat of </span>solution<span> is the</span>enthalpy<span> change associated with the dissolution of a substance in a solvent at constant pressure resulting in infinite dilution. The </span>enthalpy<span> of </span>solution<span> is most often expressed in kJ/mol at constant temperature. </span>
4 0
2 years ago
Read 2 more answers
Which pair of aqueous solutions can create a buffer solution if present in the appropriate concentrations?.
Julli [10]

HF and NaF - If the right concentrations of aqueous solutions are present, they can produce a buffer solution.

<h3>What are buffer solutions and how do they differ?</h3>
  • The two main categories of buffers are acidic buffer solutions and alkaline buffer solutions.
  • Acidic buffers are solutions that contain a weak acid and one of its salts and have a pH below 7.
  • For instance, a buffer solution with a pH of roughly 4.75 is made of acetic acid and sodium acetate.
<h3>Describe buffer solution via an example.</h3>
  • When a weak acid or a weak base is applied in modest amounts, buffer solutions withstand the pH shift.
  • A buffer made of a weak acid and its salt is an example.
  • It is a solution of acetic acid and sodium acetate CH3COOH + CH3COONa.

learn more about buffer solutions here

<u>brainly.com/question/8676275</u>

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8 0
1 year ago
Be sure to answer all parts. Styrene is produced by catalytic dehydrogenation of ethylbenzene at high temperature in the presenc
svlad2 [7]

Answer:

a) ΔHºrxn = 116.3 kJ, ΔGºrxn = 82.8 kJ,  ΔSºrxn =  0.113 kJ/K

b) At 753.55 ºC or higher

c )ΔG =  1.8 x 10⁴ J

    K = 8.2 x 10⁻²

Explanation:

a)                                 C6H5−CH2CH3  ⇒  C6H5−CH=CH2  + H₂

ΔHf kJ/mol                    -12.5                           103.8                      0

ΔGºf kJ/K                        119.7                         202.5                      0

Sº J/mol                          255                          238                      130.6*

Note: This value was not given in our question, but is necessary and can be found in standard handbooks.

Using Hess law to calculate  ΔHºrxn we have

ΔHºrxn  = ΔHfº C6H5−CH=CH2 +  ΔHfº H₂ - ΔHºfC6H5−CH2CH3

ΔHºrxn =     103.8 kJ + 0 kJ  - (-12.5 kJ)

ΔHºrxn = 116.3 kJ

Similarly,

ΔGrxn = ΔGºf C6H5−CH=CH2 +  ΔGºfH₂ - ΔGºfC6H5CH2CH3

ΔGºrxn=   202.5 kJ + 0 kJ - 119.7 kJ  = 82.8 kJ

ΔSºrxn = 238 J/mol + 130.6 J/mol -255 J/K = 113.6 J/K = 0.113 kJ/K

b) The temperature at which the reaction is spontaneous or feasible occurs when ΔG becomes negative and using

ΔGrxn =  ΔHrxn -TΔS

we see that will happen when the term  TΔS  becomes greater than ΔHrxn since ΔS  is positive  , and so to sollve for T we will make ΔGrxn equal to zero and solve for T. Notice here we will make the assumption that  ΔºHrxn and ΔSºrxn remain constant at the higher temperature  and will equal the values previously calculated for them. Although this assumption is not entirely correct, it can be used.

0 = 116 kJ -T (0.113 kJ/K)

T = 1026.5 K  =  (1026.55 - 273 ) ºC = 753.55 ºC

c) Again we will use

                       ΔGrxn =  ΔHrxn -TΔS

to calculate ΔGrxn   with the assumption that ΔHº and ΔSºremain constant.

ΔG =  116.3 kJ - (600+273 K) x 0.113 kJ/K =  116.3 kJ - 873 K x 0.113 kJ/K

ΔG =  116.3 kJ - 98.6 kJ =  17.65 kJ = 1.8 x 10⁴ J ( Note the kJ are converted to J to necessary for the next part of the problem )

Now for solving for K, the equation to use is

ΔG = -RTlnK and solve for K

- ΔG / RT = lnK  ∴ K = exp (- ΔG / RT)

K = exp ( - 1.8 x 10⁴ J /( 8.314 J/K  x 873 K)) = 8.2 x 10⁻²

8 0
3 years ago
Two isotopes of lithium are found in nature Li6 has a mass of 6. 02u and Li7 has a mass of 7.02u . Use the atomic weight of lith
alexandr402 [8]

The isotope that is more abundant, given the data is isotope Li7

<h3>Assumption</h3>
  • Let Li6 be isotope A
  • Let Li7 be isotope B

<h3>How to determine whiche isotope is more abundant</h3>
  • Molar mass of isotope A (Li6) = 6.02 u
  • Molar mass of isotope B (Li7) = 7.02 u
  • Atomic mass of lithium = 6.94 u
  • Abundance of A = A%
  • Abundance of B = (100 - A)%

Atomic mass = [(mass of A × A%) / 100] + [(mass of B × B%) / 100]

6.94 = [(6.02 × A%) / 100] + [(7.02 × (100 - A)) / 100]

6.94 = [6.02A% / 100] + [702 - 7.02A% / 100]

6.94 = [6.02A% + 702 - 7.02A%] / 100

Cross multiply

6.02A% + 702 - 7.02A% = 6.94 × 100

6.02A% + 702 - 7.02A% = 694

Collect like terms

6.02A% - 7.02A% = 694 - 702

-A% = -8

A% = 8%

Thus,

Abundance of B = (100 - A)%

Abundance of B = (100 - 8)%

Abundance of B = 92%

SUMMARY

  • Abundance of A (Li6) = 8%
  • Abundance of B (Li7) = 92%

From the above, isotope Li7 is more abundant.

Learn more about isotope:

brainly.com/question/24311846

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4 0
2 years ago
A sample of a mineral solid weighing 28.5 g was added to a graduated cylinder filled with to a level of 20.3 mL. As a result, th
bonufazy [111]

Answer:

2.8 g/mL is the density of the mineral.

Explanation:

The mass, m=28.5\ g

Volume of the object = Volume of water level rose - Initial volume of water = 30.5 mL - 20.3 mL = 10.2 mL

The expression for the calculation of density is shown below as:-

\rho=\frac{m}{V}

\rho=\frac{28.5\ g}{10.2 mL}=2.8\ g/mL

<u>2.8 g/mL is the density of the mineral.</u>

8 0
3 years ago
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