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2 years ago

If an element has 11 protons and 11 neutrons in its nucleus it must be ?

Chemistry

1 answer:

Juliette [100K]2 years ago

7 0

Answer:

the answer is 11

Explanation:

Ok, let us understand what is an atomic number, an atomic number is simply the proton number, the atomic number is always seen at the bottom left corner in a element of a periodic table, it is the number of proton in a atom of a certain element.

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Please help, so confused!!

zhannawk [14.2K]

Can you put table S in?

5 0

2 years ago

Which of the following statements best describes the Lewis structure of CIO4 and ClO3?

mash [69]

CIO4 statements best describe the Lewis structure.

<h3>What is Lewis Structure?</h3>

A Lewis Structure is a very simplified presentation of the valence shell electrons in a molecule. It is used to show how the electrons are organized around particular atoms in a molecule. Electrons are displayed as "dots" or for bonding electrons as a line between the two atoms.

Lewis structure does NOT endeavor to explain the geometry of molecules, how the bonds compose, or how the electrons are transmitted between the atoms.
It is the easiest and most limited theory on electronic format.Lewis systems show separately atom and its position in the structure of the molecule using its chemical symbol. Lines are drawn between atoms that are connected (pairs of dots can be used instead of sequences). Excess electrons that form lone pairs are defined as pairs of dots and are positioned next to the atoms.

To learn more about Lewis Structure, reffer:

brainly.com/question/7218689

#SPJ2

4 0

2 years ago

The molecular weight of NaCl is 58.44 grams/mole. If you had a 1.0 molar solution (1.0 M), you would have to put 58.44 g of salt

Andre45 [30]

Answer:

5,844 grams of NaCl

Explanation:

Knowing the molecular weight 58,44 g/mole and saying 1 molar solution is 58,44 of NaCl in 1 liter of solution. 100 mL means 10% of the whole solution then we are going to have 10% of NaCl

58,44 x 0,1 = 5,844 grams of NaCl

7 0

3 years ago

Ammonia is produced industrially via this exothermic reaction.

Cloud [144]

Answer:

high pressure of 200-300 atm.

low temp. of between 400-500 degrees celsius:this is for continuous development of ammonia since it decomposes at high temp fathermore the reaction is exothermic

a catalyst to speed up the rate of reaction:i guess it is finely divided iron impregnated in aluminium oxide

platinum can be used as a catalyst but it is easily poisoned

hope it helps

Explanation:

6 0

2 years 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