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Nadya [2.5K]
3 years ago
11

7. An element's most stable ion forms an ionic compound with chlorine having the formula XCl2. If the ion of element X has a mas

s of 89 and 36 electrons, what is the identity of the element, and how many neutrons does it have
Chemistry
1 answer:
Serhud [2]3 years ago
4 0

Answer:

The element is strontium and the number of neutrons it have is 51.

Explanation:

Based on the given information, the ionic compound is,  

XCl₂ ⇔ X₂⁺ + 2Cl⁻

X2+ is the ion of the mentioned element

As mentioned in the given question, the number of electrons of the element X is 36 and as seen from the reaction the charge present on the ion is +2. Now the atomic number will be,  

No. of electrons = atomic number - charge

36 = atomic number - 2

Atomic number = 38

Based on the periodic table, the atomic number 38 is for strontium element, and the sign of strontium is Sr. Hence, the element X is Sr.  

Now based on the given information, the mass number of the element is 89. Now the no. of neutrons will be,  

No. of neutrons = mass number - atomic number

= 89 - 38

= 51 neutrons.  

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yaroslaw [1]

Answer:

Final molarity of iodide ion C(I-) = 0.0143M

Explanation:

n = (m(FeI(2)))/(M(FeI(2))

Molar mass of FeI(3) = 55.85+(127 x 2) = 309.85g/mol

So n = 0.981/309.85 = 0.0031 mol

V(solution) = 150mL = 0.15L

C(AgNO3) = 35mM = 0.035M = 0.035m/L

n(AgNO3) = C(AgNO3) x V(solution)

= 0.035 x 0.15 = 0.00525 mol

(AgNO3) + FeI(3) = AgI(3) + FeNO3

So, n(FeI(3)) excess = 0.00525 - 0.0031 = 0.00215mol

C(I-) = C(FeI(3)) = [n(FeI(3)) excess]/ [V(solution)] = 0.00215/0.15 = 0.0143mol/L or 0.0143M

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3 years ago
What does creativity have to do with Science?​
Lady_Fox [76]

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5 0
2 years ago
What is the limiting reagent when a 2.00 g sample of ammonia is mixed with 4.00 g of oxygen?​
UNO [17]

Answer:

Ammonia is limiting reactant

Amount of oxygen left  = 0.035 mol

Explanation:

Masa of ammonia = 2.00 g

Mass of oxygen = 4.00 g

Which is limiting reactant = ?

Balance chemical equation:

4NH₃ + 3O₂     →     2N₂ + 6H₂O

Number of moles of ammonia:

Number of moles = mass/molar mass

Number of moles = 2.00 g/ 17 g/mol

Number of moles = 0.12 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 4.00 g/ 32 g/mol

Number of moles = 0.125 mol

Now we will compare the moles of ammonia and oxygen with water and nitrogen.

                      NH₃          :            N₂

                        4             :             2

                      0.12           :           2/4×0.12 = 0.06

                      NH₃         :            H₂O

                        4            :             6

                        0.12       :           6/4×0.12 = 0.18

                       

                       O₂            :            N₂

                        3             :             2

                      0.125        :           2/3×0.125 = 0.08

                        O₂           :            H₂O

                        3              :             6

                        0.125       :           6/3×0.125 = 0.25

The number of moles of water and nitrogen formed by ammonia are less thus ammonia will be limiting reactant.

Amount of oxygen left:

                        NH₃          :             O₂

                           4            :              3

                           0.12       :          3/4×0.12= 0.09

Amount of oxygen react = 0.09 mol

Amount of oxygen left  = 0.125 - 0.09 = 0.035 mol

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