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3 years ago

Which of the following correctly lists the name of the element, the symbol for the ion, and the name of the ion?

Chemistry

2 answers:

Yanka [14]3 years ago

8 0

The correct answer is B.

Hope this helps you.

Mrac [35]3 years ago

6 0

Answer:

zinc, Zn²⁺, zincate ion

Explanation:

Let us check with all the given examples:

a) Fluorine (F). It is the most electronegative element and thus cannot form positive ion. It will gain electron and form a negative ion (F⁻) known as Fluoride ion.

b) Zinc: it is a metal and give two electrons easily to attain noble gas configuration. Hence it will form Zincate ion (Zn⁺²).

c) Copper,: it is a metal and can form positive ion easily.

It can give either two or one electron to form

Cu⁺²: Cupric ion [More common]

Cu⁺: Cuprous ion

d) Barium : it is alkaline earth metal and will give only two electron to form dipositive ion barium ion (Ba⁺²)

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Calculate the entropy change for the surroundings of the reaction below at 350K: N2(g) + 3H2(g) -> 2NH3(g) Entropy data: NH3

krek1111 [17]

Answer : The entropy change for the surroundings of the reaction is, -198.3 J/K

Explanation :

We have to calculate the entropy change of reaction $(\Delta S^o)$ .

$\Delta S^o=S_{product}-S_{reactant}$

$\Delta S^o=[n_{NH_3}\times \Delta S^0_{(NH_3)}]-[n_{N_2}\times \Delta S^0_{(N_2)}+n_{H_2}\times \Delta S^0_{(H_2)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S^0{(NH_3)}$ = standard entropy of $NH_3$

$\Delta S^0{(H_2)}$ = standard entropy of $H_2$

$\Delta S^0{(N_2)}$ = standard entropy of $N_2$

Now put all the given values in this expression, we get:

$\Delta S^o=[2mole\times (192.5J/K.mole)]-[1mole\times (191.5J/K.mole)+3mole\times (130.6J/K.mole)]$

$\Delta S^o=-198.3J/K$

Therefore, the entropy change for the surroundings of the reaction is, -198.3 J/K

4 0

2 years ago

14 points!!! Pls tell me answer :3

iren [92.7K]

I’d say the last one

8 0

3 years ago

Read 2 more answers

maksim [4K]

Answer:

can you provide a picture?

7 0

3 years ago

The combustion reaction of isopropyl alcohol is given below: C 3 H 7 O H ( l ) + 9 2 O 2 ( g ) → 3 C O 2 ( g ) + 4 H 2 O ( g ) T

jek_recluse [69]

Answer:

the heat of formation of isopropyl alcohol is -317.82 kJ/mol

Explanation:

The heat of combustion of isopropyl alcohol is given as follows;

C₃H₇OH (l) +(9/2)O₂ → 3CO₂(g) + 4H₂O (g)

The heat of combustion of CO₂ and H₂O are given as follows

C (s) + O₂ (g) → CO₂(g) = −393.50 kJ

H₂ (g) + 1/2·O₂(g) → H₂O (l) = −285.83 kJ

Therefore we have

3CO₂(g) + 4H₂O (g) → C₃H₇OH (l) +(9/2)O₂ which we can write as

3C (s) + 3O₂ (g) → 3CO₂(g) = −393.50 kJ × 3 =

4H₂ (g) + 2·O₂(g) → 4H₂O (l) = −285.83 kJ × 4

3CO₂(g) + 4H₂O (g) → C₃H₇OH (l) +(9/2)O₂ = +2006 kJ/mol

-1180.5 - 1143.32 +2006 = -317.82 kJ/mol

Therefore, the heat of formation of isopropyl alcohol = -317.82 kJ/mol.

3 0

3 years ago

How does lanthanide contraction influence the density of period 6 transition element

Kitty [74]

Answer:

For example, the atomic radius of the metal zirconium, Zr, (a period-5 transition element) is 155 pm (empirical value) and that of hafnium, Hf, (the corresponding period-6 element) is 159 pm. ... The increase in mass and the unchanged radii lead to a steep increase in density from 6.51 to 13.35 g/cm3.

Explanation:

6 0

2 years ago