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Mrrafil [7]
2 years ago
15

choose the reaction that illustrates delta H *f for Ca(NO3)2.(A) Ca (s) + N2 (g) + 3O2 ---> Ca(NO3)2 (s)(B) Ca2 (aq) + 2 NO3-

(aq) ---> Ca(NO3)2 (aq)(C) Ca (s) + 2N (g) + 6O (g) ---> Ca(NO3)2 (s)(D) Ca(NO3)2 (aq) ---> Ca2+ (aq) + 2NO3- (aq)(E) Ca(NO3)2 (s) ---> Ca(s) + N2 (g) + 3 O2 (g)
Chemistry
1 answer:
kompoz [17]2 years ago
3 0

<u>Answer:</u> The correct answer is Option A.

<u>Explanation:</u>

Standard enthalpy of formation is the change in enthalpy of  one mole of a substance present at the standard state that is 1 atm of pressure and 298 K of temperature. The substance is formed from its pure elements under the same conditions.

We are given a chemical compound having chemical formula Ca(NO_3)_2

This compound is formed by the combination of calcium, nitrogen and oxygen elements.

The chemical equation for the formation of Ca(NO_3)_2 from the components in their standard states follows:

Ca+N_2+3O_2\rightarrow Ca(NO_3)_3

Hence, the correct answer is Option A.

You might be interested in
At 10°c one volume of water dissolves 3.10 volumes of chlorine gas at 1.00 atm pressure. what is the henry's law constant of cl2
s344n2d4d5 [400]
Answer is:  0,133 mol/ l· atm.
T(chlorine) = 10°C = 283K.
p(chlorine) = 1 atm.
V(chlorine) = 3,10 l.
R - gas constant, R = 0.0821 atm·l/mol·K. 
Ideal gas law: p·V = n·R·T
n(chlorine) = p·V ÷ R·T.
n(chlorine) = 1atm · 3,10l ÷ 0,0821 atm·l/mol·K · 283K = 0,133mol.
Henry's law: c = p·k.
k - <span>Henry's law constant.
</span>c - solubility of a gas at a fixed temperature in a particular solvent.
c = 0,133 mol/l.
k = 0,133 mol/l ÷ 1 atm = 0,133 mol/ l· atm.

4 0
2 years ago
A graduated cylinder is filled to 10.0 mL with water and a piece of granite is placed in the cylinder displacing the level to 23
Rus_ich [418]

Answer:

13.7 cubic centimeters

Explanation:

23.7-10=13.7mL

8 0
2 years ago
Sodium oxide (Na2O) crystallizes in a structure in which the O2– ions are in a face - centered cubic lattice and the Na+ ions ar
melisa1 [442]

We have to know the number of Na⁺ ions in the unit cell.

The number of Na⁺ ions in the unit cell is (D) 8.

Sodium oxide (Na2O) crystallizes in a structure in which the O2– ions are in a face - centered cubic lattice and the Na+ ions are in tetrahedral holes.

O²⁻ ions are in a face centred cubic lattice, so the number of O²⁻ ions per unit cell is equal to 4. The number of  tetrahedral hole= 2 X 4=8. Na+ ions are present in tetrahedral holes, which indicates there are 8 number of Na+ ions in the unit cell.


3 0
2 years ago
Using complete subshell notation (not abbreviations, 1s 22s 22p 6 , and so forth), predict the electron configuration of each of
dybincka [34]

<u>Answer:</u> The electronic configuration of the elements are written below.

<u>Explanation:</u>

Electronic configuration is defined as the representation of electrons around the nucleus of an atom.

Number of electrons in an atom is determined by the atomic number of that atom.

For the given options:

  • <u>Option a:</u>  Carbon (C)

Carbon is the 6th element of the periodic table. The number of electrons in carbon atom are 6.

The electronic configuration of carbon is 1s^22s^22p^2

  • <u>Option b:</u>  Phosphorus (P)

Phosphorus is the 15th element of the periodic table. The number of electrons in phosphorus atom are 15.

The electronic configuration of phosphorus is 1s^22s^22p^63s^23p^3

  • <u>Option c:</u>  Vanadium (V)

Vanadium is the 23rd element of the periodic table. The number of electrons in vanadium atom are 23.

The electronic configuration of vanadium is 1s^22s^22p^63s^23p^64s^23d^3

  • <u>Option d:</u>  Antimony (Sb)

Antimony is the 51st element of the periodic table. The number of electrons in antimony atom are 51.

The electronic configuration of antimony is 1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^3

  • <u>Option e:</u>  Samarium (Sm)

Samarium is the 62nd element of the periodic table. The number of electrons in samarium atom are 62.

The electronic configuration of samarium is 1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^66s^24f^6

Hence, the electronic configuration of the elements are written above.

4 0
2 years ago
How many Cl- ions are there in 400.5 grams of AlCl3?
den301095 [7]

Answer:

The answer to your question is letter A

Explanation:

Process

1.- Calculate the molecular mass of AlCl₃

Al = 1 x 27 = 27

Cl = 3 x 35.5 = 106.5

AlCl₃ = 27 + 106.5 = 133.5

2.- Find the number of ions of Cl using proportions

               133.5 grams of AlCl₃ ------------- 3 ions of Chlorine

               400.5 grams of AlCl₃ ------------   x

                x = (400.5 x 3) / 133.5

                x = 9

6 0
2 years ago
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