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andre [41]
3 years ago
14

What is the formula equation for the reaction between sulfuric acid and dissolved sodium hydroxide if all products and reactants

are in the aqueous or liquid phase?
Chemistry
2 answers:
Ber [7]3 years ago
8 0

Explanation:

Reactants are the species that are present on left hand side of a chemical reaction. Whereas products are the species that are present on right hand side of a chemical reaction.

For example, H_{2}SO_{4} + 2NaOH \rightarrow Na_{2}SO_{4} + 2H_{2}O

Here, both sulfuric acid and sodium hydroxide are the reactants. On the other hand, both water and sodium sulfate are the products.  

sattari [20]3 years ago
5 0
H2SO4 + 2NaOH = Na2SO4 + 2H20
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Calculate the energy (in kj/mol) required to remove the electron in the ground state for each of the following one-electron spec
Bess [88]

Explanation:

E_n=-13.6\times \frac{Z^2}{n^2}ev

where,

E_n = energy of n^{th} orbit

n = number of orbit

Z = atomic number

a) Energy change due to transition from n = 1 to n = ∞ ,hydrogen atom .

Z = 1

Energy of n = 1 in an hydrogen like atom:

E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6 eV

Energy of n = ∞ in an hydrogen like atom:

E_{\infty}=-13.6\times \frac{1^2}{(\infty)^2}eV=0

Let energy change be E for 1 atom.

E=E_{\infty}-E_1=0-(-13.6  eV)=13.6 eV

1 mole = 6.022\times 10^{-23}

Energy for 1 mole = E'

E'=6.022\times 10^{-23} mol^{-1}\times 13.6 eV

1 eV=1.60218\times 10^{-22} kJ

E'=6.022\times 10^{23}\times 13.6 \times 1.60218\times 10^{-22} kJ/mol

E'=1,312.17 kJ/mol

The energy  required to remove the electron in the ground state is 1,312.17 kJ/mol.

b) Energy change due to transition from n = 1 to n = ∞ ,B^{4+} atom .

Z = 5

Energy of n = 1 in an hydrogen like atom:

E_1=-13.6\times \frac{5^2}{1^2}eV=-340 eV

Energy of n = ∞ in an hydrogen like atom:

E_{\infty}=-13.6\times \frac{5^2}{(\infty)^2}eV=0

Let energy change be E.

E=E_{\infty}-E_1=0-(-340eV)=340 eV

1 mole = 6.022\times 10^{-23}

Energy for 1 mole = E'

E'=6.022\times 10^{-23} mol^{-1}\times 340eV

1 eV=1.60218\times 10^{-22} kJ

E'=6.022\times 10^{23}\times 340\times 1.60218\times 10^{-22} kJ/mol

E'=32,804.31 kJ/mol

The energy  required to remove the electron in the ground state is 32,804.31 kJ/mol.

c) Energy change due to transition from n = 1 to n = ∞ ,Li^{2+}atom .

Z = 3

Energy of n = 1 in an hydrogen like atom:

E_1=-13.6\times \frac{3^2}{1^2}eV=-122.4 eV

Energy of n = ∞ in an hydrogen like atom:

E_{\infty}=-13.6\times \frac{3^2}{(\infty)^2}eV=0

Let energy change be E.

E=E_{\infty}-E_1=0-(-122.4 eV)=122.4 eV

1 mole = 6.022\times 10^{-23}

Energy for 1 mole = E'

E'=6.022\times 10^{-23} mol^{-1}\times 122.4 eV

1 eV=1.60218\times 10^{-22} kJ

E'=6.022\times 10^{23}\times 122.4\times 1.60218\times 10^{-22} kJ/mol

E'=11,809.55 kJ/mol

The energy  required to remove the electron in the ground state is 11,809.55 kJ/mol.

d) Energy change due to transition from n = 1 to n = ∞ ,Mn^{24+}atom .

Z = 25

Energy of n = 1 in an hydrogen like atom:

E_1=-13.6\times \frac{25^2}{1^2}eV=-8,500 eV

Energy of n = ∞ in an hydrogen like atom:

E_{\infty}=-13.6\times \frac{25^2}{(\infty)^2}eV=0

Let energy change be E.

E=E_{\infty}-E_1=0-(-8,500 eV)=8,500 eV

1 mole = 6.022\times 10^{-23}

Energy for 1 mole = E'

E'=6.022\times 10^{-23} mol^{-1}\times 8,500eV

1 eV=1.60218\times 10^{-22} kJ

E'=6.022\times 10^{23}\times 8,500 \times 1.60218\times 10^{-22} kJ/mol

E'=820,107.88 kJ/mol

The energy  required to remove the electron in the ground state is 820,107.88 kJ/mol.

4 0
3 years ago
Part C: complete the third column <br> Part D: complete the fourth column
Helga [31]

Answer:

Part C: P2 = 0.30 atm

Part D: V1 = 16.22 L.

Explanation:

Part C:

Initial pressure (P1) = 2.67 atm

Initial volume (V1) = 5.54 mL

Final pressure (P2) =.?

Final volume (V2) = 49 mL

The final pressure (P2) can be obtained as follow:

P1V1 = P2V2

2.67 x 5.54 = P2 x 49

Divide both side by 49

P2 = (2.67 x 5.54)/49

P2 = 0.30 atm

Therefore, the final pressure (P2) is 0.30 atm

Part D:

Initial pressure (P1) = 348 Torr

Initial volume (V1) =?

Final pressure (P2) = 684 Torr

Final volume (V2) = 8.25 L

The initial volume (V1) can be obtained as follow:

P1V1 = P2V2

348 x V1 = 684 x 8.25

Divide both side by 348

V1 = (684 x 8.25)/348

V1 = 16.22 L

Therefore, the initial volume (V1) is 16.22 L

6 0
3 years ago
How many hydrogen atoms are in a cycloalkane with 8 carbon atoms?
frosja888 [35]
Cycloalkanes are those saturated organic compounds which exist in the form of Rings. Their Hydrogen Deficiency Index in one. The General formula for cycloalkanes is,
                                                  CnH2n
When number of Carbons = 8
Then
                                                  C₈H₂₍₈₎

                                                  C₈H₁₆

Result:
           
Cycloalkane containing 8 carbon atoms has 16 hydrogen atoms.

8 0
3 years ago
Helium occupies a volume of 3.8 L at –45°C. What was its initial temperature when it occupied 8.3 L?
USPshnik [31]

Answer:

98.3 gradius Celsius

Explanation:

This problem is solved using the Ideal Gas Equation

pV = nRT

...

Please see the step-by-step solution in the picture attached below.

Hope this answer can help you. Have a nice day!

3 0
3 years ago
Which best describes the structure of 2-butene
lyudmila [28]
The 2 represents that it is a double carbon bond
it looks like..
C-C = C-C
6 0
3 years ago
Read 2 more answers
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