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alina1380 [7]
4 years ago
15

Chemistry help please

Chemistry
1 answer:
Lerok [7]4 years ago
5 0

balanced chemical equation:

2 Na + Cl₂ → 2 NaCl

Explanation:

We have the following chemical reaction:

Na (s) + Cl₂ (g) → NaCl (s)

where:

(s) - solid

(g) - gas

To balance the equation the number and type of atoms entering the reaction have to be equal to the number and type of atoms leaving the reaction.

2 Na (s) + Cl₂ (g) → 2 NaCl (s)

2 Na atoms are entering the reaction 2 Na atoms are leaving the reaction

2 Cl atoms are entering the reaction 2 Cl atoms are leaving the reaction

Learn more about:

balancing chemical equations

brainly.com/question/13601206

#learnwithBrainly

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Describe the orbital notation and electron configuration notation to describe the placement of all of the electrons in an atom o
andriy [413]
The arrangement of the electrons on the orbitals is governed by the principles of quantum energy. To be able to draw the correct electronic configuration of an element, they should follow these three principles.

1. Aufbau's Principle. The electrons of an element should be filled up from the highest energy level. You can use the periodic table as a guide for this. In decreasing order, the energy of the orbitals are 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d and 7p. Each 'box' of these orbital must be filled with two electrons from 1s before moving on to the next.

2. Hund's rule. This states that you must fill all boxes of one orbital with one electron first. For example, 2p has 3 boxes. You fill one electron for each box first before going back to fill it to two electrons if there are still excess.

3. Pauli's Exclusion Principle. This states that no electrons of the same element should have the same quantum numbers. To achieve this, the two electrons in each box should have opposite spins, one facing up, and the other facing down.

Using these three rules, you will be able to draw the electronic configuration of Fluorine with 9 electrons as shown in the picture.

7 0
4 years ago
Calculate the new molarity that results when 250.mL of water is added to each of the following solutions.
sergiy2304 [10]

Answer:

The answer to your question is 1) 0.037 M  2) 0.32 M  3)  0.096 M

Explanation:

a) 125 ml of 0.251 M HCl

-Calculate the moles of HCl

Molarity = moles/volume

-Solve for moles

moles = Molarity x volume

-Substitution

moles = 0.251 x 0.125

           = 0.0314

-Calculate the new molarity

Molarity = 0.0314/ (0.125 + 0.250)

-Simplification

Molarity = 0.014/0.375

-Result

Molarity = 0.037 M

2.-

445 ml of 0.499 M of H₂SO₄

-Calculate the number of moles

moles = 0.499 x 0.445

moles = 0.222

-Calculate the new molarity

Molarity = 0.222/(0.445 + 0.25)

Molarity = 0.222/0.695

Molarity = 0.32

3)

5.25 l of HCO₃ 0.101 M

Calculate the number of moles

moles = 0.101 x 5.25

moles = 0.53

-Calculate the Molarity

Molarity = 0.53 / (0.25 + 5.25)

Molarity = 0.53 / 5.5

Molarity = 0.096

8 0
3 years ago
For a particular isomer of C8H18, the combustion reaction produces 5113.3 kJ of heat per mole of C8H18(g) consumed, under standa
mart [117]

Answer: The standard enthalpy of formation of this isomer of C_8H_{18}(g) is -210.9 kJ

Explanation:

The given balanced chemical reaction is,

C_8H_{18}(g)+\frac{25}{2}O_2(g)\rightarrow 8CO_2(g)+9H_2O(g)

First we have to calculate the enthalpy of formation of  C_8H_{18}.

\Delta H^o=H_f_{product}-H_f_{reactant}

\Delta H^o=[n_{CO_2}\times \Delta H_f^0_{(CO_2)}+n_{H_2O}\times \Delta H_f^0_{(H_2O)}]-[n_{O_2}\times \Delta H_f^0_{(O_2)+n_{C_8H_{18}}\times \Delta H_f^0_{(C_8H_{18})}]

where,

We are given:

\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-241.8kJ/mol\\\Delta H^o_f_{(O_2(g)))}=0kJ/mol

Putting values in above equation, we get:

-511.3kJ/mol=[(8\times -393.5)+(9\times -241.8)]-[(\frac{25}{2}\times 0)+(1\times \Delta H_f^0_{(C_8H_{18})}

\Delta H_f^0_{(C_8H_{18})}=-210.9kJ

7 0
4 years ago
Read 2 more answers
Consider a galvanic cell consisting of the following two redox couplesAlB+ 3e Al , Mga + 2e Mg, (ag. E® = 1.676 V ) E° = 2.356 V
malfutka [58]

Answer:

0.68 V

Explanation:

For anode;

3Mg(s) ---->3Mg^2+(aq) + 6e

For cathode;

2Al^3+(aq) + 6e -----> 2Al(s)

Overall balanced reaction equation;

3Mg(s) + 2Al^3+(aq) ----> 3Mg^2+(aq) + 2Al(s)

Since

E°anode = -2.356 V

E°cathode = -1.676 V

E°cell=-1.676 -(-2.356)

E°cell= 0.68 V

4 0
3 years ago
I need help as soon as posible
In-s [12.5K]
2. CO
3.O2
4.NH3
5.NO
6.NO2
7.CO2
8.H2
3 0
4 years ago
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