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larisa86 [58]
3 years ago
9

Fireworks that contain metallic salts such as sodium, strontium, and barium can generate bright colors. a technician investigate

s what colors are produced by the metallic salts by performing flame tests. during a flame test, a metallic salt is heated in the flame of a gas burner. each metallic salt emits a characteristic colored light in the flame. 72 explain why the electron configuration of 2-7-1-1 represents a sodium atom in an excited state. [1]
Chemistry
1 answer:
Tasya [4]3 years ago
7 0

Answer is: The electrons moved from the second energy level to the fourth.

Atomic number (Z) of sodium is 11, it means that it has 11 protons and 11 electrons, so atom of sodium is neutral.

Electron configuration of sodium atom in a ground state:

₁₁Na 1s² 2s² 2p⁶ 3s¹ or 2-8-1-0.

In a ground state of sodium atom, fourth energy level is empty.

In a excited state electron from second energy level (2s² 2p⁶) moves to fourth energy level.

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Komok [63]

1. a. 3

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b. 6

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4 0
2 years ago
What is the electric field (in N/C) at a point 5.0 cm from the negative charge and along the line between the two charges?
Natali [406]

Answer: E = 2.455 x 10^5 N/C

Explanation:

q1 = 1.2x10^-7C

q2 = 6.2x10^-8C

Electric field, E = kQ/r²

where k = 9.0x10^9

since the location is (27 - 5)cm from q1

hence electric field, E1 = k*q1/r²

E1= (9x10^9 x 1.2x10^-7)/(0.22)² = 22314.05 N/C

for q2:

E1 = k*q2/r²

E2 at 5cm

E2 = (9x10^9 x 6.2x10^-8)/(0.05)² = 223200 N/C

Hence, the total electric field at 5cm position is

E = E1 + E2

E = 22314.05 + 223200 = 245514.05 N/C

E = 2.455 x 10^5 N/C

4 0
3 years ago
Explain the difference between complete dominance and codominance.Give an example of each one in order to explain it.
Ira Lisetskai [31]
In complete dominance, only one allele in the genotype is seen in the phenotype. In codominance, both alleles in the genotype are seen in the phenotype. In incomplete dominance, a mixture of the alleles in the genotype is seen in the phenotype.

I hope it helps
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8 0
3 years ago
Rank these real gases according to how closely they resemble an ideal gas. CO, N2, Ne, He, NH3
Gemiola [76]

He is the closest. Then:

Ne, N2, CO, NH3.

NH3 is the least closest to ideal.

8 0
3 years ago
If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of H2SO4(aq)? For your ans
Alik [6]

Question is incomplete, complete question is;

A 34.8 mL solution of H_2SO_4 (aq) of an unknown concentration was titrated with 0.15 M of NaOH(aq).

H_2SO_4(aq)+2NaOH(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)

If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of H_2SO_4(aq)? For your answer, only type in the numerical value with two significant figures. Do NOT include the unit.

Answer:

0.044 M is the molarity of H_2SO_4(aq).

Explanation:

The reaction taking place here is in between acid and base which means that it is a neutralization reaction .

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_1,M_2\text{ and }V_2  are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=?\\V_1=34.8 mL\\n_2=1\\M_2=0.15 M\\V_2=20.4 mL

Putting values in above equation, we get:

2\times M_1\times 34.8 mL=1\times 0.15 M\times 20.4 mL\\\\M_1=\frac{1\times 0.15 M\times 20.4 mL\times 10}{2\times 34.8 mL}=0.044 M

0.044 M is the molarity of H_2SO_4(aq).

4 0
3 years ago
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