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Norma-Jean [14]
2 years ago
15

A green powder was placed into a beaker of water. After it was stirred the water looked cloudy and lumps of powder could still b

e seen. Has a solution been formed? Explain your answer?
Chemistry
1 answer:
sukhopar [10]2 years ago
7 0

Answer:

it make poisionus police come and took away

You might be interested in
The Balmer series, named after Johann Balmer, is a portion of the hydrogen emission spectrum produced from the transitions betwe
horrorfan [7]

Explanation:

The wavelength of the balmer series is calculated using the following steps;

- Find the Principle Quantum Number for the Transition

- Calculate the Term in Brackets

- Multiply by the Rydberg Constant

- Find the Wavelength

The Balmer series in a hydrogen atom relates the possible electron transitions down to the n = 2 position to the wavelength of the emission that scientists observe.

The λ symbol represents the wavelength, and RH is the Rydberg constant for hydrogen, with RH = 1.0968 × 107 m−1

n=7 to n=2

- The principal quantum numbers are 2 and 7.

-  (1/2²) − (1 / n²₂)

For n₂ = 7, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 7²)

= (1/4) − (1/49)

= 0.2230

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 0.2230

= 2445864 m−1

- λ = 1 / 2445864 m−1

= 4.08 × 10−7 m

= 408 nanometers

≈ 410nm

n=6 to n=2

- The principal quantum numbers are 2 and 6.

-  (1/2²) − (1 / n²₂)

For n₂ = 6, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 6²)

= (1/4) − (1/36)

=  0.2222

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 3/16

= 2437090 m−1

- λ = 1 / 2437090 m−1

= 4.10 × 10−7 m

= 410 nanometers

n=5 to n=2

- The principal quantum numbers are 2 and 5.

-  (1/2²) − (1 / n²₂)

For n₂ = 5, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 5²)

= (1/4) − (1/25)

= 0.21

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 0.21

= 2303280 m−1

- λ = 1 / 2303280 m−1

= 4.34 × 10−7 m

= 434 nanometers

n=4 to n=2

- The principal quantum numbers are 2 and 4.

-  (1/2²) − (1 / n²₂)

For n₂ = 4, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 4²)

= (1/4) − (1/16)

= 0.1875

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 0.1875

= 2056500 m−1

- λ = 1 / 2056500 m−1

= 4.86 × 10−7 m

= 486 nanometers

n=3 to n=2

- The principal quantum numbers are 2 and 3.

-  (1/2²) − (1 / n²₂)

For n₂ = 3, you get:

(1/2²) − (1 / n²₂) = (1/2²) − (1 / 3²)

= (1/4) − (1/9)

= 0.13889

- RH = 1.0968 × 107 m−1, to find a value for 1/λ. The formula and the example calculation gives:

1/λ = RH [(1/2²) − (1 / n²₂)]

= 1.0968 × 107 m−1 × 0.13889

= 1523345 m−1

- λ = 1 / 1523345 m−1

= 6.56 × 10−7 m

= 656 nanometers

7 0
2 years ago
Without using Appendix B, predict the sign of ΔS° for(b) KBr(s)→ KBr(aq)
Volgvan

Since the sign is positive, the entropy increased by 88.48 J/K.

Examine the phases of the species present to determine whether a physical or chemical process will cause an increase or decrease in entropy. Keep in mind "Silly Little Goats" to aid you in telling.

[1 Sf K+1 + 1 Sf Br-1 (aq)] ([1Sf(KBr (s))])

[1(102.5) + 1(82.42)] - [1(96.44)] = 88.48 J/K

If the entropy has grown, we say that Delta S is positive, and if it has dropped, we say that Delta S is negative. Due to its ionic nature, KBr is soluble in water and causes the 'K(+)' ions to hydrate.

Learn more about Entropy here-

brainly.com/question/13146879

#SPJ4

8 0
1 year ago
Complete and balance the following acid-base equations:
Alchen [17]

Answer:

a) HClO_4 (aq) + LiOH \rightarrow LiClO_4 + H_2O

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

c) Ba(OH)2 + 2HF\rightarrow BaF_2 + 2H_2O

Explanation:

a)

when HClO_4 is added to LiOH, lithium chlorate and water is formed.

HClO_4 (aq) + LiOH \rightarrow LiClO_4 + H_2O

Balancing of above reaction,

It can be seen that all the atoms in both the sides are balanced. so it is a balanced reaction.

(b) When aqueous H_2SO_4 is added to NaOH, Na_2SO_4 and H_2O is formed. It is a neutrilization reaction.

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

Balancing of above reaction,

First balance all the atoms except O and H

S atom is already balanced in either side

No. of Na atom in left hand side = 1

No. of Na atom in right hand side = 2

So multiply NaOH by 2, now the reaction becomes:

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

No. of O atoms in left hand side = 6

No. of O atoms in right hand side = 5

So multiply H2O by 2, now the reaction becomes:

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

Now, it can been seen that all the atoms are balanced.

c) When Ba(OH)2 reacts with HF, baroum fluoride and water is formed.

Ba(OH)2 + HF\rightarrow BaF_2 + H_2O

Balancing of above reaction,

Barium atoms are already balanced on either side.

No. of F atoms on left hand side = 1

No. of F atoms on right hand side = 2

So, multiply HF by 2, now reaction becomes

Ba(OH)2 + 2HF\rightarrow BaF_2 + H_2O

In order to balance H and O on either side, multiply H2O by 2.

Ba(OH)2 + 2HF\rightarrow BaF_2 + 2H_2O

7 0
3 years ago
Which of the following best defines molar mass?
coldgirl [10]

Answer:

D. the mass of one mole of a substance

Explanation:

The molar mass of a substance is the mass in grams of one mole of the substance.

  • For an element, the molar mass is the relative atoms mass expressed in grams.
  • For example 23g of Na, 40g of oxygen
  • For compounds, molar mass is the gram -formula or gram - molecular weight.
  • This is determined by the addition of its component atomic masses and then expressed in grams.
7 0
3 years ago
What is the cell potential of an electrochemical cell that has the half-reactions shown below?
Karo-lina-s [1.5K]

Answer:

E°(Ag⁺/Fe°) = 0.836 volt

Explanation:

3Ag⁺ + 3e⁻ => Ag°;            E° = +0.800 volt

Fe° => Fe⁺³ + 3e⁻ ;             E° = -0.036 volt

_________________________________

Fe°(s) + 3Ag⁺(aq) => Fe⁺³(aq) + 3Ag°(s) ...    

E°(Ag⁺/Fe°) = E°(Ag⁺) - E°(Fe°) = 0.800v - ( -0.036v) = 0.836 volt

4 0
2 years ago
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