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Nuetrik [128]
3 years ago
9

1. I am a solution with a pOH of 6. What am I?

Chemistry
1 answer:
ivolga24 [154]3 years ago
8 0

Explanation:

I think 1) acidic solution

2) basic solution.

You might be interested in
determine the frequency and wavelength (in nm) of the light emitted when the e- fell from n=4 and n=2
Lostsunrise [7]

Answer:

Frequency = 6.16 ×10¹⁴ Hz

λ = 4.87×10² nm

Explanation:

In case of hydrogen atom energy associated with nth state is,

En =  -13.6/n²

For n = 2

E₂ = -13.6 / 2²

E₂ = -13.6/4

E₂ = -3.4 ev

Kinetic energy of electron = -E₂ = 3.4 ev

For n = 4

E₄ = -13.6 / 4²

E₄ = -13.6/16

E₄ = -0.85 ev

Kinetic energy of electron = -E₄ = 0.85 ev

Wavelength of radiation emitted:

E = hc/λ = E₄ - E₂

hc/λ = E₄ - E₂

by putting values,

6.63×10⁻³⁴Js × 3×10⁸m/s / λ = -0.85ev   - (-3.4ev )

6.63×10⁻³⁴ Js× 3×10⁸m/s / λ = 2.55 ev

λ = 6.63×10⁻³⁴ Js× 3×10⁸m/s  /2.55ev

λ = 6.63×10⁻³⁴ Js× 3×10⁸m/s  /2.55× 1.6×10⁻¹⁹ J

λ = 19.89 ×10⁻²⁶ Jm / 2.55× 1.6×10⁻¹⁹ J

λ = 19.89 ×10⁻²⁶ Jm / 4.08×10⁻¹⁹ J

λ = 4.87×10⁻⁷ m

m to nm:

4.87×10⁻⁷ m ×10⁹nm/1 m

4.87×10² nm

Frequency:

Frequency = speed of electron / wavelength

by putting values,

Frequency = 3×10⁸m/s /4.87×10⁻⁷ m

Frequency = 6.16 ×10¹⁴ s⁻¹

s⁻¹ = Hz

Frequency = 6.16 ×10¹⁴ Hz

3 0
3 years ago
How many moles of Al would be produced from 20 moles of Al2O3?<br> 2Al2O3<br> -&gt;<br> 4A1 + 302
Evgesh-ka [11]
<h3>Answer:</h3>

\displaystyle 40 \ mol \ Al

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 2Al₂O₃ → 4Al + 3O₂

[Given] 20 mol Al₂O₃

<u>Step 2: Identify Conversions</u>

[RxN] 2 mol Al₂O₃ → 4 mol Al

<u>Step 3: Stoich</u>

  1. [DA] Set up:                                                                                                     \displaystyle 20 \ mol \ Al_2O_3(\frac{4 \ mol \ Al}{2 \ mol \ Al_2O_3})
  2. [DA] Multiply/Divide [Cancel out units]:                                                         \displaystyle 40 \ mol \ Al

<u>Step 4:Check</u>

<em>Follow sig fig rules and round. We are given 1 sig fig.</em>

Since our final answer already has 1 sig fig, there is no need to round.

4 0
3 years ago
ch sulfide (ZnS) occurs in the zinc blende crystal structure, (a) If 254 g of ZnS contains 170 g of Zn, what is the mass ratio o
Aleonysh [2.5K]

Answer:

The mass ratio of zinc to sulfide is 85:42.

2.5559 kg of Zn are in 3.82 kg of ZnS.

Explanation:

a) Mass of zinc sulfide = 254 g

Mass of zinc in a zinc sulfide sample = 170 g

Mass of sulfide in zinc sulfide sample = x

254 g = 170 g+ x

x = 84 g

The mass ratio of zinc to sulfide:

\frac{170 g}{84 g}=\frac{85}{42}

b) Mass of zincsulfide sample = 3.83 kg

The mass ratio of zinc to sulfide is 85:42.

Let the mass of zinc and sulfide be 85x and 42x respectively:

85 x+ 42 x=3.82 kg

x =0.03007 kg

Mass of an zinc= 85x=85 × 0.03007 kg= 2.5559 kg

8 0
3 years ago
How many moles of water as a gas can be formed 2.45 L
Alex17521 [72]

Answer:

0.11mole

Explanation:

Let us assume that the condition is at standard temperature and pressure(STP);

 Given parameters:

        Volume of water  = 2.45L

   Unknown:

       Number of moles found in this volume of water  = ?

Solution;

 At STP;

                 Number of moles  = \frac{volume of gas}{22.4}

 Input the parameters and solve;

                  Number of moles of water  = \frac{2.45}{22.4}   = 0.11mole

The number of moles of water found is  0.11mole

4 0
3 years ago
The substance argon has the following properties: normal melting point: 83.9 K normal boiling point: 87.4 K triple point: 0.68 a
Vsevolod [243]

Answer:

The final state of the substance is a gas.

The sample is initially a liquid. One or more phase changes will occur.

Explanation:

Let's consider the phase diagram for Argon (not to scale).

<em>A sample of argon is initially at a pressure of 49.6 atm and a temperature of 101.4 K. The pressure on the sample is reduced to 0.680 atm at a constant temperature of 101.4 K. Which of the following are true? Choose all that apply </em>

<em>The final state of the substance is a gas.</em> TRUE. At 0.680 atm and 101.4 K, the substance is a gas.

<em>The gas initially present will solidify.</em> FALSE. Initially, Ar is present as a liquid.

<em>The final state of the substance is a solid.</em> FALSE.

<em>The sample is initially a liquid. One or more phase changes will occur.</em> TRUE. The sample is initially liquid and only one phase change will occur.

5 0
3 years ago
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