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Shtirlitz [24]
3 years ago
14

Which is the name for the mixture of chemicals that forms as a gray brown haze in the atmosphere?

Chemistry
1 answer:
almond37 [142]3 years ago
7 0
The answer is Smog, hope this helps! pls mark brainliest! thanks!
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If you start with 0.30 m mn2 , at what ph will the free mn2 concentration be equal to 4.6 x 10-11 m?
aksik [14]

If you start with 0.30 m Mn₂ , at 12.5 pH, free Mn₂ concentration be equal to 4.6 x 10⁻¹¹ m

Initial molarity of Mn₂ = 0.30 M

Final molarity of Mn₂ = 4.6 x 10⁻¹¹

pH = ?

Ksp [Mn(OH)₂] = 4.6 x 10⁻¹⁴ (standard value)

Write the ionic equation

    Mn(OH)₂   →    Mn⁺² + 2OH⁻

    [Mn⁺²] = 4.6 x 10⁻¹¹

We will calculate the concentration of OH⁻ by using Ksp expression

    Ksp = [Mn⁺²][OH-]²

    [Mn⁺²][OH⁻]² = 4.6 x 10⁻¹⁴

    [OH⁻]² = 4.6 x 10⁻¹⁴ / 4.6 x 10⁻¹¹

    [OH⁻]² = 10⁻³

    [OH⁻] = (10⁻³)¹⁽²

    [OH⁻] = 0.0316 M

Calculate the pOH

    pOH = -log [OH⁻]

    pOH =  -log [0.0316]

    pOH = 1.5

Now calculate pH

   pH = 14 - pOH

   pH = 14 - 1.5

   pH = 12.5

You can also learn about molarity from the following question:

brainly.com/question/14782315

#SPJ4

7 0
1 year ago
I need help I’ll give you brainliest
Svetlanka [38]

Answer:

2 or 3 i think

Explanation:

5 0
3 years ago
Read 2 more answers
Using dimensional analysis convert 238.0 cm3 to L
Bezzdna [24]

Answer:

=  0.238 L

Explanation:

From the given information:

To convert cm³ to L(liter), we utilize the advent of a unit fraction that comprises a liter in the upper part of the fraction(numerator) and a cubic centimeter in the lower part of the fraction(denominator)

Recall that:

1000 cm³ = 1 liter

∴

238.0 cm³ = \dfrac{238.0 \ cm^3}{1} \times \dfrac{1 \ L}{1000 \ cm^3}

=  0.238 L

7 0
3 years ago
Elements in the same period have the same______. Every element in the first period has______ shell/orbital for its______. Every
vodomira [7]

Answer:

1. Number of shell

2. 1 shell

3. Atom

4. 2 shells

5. Atom

Explanation:

5 0
3 years ago
How many moles of N2 are in a flask with a volume of 0.5 L at a pressure of 400.0 kPa and a temperature of 300.0 K?
mylen [45]

Answer:

0.08moles

Explanation:

Given parameters:

Volume of gas  = 0.5L

Pressure of gas  = 400kPa

Temperature of gas  = 300K

Unknown:

Number of moles of N₂ = ?

Solution:

Applying the ideal gas law which is a combination of the three gas law: Boyle's law, Charles's law and Avogadro's law will solve this problem.

The ideal gas law is stated as;

              PV  = nRT

P is the pressure of the gas

V is the volume of the gas

n is the number of moles

R is the gas constant

T is the temperature of the gas

  We need to convert kPa of the pressure to atm which is a more comfortable unit to work with.

      400kPa will be \frac{400}{101.325}   = 3.95atm

Input the variables in the equation;

             3.95 x 0.5  = n x 0.082 x 300

                            n  = 0.08moles

5 0
3 years ago
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