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chubhunter [2.5K]
3 years ago
6

What is the molarity of a solution made with 400 grams of copper II chloride in 3.5 liters of water?

Chemistry
1 answer:
Leni [432]3 years ago
4 0

Answer:

0.85M

Explanation:

Given parameters:

Mass of CuCl₂  = 400g

Volume of water  = 3.5L

Unknown:

Molarity of the solution = ?

Solution:

Molarity is one of the ways of expressing the concentrations of a solution. It is defined as the number of moles per unit volume of a solvent.

          Molarity  = \frac{number of moles}{volume}

To find the number of moles of CuCl₂;

    Number of moles  = \frac{mass}{molar mass}

  Molar mass of CuCl₂ = 63.6 + 2(35.5)  = 134.6g/mol

   Number of moles  = \frac{400}{134.6}   = 2.97moles

      Molarity  = \frac{2.97}{3.5}    = 0.85M

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Visible light has an average wavelength of 550 nm. How much energy would be in this wavelength of light? 5 x 10-19​
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Answer:

E = 3.6×10⁻¹⁹ J

Explanation:

Given data:

Wavelength = 550 nm  (550 ×10⁻⁹ nm)

Energy of wave = ?

Solution:

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E = h c/λ

c = 3×10⁸ m/s

h = 6.63×10⁻³⁴ Js

Now we will put the values in formula.

E = 6.63×10⁻³⁴ Js × 3×10⁸ m/s /550 ×10⁻⁹ nm

E = 19.89×10⁻²⁶ J.m /550 ×10⁻⁹ nm

E = 0.036×10⁻¹⁷ J

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Read 2 more answers
The total mass of the atmosphere is about 5.00 x 1018 kg. How many moles each of air, O2, and CO2 are present in the atmosphere?
n200080 [17]

<u>Answer:</u> The moles of oxygen and carbon dioxide in air is 3.63\times 10^{19}mol and 7.18\times 10^{16}mol respectively

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of atmosphere = 5.00\times 10^{18}kg=5.00\times 10^{21}g

Average molar mass of atmosphere = 28.96 g/mol

Putting values in above equation, we get:

\text{Moles of atmosphere}=\frac{5.00\times 10^{21}g}{28.96g/mol}=1.73\times 10^{20}mol

We know that:

Percent of oxygen in air = 21 %

Percent of carbon dioxide in air = 0.0415 %

Moles of oxygen in air = \frac{21}{100}\times 1.73\times 10^{20}=3.63\times 10^{19}mol

Moles of carbon dioxide in air = \frac{0.0415}{100}\times 1.73\times 10^{20}=7.18\times 10^{16}mol

Hence, the moles of oxygen and carbon dioxide in air is 3.63\times 10^{19}mol and 7.18\times 10^{16}mol respectively

6 0
3 years ago
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