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laiz [17]
2 years ago
9

A gas occupies 3.5L at 2.5 kPa pressure. What is the volume at 100 mmHg at the same temperature? Be sure to

Chemistry
1 answer:
nataly862011 [7]2 years ago
7 0

Answer:

V₂ = 0.656 L

Explanation:

Given data:

Initial volume = 3.5 L

Initial pressure = 2.5 KPa

Final volume = ?

Final pressure = 100 mmHg (100/7.501=13.33 KPa)

Solution:

The given problem will be solved through the Boyle's law,

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume  

Now we will put the values in formula,

P₁V₁ = P₂V₂

2.5 KPa × 3.5 L = 13.33 KPa × V₂

V₂ = 8.75 KPa. L/13.33 KPa

V₂ = 0.656 L

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3 years ago
Calculate the solubility of copper(II) hydroxide, Cu(OH)2, in g/L​
Oduvanchick [21]

Answer:

Ksp = [ Cu+² ] [ OH-] ²

molar mass Cu(oH )2 ==> M= 63.546 (1) + 16 (2) + 1 (2) = 97.546 g/mol

Ksp = [ Cu+² ] [ OH-] ²

Ksp [ cu (OH)2 ] = 2.2 × 10-²⁰

|__________|___<u>Cu</u><u>+</u><u>²</u><u> </u>__|_<u>2</u><u>OH</u><u>-</u>____|

|<u>Initial concentration(M</u>)|___<u>0</u>__|_<u>0</u>______|

<u>|Change in concentration(M)</u>|_<u>+S</u><u> </u>|__<u>+2S</u>__|

|<u>Equilibrium concentration(M)|</u><u>_S</u><u> </u><u>_</u><u>|</u><u>2S___</u><u>|</u>

Ksp = [ Cu+² ] [ OH-] ²

2.2 ×10-²⁰ = (S)(2S)²= 4S³

s =  \sqrt[3]{ \frac{2.2 \times  {10}^{ - 20} }{4} }  = 1.8 \times  {10}^{ - 7}

S = 1.8 × 10-⁷ M

The molar solubility of Cu(OH)2 is 1.8 × 10-⁷ M

Solubility of Cu (OH)2 =

Cu (OH)2 =  \frac{1.8 \times  {10}^{ - 7} mol \:Cu (OH)2 }{1L}  \times  \frac{97.546 \: g \: Cu (OH)2}{1 \: mol \: Cu (OH)2}  \\  = 1.75428 \times 10 ^{ - 5}

<h3>Solubility of Cu (OH)2 = 1.75428 × 10 -⁵ g/ L</h3>

I hope I helped you^_^

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daser333 [38]

In 1869 he published a table of the elements organized by increasing atomic mass.

Mendeleev is called the "father of the modern periodic table

stated that if the atomic weight of an element caused it to be placed in the wrong group, then the weight must be wrong. (He corrected the atomic masses of Be, In, and U)

was so confident in his table that he used it to predict the physical properties of three elements that were yet unknown.

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However, in spite of Mendeleev's great achievement, problems arose when new elements were discovered and more accurate atomic weights determined.

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