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Levart [38]
3 years ago
12

The area of the circular base of a can of

Chemistry
1 answer:
zlopas [31]3 years ago
6 0

Answer:

Volume = 75 inch³

Explanation:

Given data:

Area = 12.5 inch

Height = 6 inch

Volume =?

Solution:

Volume = height × length × width

Volume =  Area × height

Now we will put the values in formula;

Volume = 12.5 inch² ×  6 inch

Volume = 75 inch³

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Freezing - Liquid to Solid

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Sublimation - Solid to Gas & Beleive It Or Not There Is One Called

Deposition (sometimes called Desublimation) - Gas to Solid

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Metallic bonds are found between two?
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Are found between two metals

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The number of outer shell electrons determines the chemical properties of an element?
Mandarinka [93]
The number of outer electrons determines the chemical properties of an element, which is a correct statement. Because the number of valence electrons determines the bonding. 

Answer: True
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ILL GIVE 30 POINTS TO THE BRAINIEST
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Every chemical equation follows to the law of conservation of mass, which states that matter cannot be created or destroyed. When you have an equal number of atoms of an element is present on both sides of a chemical equation, the equation is balanced therefore it is following the law of conservation of mass.

Explanation:

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1.674×10-4 mol of an unidentified gaseous substance effuses through a tiny hole in 86.6 s. Under identical conditions, 1.715×10-
siniylev [52]
<h2>Answer:</h2>

44.06 g/mol

<h3>Explanation:</h3>

We are given;

  • Number of moles of unidentified gas as 1.674×10^-4 mol
  • Time of effusion of unidentified gas 86.6 s
  • Number of moles of Argon gas as 1.715×10^-4 mol
  • Time of effusion of Argon gas is 84.5 s

We are supposed to calculate the molar mass of unidentified gas

<h3>Step 1: Calculate the effusion rates of each gas</h3>

Effusion rate = Number of moles/time

Effusion rate of unidentified gas (R₁)

 =  1.674×10^-4 mol ÷ 86.6 s

 = 1.933 × 10^-6 mol/s

Effusion rate of Argon gas (R₂)

 = 1.715×10^-4 mol ÷ 84.5 sec

= 2.030 × 10^-6 mol/s

<h3>Step 2: Calculate the molar mass of unidentified gas</h3>
  • Assuming the molar mass of unidentified gas is x;
  • We can use the Graham's law of effusion to find x;
  • According to Graham's law of diffusion;

\frac{R_{1}}{R_{2}}}=\frac{\sqrt{MM_{Ar}}}{\sqrt{X}}

But, Molar mass of Argon is 39.948 g/mol

Therefore;

\frac{1.933*10^-6mol/s}{2.030*10^-6mol/s}}=\frac{\sqrt{39.948}}{\sqrt{X}}

0.9522=\frac{\sqrt{39.948}}{\sqrt{X}}

Solving for X

x = 44.06 g/mol

Therefore, the molar mass of the identified gas is 44.06 g/mol

3 0
4 years ago
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