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Katyanochek1 [597]
2 years ago
11

A sample of a pure element has a mass of 45.6g and contains 4.19 x 10 23 atoms. Identify the element.

Chemistry
1 answer:
max2010maxim [7]2 years ago
7 0
<h2>Answer:</h2>

    ZINC

<h2>Explanation:</h2>

<em>To identify the element based on the informartion given, we have to find the molar mass since this mass is unique to each element.</em>

            Molar mass = mass ÷ moles

<em>We already know the mass based on the question, as such we now need to find the # of moles.</em>

           Since 1 mole contains 6.02214 × 10²³ atoms

      then let  x moles contain 4.19 × 10²³ atoms <em>(given in the question)</em>

<em>     </em><em> </em>   ⇒   x =  (4.19 × 10²³ atoms  ×  1 mol) ÷ 6.02214 × 10²³ atoms

              x  =  0.69577 mol

<em>Now that we have the moles we can substitute it into the molar mass equation and solve for the molar mass.</em>

           ⇒ molar mass = 45.6 g ÷ 0.69577 mol

           ⇒ molar mass ≈  65.54 g/mol

                    This molar mass is closest to that of ZINC.

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Explanation:

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3 years ago
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Answer:

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3 0
2 years ago
How does positive catalyst increase the rate of chemical reaction?​
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6 0
3 years ago
In a certain industrial process involving a heterogeneous catalyst, the volume of the catalyst (in the shape of a sphere) is 10.
olga nikolaevna [1]

Answer:

a. The second run will be faster.

d. The second run has twice the surface area.

Explanation:

The rate of a reaction is proportional to the surface area of a catalyst. Given the volume (V) of a sphere, we can find its surface area (A) using the following expression.

A=\pi ^{1/3} (6V)^{2/3}

The area of the 10.0 cm³-sphere is:

A=\pi ^{1/3} (6.10.0)^{2/3}=22.4cm^{2}

The area of each 1.25 cm³-sphere is:

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The ratio of  8 1.25cm³-sphere to 10.0 cm³-sphere is 44.9 cm²/22.4 cm² = 2.00

Since the surface area is doubled, the second run will be faster.

6 0
3 years ago
(03.03 LC)
scoundrel [369]
I believe the answer is option B. The bonded pair of valence electrons are shown using circles
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3 years ago
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