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Papessa [141]
2 years ago
5

Tracie measured 87.47 mg of cholesterol in 0.03 mL of blood. What is the density of this mixture in g/mL?

Chemistry
1 answer:
Lelu [443]2 years ago
5 0

We can use a simple equation to solve this problem.


d = m/v


Where d is the density, m is the mass and v is the volume.


d = ?

m = 87.47 mg = 87.47 x 10⁻³ g

v = 0.03 mL


By applying the equation,

  d = 87.47 x 10⁻³ g / 0.03 mL

  d = 2.92 g/mL



Hence, the density of the mixture is 2.92 g/mL.


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Explain why groups 14 and 15 are better representatives of mixed groups than groups 13 and 16
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<h3><u>Answer</u>;</h3>

Groups 14 and 15 each contain metals, nonmetals, and metalloids while Group 13 contains metals and a metalloid, and Group 16 contains metalloids and nonmetals.

<h3><u>Explanation;</u></h3>
  • Groups 13–16 of the periodic table contain one or more metalloids, in addition to metals, nonmetals, or both.
  • Unlike other groups of the periodic table, which contain elements in one class, groups 13–16 referred to as mixed groups contain elements in at least two different classes. In addition to metalloids, they also contain metals, nonmetals, or both.
  • <em><u>Group 14 also known as the carbon group contains carbon which is a non metal, silicon and germanium which are metalloids and tin and lead which are metals.</u></em>
  • <em><u>Group 15 also known as the Nitrogen group contains non metals such as oxygen, metalloid tellurium and a metal polonium.</u></em>
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A chemist is working on a reaction represented by this chemical equation:
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Answer is: A. 1.81 mol.

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n(FeCl₂) = 4.15 mol; amount of iron(II) chloride.

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From chemical reaction: n(KOH) : n(Fe(OH)₂) = 2 : 1.

n(Fe(OH)₂) = n(KOH) ÷ 2.

n(Fe(OH)₂) = 3.62 mol ÷ 2.

n(Fe(OH)₂) = 1.81 mol; amount of iron(II) hydroxide.

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Answer:

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Equilibrium constant for decomposition of NO_2 is 1.79 \times 10^{-14}

Explanation:

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                    PCl_5 \rightleftharpoons PCl_3+Cl_2

initial          0.72 mol     0         0

at eq.     0.72 - 0.40   0.40      0.40

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It is given that 3.3 \times 10^{-3} \% is decomposed.

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                                  2NO_2 \rightleftharpoons 2NO + O_2

initial                            1.0 M       0           0

at eq. concentration of  NO_2   is:

[NO_2]_{eq}=1-(0.000066) = 0.999934\ M

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Expression for equilibrium constant is as follows:

K=\frac{[NO]^2[O_2]}{[NO_2]^2}

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Equilibrium constant for decomposition of NO_2 is 1.79 \times 10^{-14}

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