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3 years ago

A mothball, composed of naphthalene (c10h8), has a mass of 1.64 g . part a how many naphthalene molecules does it contain?

1 answer:

6 0

Answer is: 7,5·10²¹ <span>naphthalene molecules.
</span>m(C₁₀H₈) = 1,64 g.
n(C₁₀H₈) = m(C₁₀H₈) ÷ M(C₁₀H₈).
n(C₁₀H₈) = 1,64 g ÷ 128,17 g/mol.
n(C₁₀H₈) = 0,0125 mol.
N(C₁₀H₈) = n(C₁₀H₈) · Na.
N(C₁₀H₈) = 0,0125 mol · 6,022·10²³ 1/mol.
N(C₁₀H₈) = 7,5·10²¹.
Na - Avogadro number.

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If a red blood cell is 100% saturated, how many molecules of O2 are bound to it?1 billion molecules of O24 molecules of O2250 mi

kondor19780726 [428]

Answer:

1 billion molecules O₂

Explanation:

From my research, a human red blood cell contains approximately 270 million hemoglobin molecules.

A hemoglobin molecule contains four heme groups, <em>each of which has an iron ion forming a coordination complex that carries every dioxygen molecule. </em>Therefore for each hemoglobin molecule, we will have 4 dioxygen molecules. The heme groups are responsible for the transport of every dioxygen and other diatomic gases.

Hence, the number of O₂ molecules in a red blood cell saturated with 100% will be:

$\frac{270 \cdot10^{6} hemoglobine molecules}{1 red blood cell} \cdot \frac{4 heme group}{1 hemoglobine molecule} \cdot \frac{1 O_{2} molecules}{1 heme group} = 1 \cdot 10^{9} O_{2} \frac{molecules}{red blood cell}$

So, the correct answer is 1 billion of O₂ molecules.

Have a nice day!

8 0

3 years ago

How many moles are in 1.51x10^26 atoms of xenon (Xe)? Please and thank you :)!!

RoseWind [281]

<h3>Answer:</h3>

251 mol Xe

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 1.51 × 10²⁶ atoms Xe

[Solve] moles Xe

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 1.51 \cdot 10^{26} \ atoms \ Xe(\frac{1 \ mol \ Xe}{6.022 \cdot 10^{23} \ atoms \ Xe})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 250.747 \ mol \ Xe$

<u>Step 4: Check</u>

<em>Follow sig fig rule and round. We are given 3 sig figs.</em>

250.747 mol Xe ≈ 251 mol Xe

3 0

3 years ago

Why are plastic containers preferred to glass containers for storing chemicals in the laboratory?

Nookie1986 [14]

Plastics are non-corrosive and non-reactive in nature. So they are used for storing chemicals in the laboratory. They are used for strong chemicals because they do not react with chemicals neither do they corrode

7 0

3 years ago

What are the very small particles that make up all matter?

nadezda [96]

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3 0

3 years ago

Consider 0.01 m aqueous solutions of each of the following. a) NaI; b) CaCl2; c) K3PO4; and d) C6H12O6 (glucose) Arrange the sol

stealth61 [152]

Answer:

The solutions are ordered by this way (from lowest to highest freezing point): K₃PO₄ < CaCl₂ < NaI < glucose

Option d, b, a and c

Explanation:

Colligative property: Freezing point depression

The formula is: ΔT = Kf . m . i

ΔT = Freezing T° of pure solvent - Freezing T° of solution

We need to determine the i, which is the numbers of ions dissolved. It is also called the Van't Hoff factor.

Option d, which is glucose is non electrolyte so the i = 1

a. NaI → Na⁺ + I⁻ i =2

b. CaCl₂ → Ca²⁺ + 2Cl⁻ i =3

c. K₃PO₄ → 3K⁺ + PO₄⁻³ i=4

Potassium phosphate will have the lowest freezing point, then we have the calcium chloride, the sodium iodide and at the end, glucose.

7 0

3 years ago