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3 years ago

Is an apple a pure substance or a mixture?

Chemistry

2 answers:

Gnesinka [82]3 years ago

7 0

An apple is a mixture because a mixture has chemicals such as water, vitamins, sugar, and lots more.

guajiro [1.7K]3 years ago

6 0

An apple would be a mixture. While a mixture is a combination of elements and substances, a pure substance acts as an element itself, therefore, having the properties of only THAT certain element.

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After the solution reaches equilibrium, what concentration of Ni2+(aq) remains? The value of Kf for Ni(NH3)62+ is 2.0×108. Expre

Marina CMI [18]

Answer:

$\large \boxed{1.77 \times 10^{-5}\text{ mol/L}}$

Explanation:

Assume that you have mixed 135 mL of 0.0147 mol·L⁻¹ NiCl₂ with 190 mL of 0.250 mol·L⁻¹ NH₃.

1. Moles of Ni²⁺

$n = \text{135 mL} \times \dfrac{\text{0.0147 mmol}}{\text{1 mL}} = \text{1.984 mmol}$

2. Moles of NH₃

$n = \text{190 mL} \times \dfrac{\text{0.250 mmol}}{\text{1 mL}} = \text{47.50 mmol}$

3. Initial concentrations after mixing

(a) Total volume

V = 135 mL + 190 mL = 325 mL

(b) [Ni²⁺]

$c = \dfrac{\text{1.984 mmol}}{\text{325 mL}} = 6.106 \times 10^{-3}\text{ mol/L}$

(c) [NH₃]

$c = \dfrac{\text{47.50 mmol}}{\text{325 mL}} = \text{0.1462 mol/L}$

3. Equilibrium concentration of Ni²⁺

The reaction will reach the same equilibrium whether it approaches from the right or left.

Assume the reaction goes to completion.

Ni²⁺ + 6NH₃ ⇌ Ni(NH₃)₆²⁺

I/mol·L⁻¹: 6.106×10⁻³ 0.1462 0

C/mol·L⁻¹: -6.106×10⁻³ 0.1462-6×6.106×10⁻³ 0

E/mol·L⁻¹: 0 0.1095 6.106×10⁻³

Then we approach equilibrium from the right.

Ni²⁺ + 6NH₃ ⇌ Ni(NH₃)₆²⁺

I/mol·L⁻¹: 0 0.1095 6.106×10⁻³

C/mol·L⁻¹: +x +6x -x

E/mol·L⁻¹: x 0.1095+6x 6.106×10⁻³-x

$K_{\text{f}} = \dfrac{\text{[Ni(NH$_{3}$)$_{6}^{2+}$]}}{\text{[Ni$^{2+}$]}\text{[NH$_{3}$]}^{6}} = 2.0 \times 10^{8}$

Kf is large, so x ≪ 6.106×10⁻³. Then

$K_{\text{f}} = \dfrac{\text{[Ni(NH$_{3}$)$_{6}^{2+}$]}}{\text{[Ni$^{2+}$]}\text{[NH$_{3}$]}^{6}} = 2.0 \times 10^{8}\\\\\dfrac{6.106 \times 10^{-3}}{x\times 0.1095^{6}} = 2.0 \times 10^{8}\\\\6.106 \times 10^{-3} = 2.0 \times 10^{8}\times 0.1095^{6}x= 345.1x\\x= \dfrac{6.106 \times 10^{-3}}{345.1} = 1.77 \times 10^{-5}\\\\\text{The concentration of Ni$^{2+}$ at equilibrium is $\large \boxed{\mathbf{1.77 \times 10^{-5}}\textbf{ mol/L}}$}$

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4 years ago

13. Due to changes in energy, substances can change their state of matter. When

AlexFokin [52]

When’s I would like to say that it may be B because

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3 years ago

Think about how things might have been different if Becquerel had ignored the unexpected results of his experiment. Which charac

olga55 [171]

Answer: B, C, D, F

Questioning, creative, open-minded, objective

Explanation:

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4 years ago

How many grams of molybdenum(Mo) are in 2.68E24 atoms of Mo?

loris [4]

<h3>Answer:</h3>

427 g Mo

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

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table
Using Dimensional Analysis
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<h3>Explanation:</h3>

<u>Step 1: Define</u>

2.68E24 atoms Mo or 2.68 × 10²⁴ atoms Mo

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

Avogadro's Number

Molar Mass of Mo - 95.94 g/mol

<u>Step 3: Convert</u>

<u /> $2.68 \cdot 10^{24} \ atoms \ Mo(\frac{1 \ mol \ Mo}{6.022 \cdot 10^{23} \ atoms \ Mo} )(\frac{95.94 \ g \ Mo}{1 \ mol \ Mo} )$ = 426.966 g Mo

<u>Step 4: Check</u>

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

426.966 g Mo ≈ 427 g Mo

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3 years ago

What would the unit of measurement be for the density of a regular solid?

balu736 [363]

Answer:

SI unit: kg/m³

but g/cm³ is commonly used

Explanation:

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3 years ago