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2 years ago

Iron and aluminim are examples of _________. a. pure substances b. compounds

Chemistry

2 answers:

Ierofanga [76]2 years ago

8 0

Answer:

compounds

Explanation:

Sindrei [870]2 years ago

8 0

A. Pure substances

Iron aluminum, silver, and gold are all examples of a pure substance. Substances that are free from any kind of mixture and contain only one kind of particle are pure substances.

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When 1.2383 g of an organic iron compound containing Fe, C, H, and O was burned in O2, 2.3162 g of CO2 and 0.66285 g of H2O were

uysha [10]

Answer:

$\boxed{\text{C$_{15}$H$_{21}$FeO$_{6}$}}$

Explanation:

Let's call the unknown compound X.

1. Calculate the mass of each element in 1.23383 g of X.

(a) Mass of C

$\text{Mass of C} = \text{2.3162 g } \text{CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{0.632 07 g C}$

(b) Mass of H

$\text{Mass of H} = \text{0.66285 g }\text{H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g } \text{{H$_{2}$O}}} = \text{0.074 157 g H}$

(c)Mass of Fe

(i)In 0.4131g of X

$\text{Mass of Fe} = \text{0.093 33 g Fe$_{2}$O$_{3}$}\times \dfrac{\text{111.69 g Fe}}{\text{159.69 g g Fe$_{2}$O$_{3}$}} = \text{0.065 277 g Fe}$

(ii) In 1.2383 g of X

$\text{Mass of Fe} = \text{0.065277 g Fe}\times \dfrac{1.2383}{0.4131} = \text{0.195 67 g Fe}$

(d)Mass of O

Mass of O = 1.2383 - 0.632 07 - 0.074 157 - 0.195 67 = 0.336 40 g

2. Calculate the moles of each element

$\text{Moles of C = 0.63207 g C}\times\dfrac{\text{1 mol C}}{\text{12.01 g C }} = \text{0.052 629 mol C}\\\\\text{Moles of H = 0.074157 g H} \times \dfrac{\text{1 mol H}}{\text{1.008 g H}} = \text{0.073 568 mol H}\\\\\text{Moles of Fe = 0.195 67 g Fe} \times \dfrac{\text{1 mol Fe}}{\text{55.845 g Fe}} = \text{0.003 5038 mol Fe}\\\\\text{Moles of O = 0.336 40} \times \dfrac{\text{1 mol O}}{\text{16.00 g O}} = \text{0.021 025 mol O}$

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

$\text{C: } \dfrac{0.052629}{0.003 5038}= 15.021\\\\\text{H: } \dfrac{0.073568}{0.003 5038} = 20.997\\\\\text{Fe: } \dfrac{0.003 5038}{0.003 5038} = 1\\\\\text{O: } \dfrac{0.021025}{0.003 5038} = 6.0006$

4. Round the ratios to the nearest integer

C:H:O:Fe = 15:21:1:6

5. Write the empirical formula

$\text{The empirical formula is } \boxed{\textbf{C$_{15}$H$_{21}$FeO$_{6}$}}$

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3 years ago

Why is atmospheric pressure much lower on the top of a mountain than it is as sea level?

Vika [28.1K]

Because the gravity levels decrese and so does the oxygen levels

3 0

3 years ago

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mash [69]

Answer:

ans : C

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8 0

3 years ago

If a 750 mL of a gas at a pressure of 100.7 kPa has a decrease of pressure to 99.8 kPa, what is the new volume? Show work

SVEN [57.7K]

Explanation:

P1V1 = P2V2

(100.7 kPa)(0.75 L) = (99.8 kPa)V2

V2 = (100.7 kPa)(0.75 L)/(99.8 kPa)

= 0.757 L

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3 years ago

21 mL It required 42.35 mL of H2SO4 to neutralize 21.17 mL of 0.5000 M NaOH. Calculate the concentration of H2SO4

bija089 [108]

The balanced equation for the above reaction is
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
number of NaOH moles required-0.5000 M / 1000 mL/L x 21.17 mL = 0.010585 mol
According to stoichiometry, acid moles required are 1/2 of the base moles reacted
Therefore number of H₂SO₄ moles reacted - 0.010585 /2 mol
Number of moles in 42.35 mL of H₂SO₄ - 0.010585 /2 mol
Therefore in 1 L solution - (0.010585) /2 / 42.35 mL x 1000 mL/L = 0.125 M
Molarity of H₂SO₄ - 0.125 M

5 0

3 years ago