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Serga [27]
3 years ago
11

Manganese forms several oxides when combined with oxygen. One of the oxides (Oxide 1) contains 63.2% of Mn and another oxide (Ox

ide 2) contains 77.5% Mn. You will show that these compounds obey the law of multiple proportions. In the first sample, what is the mass of Manganese per 1 gram of oxygen (O)? Mass Mn/1go
Chemistry
1 answer:
Nina [5.8K]3 years ago
3 0

Explanation:

Defining law of definite proportions, it states that when two elements form more than one compound, the ratios of the masses of the second element which combine with a fixed mass of the first element will always be ratios of small whole numbers.

A. One of the oxides (Oxide 1) contains 63.2% of Mn.

Mass of the oxide = 100g

Mass of Mn = 63.2 g

Mass of O = 100 - 63.2

= 36.8 g

Ratio of Mn to O = 63.2/36.8

= 1.72

Another oxide (Oxide 2) contains 77.5% Mn.

Mass of oxide = 100 g

Mass of Mn = 77.5 g

Mass of O = 100 - 77.5

= 22.5 g

Ratio of Mn to O = 77.5/22.5

= 3.44

Therefore, the ratio of the masses of Mn and O in Oxide 1 and Oxide 2 is in the ratio 1.72 : 3.44, which is also 1 : 2. So the law of multiple proportions is obeyed.

B.

Oxide 1

Mass of Mn per 1 g of O = mass of Mn/mass of O

= 77.5/22.5

= 3.44 g/g of Oxygen.

Oxide 2

Mass of Mn per 1 g of O = mass of Mn/mass of O

= 77.5/22.5

= 3.44 g/g of Oxygen.

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a_sh-v [17]

Answer:

8.72 mol PbO₂

General Formulas and Concepts:

<u>Chemistry - Atomic Structure</u>

  • Using Dimensional Analysis
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Explanation:

<u>Step 1: Define</u>

5.25 × 10²⁴ formula units PbO₂

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

Avogadro's Number

<u>Step 3: Convert</u>

<u />5.25 \cdot 10^{24} \ formula \ units \ PbO_2(\frac{1 \ mol \ PbO_2}{6.022 \cdot 10^{23} \ formula \ units \ PbO_2} ) = 8.71803 mol PbO₂

<u>Step 4: Check</u>

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

8.71803 mol PbO₂ ≈ 8.72 mol PbO₂

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Convert 614 Cal to J. Only include the numerical value in your answer. Respond with the correct number of significant figures in
Kruka [31]

Answer:

Explanation:

Given parameters:

Number of given calories = 614 cal

Note that  1cal = 4.18J

Now to convert the given calories to joule, we use the relationship below:

         Since 1Cal gives 4.18Joule of energy

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This written as 2.57 x 10³J because it has three significant numbers of 2, 5 and 7.

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13.50 grams of Pb(NO3)4 are dissolved in enough water to make 250 mL of solution. What is the molar its of the resulting solutio
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Data:
M (molarity) = ? (M or Mol/L)
m (mass) = 13.50 g
V (volume) = 250 mL → 0.25 L
MM (Molar Mass) of Lead(IV) Nitrate Pb(NO_3)_4
Pb = 1*207 = 207 amu
N = (1*14)*4 = 14*4 = 56 amu
O = (3*16)*4 = 48*4 = 192 amu
------------------------------------
MM of Pb(NO_3)_4 = 207+56+192 = 455 g/mol

Formula:
M =  \frac{m}{MM*V}

Solving:
M = \frac{m}{MM*V}
M =  \frac{13.50}{455*0.25}
M =  \frac{13.50}{113.75}
M = 0.118681318...\:\:\to\:\:\boxed{\boxed{M \approx 0.119\:Mol/L}}\end{array}}\qquad\quad\checkmark

Answer:
<span>B. 0.119 M</span>
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