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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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Answer:

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Explanation:

We are asked to find the mass of a sample of metal. We are given temperatures, specific heat, and joules of heat, so we will use the following formula.

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The heat added is 4500.0 Joules. The mass of the sample is unknown. The specific heat is 0.4494 Joules per gram degree Celsius. The difference in temperature is found by subtracting the initial temperature from the final temperature.

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Substitute these values into the formula.

4500.0 \ J = m (0.4494 \ J/g \textdegree C)(30.1 \textdegree C)

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We are solving for the mass, so we must isolate the variable m. It is being multiplied by 13.52694 Joules per gram. The inverse operation of multiplication is division, so we divide both sides by 13.52694 J/g

\frac {4500.0 \ J }{13.52694 J/g}= \frac{m (13.52694 J/g)}{13.52694 J/g}

The units of Joules cancel.

\frac {4500.0 \ J }{13.52694 J/g}= m

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The original measurements have 5,4, and 3 significant figures. Our answer must have the least number or 3. For the number we found, that is the ones place. The 6 in the tenth place tells us to round the 2 up to a 3.

333 \ g \approx m

The mass of the sample of metal is approximately <u>333 grams.</u>

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