What mass of carbon dioxide (co2) can be produced from 86.17 grams of c6h14 and excess oxygen?

1 answer:

3 0

2C6H14 + 13O2 ---> 6CO2 +14H2O

Calculating the molar mass of C6H14: M(C6H14)=12.011*6 +1.008*14 ≈ 86.17 g/mol

Thus, 86.17 g of C6H14 corresponds to 1 mole.

                                  2C6H14 + 13O2 ---> 6CO2 +14H2O
based on the equation        2 mol                            6 mol
according to the question    1 mol                            3 mol

To determine M(CO2): M(CO2)= 12.011 + 2*15.999= 44.009 g/mol
Therefore, 3 mol CO2*44.009 g/1 mol CO2 ≈ 132.0 g CO2
Final answer: 132.0 g CO2

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Sodium only has one naturally occuring isotope, 23 Na , with a relative atomic mass of 22.9898 u . A synthetic, radioactive isot

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Answer:

The mass of 22-Na included in the sample amounts to 0.0599 g

Explanation:

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It has an apparent mass of 22.9573 u.

For 23-Na, the relative atomic mass is 22.9898 u, while for 22-Na it is 21.9944 u.

Let the relative abundance of 23-Na be denoted as X.

This means that the relative abundance of 22-Na can be expressed as (1-X).

The equation formed is 21.9944 (1-X) + 22.9898 X = 22.9573.

Rearranging gives: 21.9944 - 21.9944X + 22.9898X = 22.9573.

Which simplifies to 22.9898X - 21.9944X = 22.9573 - 21.9944.

Hence, 0.9954X = 0.9639, leading to X = 0.9674.

The relative abundance of 23-Na is now identified as 0.9674.

Consequently, the relative abundance of 22-Na is 1 - 0.9674 = 0.0326.

Now, the mass of 22-Na contained within the 1.8385g sample is determined by

Relative abundance of 22-Na multiplied by the mass of the total sample = 0.0326 × 1.8385g = 0.0599 g.

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What is the symbol for the isotope of 58 co that possesses 33 neutrons?

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The element with atomic number 58 is Cerium, meaning its symbol should be Ce rather than Co, which belongs to Cobalt with atomic number 27. Therefore, the notation for isotopes consists of the element's symbol accompanied by a superscript and a subscript, properly aligned. The superscript indicates the mass number.

Mass number = protons + neutrons = 58 + 33 = 91

The subscript denotes the atomic number, which is 58. This notation is illustrated in the attached image.