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3 years ago

The combustion of 14 grams of co according to the reaction co(g) ½o2(g) co2(g) 67.6 kcal gives off how much heat?

1 answer:

3 0

The balanced reaction that describes the reaction between carbon monoxide and oxygen to produce carbon dioxide is expressed co(g)+ ½o2(g) <span>→ </span>co2(g). The heat of combustion is 67.6 kcal per mole CO. We are given with 14 grams CO. Convert this to mole equal to 0.5 mole. Hence the total heat is 33.8 kcal.

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A piece of wood has a labeled length value of 63.2 cm. You measure its length three times and record the following data: 63.1 cm

Ulleksa [173]

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

Accepted value is true value.

Measured values is calculated value.

In the question given Accepted value (true value) = 63.2 cm

Given Measured(calculated values) = 63.1 cm , 63.0 cm , 63.7 cm

1) Percent error (%) for first measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.1 cm

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

$Percent error = \frac{\left | 63.2 - 63.1 \right |}{63.2}\times 100$

$Percent error = \frac{0.1}{63.2}\times 100$

$Percent error = 0.00158\times 100$

Percent error = 0.158 %

2) Percent error (%) for second measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.0 cm

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

$Percent error = \frac{\left | 63.2 - 63.0 \right |}{63.2}\times 100$

$Percent error = \frac{0.2}{63.2}\times 100$

$Percent error = 0.00316\times 100$

Percent error = 0.316 %

3) Percent error (%) for third measurement.

Accepted value (true value) = 63.2 cm, Measured(calculated values) = 63.7 cm

Percent error (%)= $\frac{\left | Accepted value - Measured value \right |}{Accepted value}\times 100$

$Percent error = \frac{\left | 63.2 - 63.7 \right |}{63.2}\times 100$

$Percent error = \frac{\left | -0.5 \right |}{63.2}\times 100$

$Percent error = \frac{(0.5)}{63.2}\times 100$

$Percent error = 0.00791\times 100$

Percent error = 0.791 %

Percent error for each measurement is :

63.1 cm = 0.158%

63.0 cm = 0.316%

63.7 cm = 0.791%

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3 years ago

To describe the length of a classroom, a student should use

Kamila [148]

The answer is D).....

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3 years ago

Read 2 more answers

SeF6 1. Lewis Structure 2. Perspective drawing 3. Number of atoms bonded to central atom 4. Number of non-bonding electron pairs

Anarel [89]

The answer is- $SF_{6}$ is octahedral in electronic and molecular geometry with 6 Fluorine atoms bonded to central atom S.

Lewis structures are the diagrams in which the valence electrons of the atoms of a compound are arranged around the atoms showing the bonding between the atom and the lone pair of electrons existing in the molecule.

Determine the molecular geometry of $SF_{6}$ .

Valence Shell Electron Pair Repulsion theory is commonly known as VSEPR theory and it helps to predict the geometry of molecules.
According to this theory, electrons are arranged around the central atom of the molecule in such a way that there is minimum electrostatic repulsion between these electrons.
Now, calculate the total number of valence electrons in $SF_{6}$ .

$Valence\ electrons\ in\ SF_{6}= Valence\ electrons\ in\ S +\ 6(Valence\ electrons\ in\ F)$

Valence electrons of S = 6

Valence electrons of F = 7

Thus, the valence electrons in $SF_{6}$ are-

$Valence\ number\ of\ electrons\ in\ SF_{6} = (6) + 6(7) = 48\ electrons.$

The Lewis structure of $SF_{6}$ is - (Image attached).
In the structure, the number of atoms bonded to central atom (S) = 6.
Number of non-bonding electron pairs on the central atom = 0 (as all the valence electrons are bonded to F).
Electronic geometry in case of 6 bond pairs is octahedral.
Molecular geometry us also octahedral with bond angles 90°.
Central atom is sp3d2 hybridised.
$SF_{6}$ is a non-polar molecule.

To learn more about Lewis structures visit:

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