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maw [93]
3 years ago
10

If you walk 6336 feet, how many miles will you have walked?

Chemistry
2 answers:
Tamiku [17]3 years ago
8 0
You would have walked 1.2 miles
S_A_V [24]3 years ago
4 0

Answer:

1.2 miles

Explanation:

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How was president Andrew Jackson wrong for closing the bank
Paladinen [302]
Jackson, the epitome of the frontiersman, resented the bank's lack of funding for expansion into the unsettled Western territories. Jackson also objected to the bank's unusual political and economic power and to the lack of congressional oversight over its business dealings.
5 0
2 years ago
How might a molecule with two strong bond dipoles have no molecular dipole at all?
Reil [10]
 A molecule with two strong bond dipoles can have no molecular dipole if the bond dipoles cancel each other out by pointing in exactly opposite directions. For example, in carbon dioxide (a linear molecule), the carbon-oxygen bonds have a <span>large dipole moment. However, because one dipole points to the left and the other to the right the dipole is cancelled.</span>
5 0
3 years ago
A 0.1510 gram sample of a hydrocarbon produces 0.5008 gram CO2 and 0.1282 gram H2O in combustion analysis. Its
Over [174]
In a combustion of a hydrocarbon compound, 2 reactions are happening per element:

C + O₂ → CO₂
2 H + 1/2 O₂ → H₂O

Thus, we can determine the amount of C and H from the masses of CO₂ and H₂O produced, respectively.

1.) Compute for the amount of C in the compound. The data you need to know are the following:
Molar mass of C = 12 g/mol
Molar mass of CO₂ = 44 g/mol
Solution:
0.5008 g CO₂*(1 mol CO₂/ 44 g)*(1 mol C/1 mol CO₂) = 0.01138 mol C
0.01138 mol C*(12 g/mol) = 0.13658 g C

Compute for the amount of H in the compound. The data you need to know are the following:
Molar mass of H = 1 g/mol
Molar mass of H₂O = 18 g/mol
Solution:
0.1282 g H₂O*(1 mol H₂O/ 18 g)*(2 mol H/1 mol H₂O) = 0.014244 mol H
0.014244 mol H*(1 g/mol) = 0.014244 g H

The percent composition of pure hydrocarbon would be:
Percent composition = (Mass of C + Mass of H)/(Mass of sample) * 100
Percent composition = (0.13658 g + 0.014244 g)/(<span>0.1510 g) * 100
</span>Percent composition = 99.88%

2. The empirical formula is determined by finding the ratio of the elements. From #1, the amounts of moles is:

Amount of C = 0.01138 mol
Amount of H = 0.014244 mol

Divide the least number between the two to each of their individual amounts:
C = 0.01138/0.01138 = 1
H = 0.014244/0.01138 = 1.25

The ratio should be a whole number. So, you multiple 4 to each of the ratios:
C = 1*4 = 4
H = 1.25*4 = 5

Thus, the empirical formula of the hydrocarbon is C₄H₅.

3. The molar mass of the empirical formula is

Molar mass = 4(12 g/mol) + 5(1 g/mol) = 53 g/mol
Divide this from the given molecular weight of 106 g/mol
106 g/mol / 53 g/mol = 2
Thus, you need to multiply 2 to the subscripts of the empirical formula.

Molecular Formula = C₈H₁₀

4 0
3 years ago
Please help I really appreciate it thank you ❤️
Feliz [49]

II. sulfur (S) and carbon (C)

and

III. fluorine (F) and oxygen (O)

will form covalent bonds, so the answer will be:

e. II and III

Explanation:

To know is what type of bond is formed between atoms we need to look at the electronegativity difference between the atoms.

If the electronegativity difference is less than 0.4 there is a nonpolar covalent bond.

If the electronegativity difference is between 0.4 and 1.8 there is a polar covalent bond. (if is a metal involved we consider the bond to be ionic)

If the electronegativity difference is greater then 1.8 there is an ionic bond.

We have the following cases:

I. lithium (Li) and sulfur (S)

electronegativity difference = 2.5 (S) - 1 (Li) = 1.5 but because there is a metal involved the bond will be ionic

II. sulfur (S) and carbon (C)

electronegativity difference = 2.5 (S) - 2.5 (C) = 0 so the bond will be nonpolar covalent

III. fluorine (F) and oxygen (O)

electronegativity difference = 4 (F) - 3.5 (O) = 0.5 so the bond will be polar covalent bond.

Learn more about:

covalent and ionic bonds

brainly.com/question/1802971

#learnwithBrainly

3 0
3 years ago
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:

brainly.com/question/12307841?referrer=searchResults

#SPJ4

7 0
1 year ago
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