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miss Akunina [59]
3 years ago
6

Classify each structure as alkane, alcohol, carboxylic acid, or other. click on a structure to access the rotational controls.

Chemistry
1 answer:
noname [10]3 years ago
4 0
Since the structures are not given, I'll just provide the structure for these functional groups.

Alkane: You know the compound is an alkane if it conforms to the general formula CₓH₂ₓ₊₂, where x is any positive integer.
Alcohol: A compound is an alcohol if the compound contains an -OH functional group.
Carboxylic acid: The general formula for carboxylic acid is RC=OOH. The R is any hydrocarbon chain. The C atom is attached to one oxygen atom by a double bond, and one O atom by a single bond.
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How many titanium atoms does it contain? titanium has a density of 4.50g/cm3, a pure titanium cube has an edge length of 2.82 in
yKpoI14uk [10]
2.82 inches is are equivalent to 7.16 cm.
The volume of the cube ;
 7.16 × 7.16 ×  7.16 = 367.06 cm³
The density of titanium is 4.5 g/cm³
Therefore; mass will be 4.5 × 367.06 = 1651.77 g
1 mole of titanium has 47.867 g
thus, 1651.77 g is equivalent to  1651.77/ 47.867 = 34.507 moles
According to the Avagadro's constant 1 mole = 6.022 × 10∧23 atoms 
 Therefore, we get 6.022 × 10∧23 × 34.507 = 2.078 ×10∧25 atoms
5 0
3 years ago
Read 2 more answers
Determine the empirical formula for the compound represented by each molecular formula. c2h4
dybincka [34]
Carbon and hydrogen and the number next to is the number of moles
6 0
4 years ago
Convert 15.5m³ to mL
ziro4ka [17]

Answer:

Explanation:

1.55e+7

3 0
1 year ago
27.8 mL solution of 0.797 M HCHO2 with 0.928 M NaOH. What is the pH for the solution at the equivalence point in the titration?
kati45 [8]

Answer:

8.69 is the pH at the equivalence point

Explanation:

Formic acid, HCHO₂, reacts with NaOH as follows:

HCHO₂ + NaOH → NaCHO₂ + H₂O.

At the equivalence point you will have in the reaction just NaCHO₂ and H₂O. The concentration of NaCHO₂ will be:

<em />

<em>Moles: </em>0.0278L * 0.797mol/L = 0.02216moles

To reach the equivalence point it is necessary to add:

0.02216mol * (1L / 0.928mol) = 0.0239L

Total volume in the equivalence point:

0.0278L + 0.0239L = 0.0517L

Concentration: 0.02216moles / 0.0517L = 0.429M

The equilibrium of NaCHO₂, CHO₂⁻, in water is:

CHO₂⁻(aq) + H₂O(l) ⇄ OH⁻(aq) + HCHO₂(aq)

Where Kb, 5.56x10⁻¹¹ is defined as:

5.56x10⁻¹¹ = [OH⁻] [HCHO₂] / [CHO₂⁻]

In the equilibrium, it is produced X OH⁻ and HCHO₂, and as concentration of NaCHO₂ is 0.429M:

5.56x10⁻¹¹ = [X] [X] / [0.429M]

2.383x10⁻¹¹ = X²

4.88x10⁻⁶ = X = [OH⁻]

As pOH = -log [OH⁻]

pOH = 5.31

And pH = 14 - pH

pH = 8.69 is the pH at the equivalence point

3 0
4 years ago
Chemistry Help:
Umnica [9.8K]
It's answer A!
Reactants on left (see on the arrow), products on right.
7 0
3 years ago
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