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dangina [55]
2 years ago
10

During a titration the following data were collected. A 20.0 mL portion of solution of an unknown acid HX was titrated with 2.0

M KOH. It required 60.0 mL of the base to neutralize the sample. What is the molarity of the acid HX?
Chemistry
1 answer:
Dafna1 [17]2 years ago
8 0

Answer:

The molarity of the acid HX is 6.0 M.

Explanation:

We determine the amount of moles of KOH used to neutralize the acid:

\frac{2.0moles_{KOH}}{1000ml} *60ml=0.12 moles KOH

Then, we calculate the amount of moles of acid:

0.12 moles KOH×\frac{1 mole HX}{1 moles KOH}=0.12 moles HX

The molarity of HX is:

\frac{0.12 moles HX}{20ml} *\frac{1000ml}{1l}=6.0 M

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What is the hydronium ion concentration of a solution with a pOH of 7.20?
-Dominant- [34]

Answer:

[H⁺] = 1.58 x 10⁻⁷ M.

Explanation:

∵ pOH = - log[OH⁻]

7.20 = - log[OH⁻]

log[OH⁻] = - 7.20

∴ [OH⁻] = 6.31 x 10⁻⁸.

∵ [H⁺][OH⁻] = 10⁻¹⁴.

∴ [H⁺] = 10⁻¹⁴/[OH⁻] = 10⁻¹⁴/(6.31 x 10⁻⁸) = 1.585 x 10⁻⁷ M.

3 0
2 years ago
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Functional group and bond hybridization of vanillin
lana66690 [7]

Vanillin is the common name for 4-hydroxy-3-methoxy-benzaldehyde.

See attached figure for the structure.

Vanillin have 3 functional groups:

1) aldehyde group:  R-HC=O, in which the carbon is double bonded to oxygen

2) phenolic hydroxide group: R-OH, were the hydroxyl group is bounded to a carbon from the benzene ring

3) ether group: R-O-R, were hydrogen is bounded through sigma bonds to carbons

Now for the hybridization we have:

The carbon atoms involved in the benzene ring and the red carbon atom (from the aldehyde group) have a <u>sp²</u> hybridization because they are involved in double bonds.

The carbon atom from the methoxy group (R-O-CH₃) and the blue oxygen's have a  <u>sp³</u>  hybridization because they are involved only in single bonds.

7 0
2 years ago
Phosphorus has the molecular formula p4, and sulfur has the molecular formula s8. how many grams of phosphorus contain the same
Alinara [238K]
1) Find the number of molecules in 7.88 g of sulfur

molar mass of S8 = 8*atomic mass of S = 8 * 32.0 g / mol = 256.0 g/mol

Number of moles  = mass in grams / atomic mass = 7.88 g / 256.0 g / mol = 0.0308 moles

2) Find the mass of 0.0308 moles of P4

mass = number of moles * molar mass

molar mass of P4 = 4 * atomic mass of P = 4 * 31 g/mol = 124 g/mol

mass of P4 = 0.0308 moles * 124 g/mol = 3.8192g ≈ 3.82 g.

Answer: 3.82 grams of P4 will have the same number of molecules as 7.88 g of S8 (that is 0.0308 moles of molecules)
6 0
2 years ago
A 0.216 g sample of an aluminium compound X reacts with an excess of water to produce a single hydrocarbon gas. This gas burns c
makkiz [27]
0.216g of aluminium compound X  react with an excess of water water to produce gas. this gas burn completely  in O2  to form H2O and 108cm^3of CO2 only . the volume of CO2 was measured at room temperature and pressure

0.108 / n  =  24 / 1 
n = 0.0045 mole ( CO2 >>0.0045 mole 
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C = 0.0045 * 1000 => 4.5    and Al  = 0.0078 * 1000 = 7.8 

7 0
2 years ago
How much energy is required to heat 0.24 KG lutetium from 296.2K to 373.5 K? The specific heat for lutetium is 0.154 J/g-K
Olenka [21]

I’m not sure I need help with this question

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