answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Step2247 [10]
2 years ago
13

. Citric acid, which can be obtained from lemon juice, has the molecular formula C6H8O7. A 0.250-g sample of citric acid dissolv

ed in 25.0 mL of water requires 37.2 mL of 0.105 M NaOH for complete neutralization. What number of acidic hydrogens per molecule does citric acid have?
Chemistry
1 answer:
marysya [2.9K]2 years ago
3 0

Answer:

3 acidic hydrogens per molecule of citric acid

Explanation:

In a sample of 37.2 mL(0.0372 L) of 0.105 mol/L of NaOH, will have:

n = 0.105x0.0372 = 0.0039 mol of NaOH

The dissociation of NaOH will give the same number of moles of Na⁺ and OH⁻.

The molar mass of citric acid is:

C: 12g/mol x 6 = 72 g/mol

H: 1g/mol x 8 = 8g/mol

O: 16 g/mol x 7 = 112 g/mol

192 g/mol

So, 0.250g of the acid has

n = mass/molar mass

n = 0.250/192

n = 0.0013 mol.

To be neutralized, it will be necessary 0.0039 mol of acidic hydrogens to react with the 0.0039 mol of OH⁻.

The dissociation reaction of one molecule of  the acid will give the stoichiometry:

1 mol of acid ----------------------- x mol of acidic hydrogens

0.0013 mol --------------------------- 0.0039

For a simple direct three rule:

0.0013x = 0.0039

x = 3 acidic hydrogens per molecule of citric acid.

You might be interested in
Challenge question: This question is worth 6 points. As you saw in problem 9 we can have species bound to a central metal ion. T
GenaCL600 [577]

Answer:

CN^- is a strong field ligand

Explanation:

The complex, hexacyanoferrate II is an Fe^2+ specie. Fe^2+ is a d^6 specie. It may exist as high spin (paramagnetic) or low spin (diamagnetic) depending on the ligand. The energy of the d-orbitals become nondegenerate upon approach of a ligand. The extent of separation of the two orbitals and the energy between them is defined as the magnitude of crystal field splitting (∆o).

Ligands that cause a large crystal field splitting such as CN^- are called strong field ligands. They lead to the formation of diamagnetic species. Strong field ligands occur towards the end of the spectrochemical series of ligands.

Hence the complex, Fe(CN)6 4− is diamagnetic because the cyanide ion is a strong field ligand that causes the six d-electrons present to pair up in a low spin arrangement.

5 0
2 years ago
Circle the molecule(s) in the atmosphere that come from animals. (circle all that apply)
bagirrra123 [75]

Answer:

A and C

Explanation:

A; Animals inhale and exhale, breath oxygen, leave carbon dioxide.

C; We fart, so do they! (farts are a release of methane gas from our bodies) Just ask my dog...

Hope this helped!

6 0
2 years ago
How much maleic anhydride would you need to react 178 mg of anthracene? Assume 1:1 ratio from maleic anhydride to anthracene.
Licemer1 [7]

Answer:

(1) 0.10      (2) 17.8 g

Explanation:

Since the reaction ratio is 1:1 what we need is to convert the given masses to moles and you will have the answer:

MW anthracene = 178.23 g/mol

MW maleic anhydride = 98.06 g/mol

a) mass anthracene = 178 mg x 1 g/ 1000 mg = 0.178 g anthracene

Moles anthracene = 0.178 g anthracene/ 178.23 g/mol

= 0.001 mol anthracene

0.001 mol anthracene x 1 mol maleic acid/mol anthracene

= 0.001 mol maleic anhydride

mass maleic anhydride  = 0.001 mol x 98.06 g/mol =  0.10 g

b) moles maleic anhydride = 9.8 g/ 98.06 g/mol = 0.099 moles

0.099 moles maleic anhydride x 1 mol anthracene/mol  maleic anhydride =

0.099 mol anthracene

g anthracene = 0.10mol x 178 g/mol = 17.8 g

8 0
2 years ago
A student measured the mass of some sodium chloride for a lab. the lab called for 6.25 g (the accurate measurement ) , but the s
Yuri [45]

Answer:

The answer is

\huge 4\%

Explanation:

The percentage error of a certain measurement can be found by using the formula

P(\%) =  \frac{error}{actual \:  \: number}  \times 100\% \\

From the question

actual mass = 6.25 g

error = 6.25 - 6 = 0.25

So we have

p(\%) =  \frac{0.25}{6.25}  \times 100

We have the final answer as

<h3>4 %</h3>

Hope this helps you

3 0
2 years ago
A substance is 89.2% carbon by mass. how much of the substance would be needed to recover 34.6 mol of pure carbon?
kari74 [83]
1 mole of carbon contains 12 g
Thus, 34.6 moles will contain; 34.6 × 12 = 415.2 g
If a substance contains 89.2 % carbon, 
then, (415.2/89.2) ×100 =  465.47 g of the substance will be required to yield 34.6 moles of carbon.
8 0
2 years ago
Read 2 more answers
Other questions:
  • the air we breathe is approximately 21% oxygen.atypical breath has a volume of 450 ml.how many grams of O2 are in a breath of ai
    12·1 answer
  • How many structures are possible for a trigonal bipyramidal molecule with a formula of ax3y2?
    15·1 answer
  • Based on a kc value of 0.200 and the given data table, what are the equilibrium concentrations of xy, x, and y, respectively? ex
    5·1 answer
  • A 2.20 g sample of a compound gave 5.63 g CO2 and 2.30 g H2O on combustion in air. The compound is known to contain only C, H, O
    8·1 answer
  • A compound is 40.0% C, 6.70% H, and 53.3% O by mass. Assume that we have a 100.-g sample of this compound. Part A What are the s
    10·2 answers
  • A 5.36–g sample of NH4Cl was added to 25.0 mL of 1.00 M NaOH and the resulting solution diluted to 0.100 L. (a) What is the pH o
    15·1 answer
  • Calculate the grams of so2 gas present at stp in a 5.9 l container. (r = 0.0821 l·atm/k·mol)
    15·1 answer
  • What is the mole fraction of O2O2 in a mixture of 15.1 gg of O2O2, 8.19 gg of N2N2, and 2.46 gg of H2H2
    12·1 answer
  • Commercially available hot packs are simple in design: a pouch with water on one side, isolated by a barrier from a specific sal
    7·1 answer
  • How many grams are in 2.0 x 10-5 moles of uranium?
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!