1answer.
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
Anton [14]
3 years ago
5

Help on part "c": The forensic technician at a crime scene has just prepared a luminol stock solution by adding 19.0g of luminol

into a total volume of 75.0mL of H2O.
a)What is the molarity of the stock solution of luminol?
anwer I got: molarity of luminol solution = 1.43M b)Before investigating the scene, the technician must dilute the luminol solution to a concentration of 6.00×10−2 M. The diluted solution is then placed in a spray bottle for application on the desired surfaces.
I cannot get the correct answer for "c"...I have tried: 172mL,11.9mL, and 1.19*10^4. The only other possibility that I can come up with is: 83.9mL. Would this one be correct?...Or...am I still completely out to lunch???
c)How many moles of luminol are present in 2.00 L of the diluted spray?
anwer I got: moles of luminol = 0.120mol What volume of the stock solution (Part A) would contain the number of moles present in the diluted solution (Part B)?
Express your answer in milliliters.

Chemistry
2 answers:
Contact [7]3 years ago
6 0

1. The molarity of the stock solution of luminol = 1,431 M

2. 0.12 moles of luminol is present in 2.00 L of the diluted spray

3. The volume of the stock solution (Part A) would contain the number of moles present in the diluted solution (Part B): 83.86 ml

<h3>Further explanation</h3>

Stoichiometry in Chemistry studies about chemical reactions mainly emphasizing quantitative, such as the calculation of volume, mass, amount, which is related to the number of ions, molecules, elements, etc.

In stoichiometry included :

  • 1. Relative atomic mass
  • 2. Relative molecular mass

is the relative atomic mass of the molecule

  • 3. mole

1 mole is the number of particles contained in a substance with the same number of atoms in 12 gr C-12

1 mole = 6.02.10²³ particles

While the number of moles can also be obtained by dividing the mass (in grams) by the relative mass of the element or the relative mass of the molecule

\large{\boxed{\bold{mol\:=\:\frac{grams}{ relative\:mass} }}}

Luminol (C₈H₇N₃O₂) is a substance used to detect traces of blood in the scene of a crime because it reacts with iron in the blood

a. a luminol stock solution by adding 19.0g of luminol into a total volume of 75.0mL of H2O.

So the molarity is

  • 1. Luminol mole

- the relative molecular mass of Luminol

= 8. C + 7.H + 3.N + 2.16

= 8.12 + 7.1 + 3.14 + 2.16

= 177 grams / mol

so the mole:

mol = gram / relative molecular mass

mole=\frac{19}{177}

mole = 0.1073

2. Molarity (M)

M = mole / volume

M\:=\:{\frac{ 0.1703 }{75.10^{-3} L}

M = 1,431

  • b. luminol concentration in a spray bottle 6.00 × 10⁻² M. So that in 2 L of solution, the number of moles is:

mole = M x volume

mole = 6.10⁻² x 2

mole = 0.12

  • c. Molarity of the stock solution (Part A) = 1,431 M

the number of moles present in the diluted solution (Part B) = 0.12

So the volume of the stock solution (Part A) would contain the number of moles present in the diluted solution (Part B) is:

volume = mol / M

volume\:=\:\frac{0.12}{1.431}

volume = 0.08386 L = 83.86 ml

<h3>Learn more</h3>

moles of water you can produce

brainly.com/question/1405182

the number of each atom present in the compound's formula

brainly.com/question/5303004

the ratio of hydrogen atoms (H) to oxygen atoms (O) in 2 l of water

brainly.com/question/10861183

Keywords: mole, volume, molarity, Luminol, the relative molecular mass

agasfer [191]3 years ago
5 0

Answer:

I didn't solve part a) because you said in the statement that you needed only the answer from part c), therefore the answer equals:

n = 0.120 moles

V = 83.9 mL

Explanation:

The dilution of a 1.43 M to 0.06 M solution is 22.8 to 1 relative. Therefore, 22.8 parts of water must be added to the original solution. The amount of moles is calculated as follows:

n = M * V = 2 L * 0.06 M = 0.120 moles

The volume is calculated as follows:

V = n/M = 0.120 moles/1.43 M = 83.9 mL

You might be interested in
Bob walks 81m East and then he walks 103m West.
kogti [31]

The Answer your looking for is

He walked  the distance of: = 184 m

4 0
2 years ago
Read 2 more answers
Please help me. How do I do ideal gas law?
Tcecarenko [31]
The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).
4 0
3 years ago
When 50 ml of 1.000x10^-1m pb(no3)2 solution was added to 50 ml of 1.000x10^-1m nai solution?
podryga [215]

Balanced chemical reaction: Pb(NO₃)₂ (aq) + 2NaI(aq) → 2PbI₂(s) + 2NaNO₃(aq).

V(Pb(NO₃)₂) = 50 mL ÷ 1000 mL = 0.05 L, volume of solution.

c(Pb(NO₃)₂) = 0.1 mol/L; concentration of solution.

n(Pb(NO₃)₂) = c(Pb(NO₃)₂) · V(Pb(NO₃)₂).

n(Pb(NO₃)₂) = 0.1 mol/L · 0.05 L.

n(Pb(NO₃)₂) = 0.005 mol.

n(NaI) = c(NaI) · V(NaI).

n(NaI) = 0.1 mol/L · 0.05 L.

n(NaI) = 0.005 mol; amount of substance.

From chemical reaction: n(Pb(NO₃)₂) : n(NaI) = 1 : 2.

n(Pb(NO₃)₂) = 0.005 mol ÷ 2.

n(Pb(NO₃)₂) = 0.0025 mol; number of moles Pb(NO₃)₂ used.

n(NaI) = 0.005 mol; number of moles NaI used.

The limiting reagent is Pb(NO₃)₂.

n(PbI₂) = 0.005 mol.

m(PbI₂) = n(PbI₂) · M(PbI₂).

m(PbI₂) = 0.005 mol · 461 g/mol.

m(PbI₂) = 2.305 g; the theoretical yield of PbI₂.

3 0
3 years ago
At what temperature would 2.10moles of N2 gas have a pressure of 1.25atm and fill a 25.0 L tank
hodyreva [135]

Answer:

\large \boxed{\text{-92 $^{\circ}$C}}

Explanation:

We can use the Ideal Gas Law and solve for T.

pV = nRT

Data  

p = 1.25 atm

V = 25.0 L

n = 2.10 mol

R = 0.082 06 L·atm·K⁻¹mol⁻¹

Calculations

1. Temperature in kelvins

\begin{array} {rcl}pV & = & nRT\\\text{1.25 atm} \times \text{25.0 L} & = & \rm\text{2.10 mol} \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times T\\31.25&=&0.09847T\text{ K}^{-1}\\T& = &\dfrac{31.25}{\text{0.098 47 K}^{-1}}\\\\& = &\text{181 K}\end{array}

2. Temperature in degrees Celsius

\begin{array} {rcl}T & = & (181 - 273.15) \, ^{\circ}\text{C}\\& = & -92 \, ^{\circ}\text{C}\\\end{array}\\\text{The temperature of the gas is $\large \boxed{\mathbf{-92 \, ^{\circ}}\textbf{C}}$}

8 0
3 years ago
Can someone help me with this please
never [62]

Answer: B. 1:2

Explanation: Beryllium and chlorine forms a binary ionic compound. Ionic compound is formed when a metal loses its electrons to a receiving non metal. Beryllium (metal) has two valence electrons while chlorine (nonmetal) has seven valence electrons, and so a beryllium atom has to give out its two valence electrons to attain a duplet stable structure while a chlorine atom will gain one electron to attain its stable octet structure. In the reaction between beryllium and chlorine, two atoms of chlorine have to accept the two electrons from one beryllium atom to attain their stable octet structure.

The formula of the compound formed is BeCl2.

3 0
3 years ago
Other questions:
  • Can someone please help me on #2
    10·1 answer
  • FOR 100 PTS
    6·1 answer
  • Draw the structure of two alkenes that would yield 1-methylcyclohexanol when treated with Hg(OAc)2 in water, then NaBH4:
    12·2 answers
  • Which pairing accurately matches a biomolecules polymer with its monomer subunit
    13·2 answers
  • For the equilibrium reaction 2NH3(g) + 22 kJ ↔ N2(g) + 3H2(g), which of the following changes would result in the formation of m
    14·1 answer
  • Which of the following processes is spontaneous?
    13·2 answers
  • I really need help with this!
    15·1 answer
  • A gas initially has a volume of 300. mL at a pressure of 1.0 atm, what will the
    13·1 answer
  • Predict the final temperature of 3.50 kg of water in a calorimeter if the water is at 27.5°C before 0.77 oz of noodles containin
    12·1 answer
  • Find the volume of a box with length 25 cm, height 25 cm and width 1.0 m. Volume
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!