To prepare a standard (calibration) curve for a spectroscopy experiment, start by preparing ___________ with ______________ Then, measure the ______________ of each solution at _____________ and create a plot of ____________ for the measured values. Finally, find the best-fit line of the data set.

Answers

Answer 1

Answer: See explanation

Explanation:

The calibration curve is the method used for the determination of the concentration of a substance such that the unknown sample will be compared to some standard samples of the known concentration.

To prepare a standard (calibration) curve for a spectroscopy experiment, start by preparing (multiple solutions) with (different known concentrations). Then, measure the (absorbance) of each solution at (thesame wavelength) and create a plot of (absorbance vs. concentration) for the measured values. Finally, find the best-fit line of the data set.


Related Questions

You will observe a weak acid-strong base titration in this experiment. Select all statements that are true about weak acid-strong base titrations.
A. Weak acid-strong base titrations always start at a higher pH than strong acid-strong base titrations, no matter the initial concentration.
B. The pH is less than 7 at the equivalence point.
C. The pH is greater than 7 at the equivalence point.
D. Half way to the equivalence point, a buffer region is observed.

Answers

Answer:

The pH is greater than 7 at the equivalence point.

Explanation:

Equivalence point is the point where the acid reacts with the base as stipulated in the equation of the reaction.

When a weak acid and a strong base are titrated, the pH of the solution at equivalence point is actually found to be around about pH ~ 9.

Hence, for a weak acid and strong base titration, The pH is greater than 7 at the equivalence point.

A titration between a weak acid and a strong base yields a solution whose pH is greater than 7 at the equivalence point.

What are weak acids?

Weak acids are acids which only ionize partially in aqueous solutions.

When weak acids are dissolved in water, they produce only few hydrogen ions.

A strong base on the other hand ionizes completely to produce hydroxide ions in aqueous solutions.

The titration of a weak acid and a strong base gives a solution whose pH is greater than 7 at equivalence point.

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g A high altitude balloon is filled with 1.41 x 104 L of hydrogen gas (H2) at a temperature of 21oC and a pressure of 745 torr. What is the volume of the balloon at a height of 20 km, where the temperature is -48oC and the pressure is 63.1 torr

Answers

Answer:

1.27 × 10⁵ L

Explanation:

Step 1: Given data

Initial pressure (P₁): 745 TorrInitial volume (V₁): 1.41 × 10⁴ LInital temperature (T₁): 21 °CFinal pressure (P₂): 63.1 TorrFinal volume (V₂): ?Final temperature (T₂): -48 °C

Step 2: Convert the temperatures to the Kelvin scale

We will use the following expression.

K = °C + 273.15

K = 21 °C + 273.15 = 294 K

K = -48 °C + 273.15 = 225 K

Step 3: Calculate the final volume of the balloon

We will use the combined gas law.

P₁ × V₁ / T₁ = P₂ × V₂ / T₂

V₂ = P₁ × V₁ × T₂/ T₁ × P₂

V₂ = 745 Torr × 1.41 × 10⁴ L × 225 K/ 294 K × 63.1 Torr

V₂ = 1.27 × 10⁵ L

2. For each of the ionic compounds in the table below, name the compound and explain the rule that you
used in formulating your name for the compound.
Name:
Rule for naming compound:
-PbF4
-NH4NO3
-Li2S

Answers

Answer:

2

Explanation:

Lead(|V) fluoride

Ammonium Nitrate

Lithium sulfide

For the rules, I don't know what you were taught. I just do it intuitively since I have done so much chemistry.

The first one the roman numerals represents the charge of the lead which much match the 4- charge from the 4 fluorides.

The second one is just two polyatomic ions which you just have to remember.

The last one is the typical ionic compound naming technique i guess.

A sample of oxygen gas has a volume of 89.6 L at STP. How many moles of oxygen gas are present ?

Answers

Answer:

89,6/22,4 =4(mol)

Explanation:

There are approximately 1.089 moles of oxygen gas present in the sample at STP.

At STP (Standard Temperature and Pressure), the conditions are defined as follows:

Temperature (T) = 0 degrees Celsius = 273.15 Kelvin

Pressure (P) = 1 atmosphere (atm) = 101.325 kPa = 1013.25 hPa

Now, to find the number of moles of oxygen gas (O2) present in the sample, we can use the ideal gas law:

PV = nRT

Where:

P = pressure (in atm)

V = volume (in liters)

n = number of moles

R = ideal gas constant = 0.0821 L.atm/(mol.K)

T = temperature (in Kelvin)

Given:

V = 89.6 L (volume at STP)

T = 273.15 K (STP temperature)

Let's plug in the values and solve for n (number of moles):

n = PV / RT

n = (1 atm) × (89.6 L) / (0.0821 L.atm/(mol.K) × 273.15 K)

n = 1.089 moles

So, there are approximately 1.089 moles of oxygen gas present in the sample at STP.

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How many atom in protons

Answers

Answer:

Its atomic number is 14 and its atomic mass is 28. The most common isotope of uranium has 92 protons and 146 neutrons. Its atomic number is 92 and its atomic mass is 238 (92 + 146).

A second-order reaction has a half-life of 12 s when the initial concentration of reactant is 0.98 M. The rate constant for this reaction is ________ M-1s-1. A) 12

Answers

Answer: 0.085 (Ms)⁻¹

Explanation: Half life = 12 s

is the initial concentration = 0.98 M

Half life expression for second order kinetic is:

k = 0.085 (Ms)⁻¹

The rate constant for this reaction is 0.085 (Ms)⁻¹ .

Which statement is true with respect to standard reduction potentials?
SRP values that are greater than zero always represent a reduction reaction.
SRP values that are less than zero always represent a reduction reaction.
Half-reactions with SRP values greater than zero are spontaneous.
Half-reactions with SRP values greater than zero are nonspontaneous.

Answers

Answer:

C). Half-reactions with SRP values greater than zero are spontaneous.

Explanation:

SRPs or Standard Reduction Potentials are characterized as the ability of a probable distinction among the anode and cathode of a usual/standard cell. It aims to examine the capacity of chemicals to reduce themselves.

The third statement asserts a true claim regarding the SRPs(Standard Reduction Potentials) that the 'half-reactions which take place with the SRP possesses the values higher than zero and they are unconstrained.' The other statements are incorrect as they either show the estimation of SRPs more than 0 or display them as being restricted. Thus, option C is the correct answer.

A molecule of acetone and a molecule of propyl aldehyde are both made from 3 carbon atoms, 6 hydrogen atoms, and 1 oxygen atom. The molecules differ in their arrangement of atoms. How do formulas for the two compounds compare? Both compounds have the same molecular formula, but have unique structural formulas. Both compounds have unique molecular formulas and structural formulas. Both compounds have the same structural formula, but have unique molecular formulas.

Answers

Explanation:

The structures of both acetone and propanal are shown below:

In the formula of propanal there is -CHO functional group at the end.

In acetone -CO- group is present in the middle that is on the second carbon.

The molecular formula is C3H6O.

Both have same molecular formula but different structural formulas.

2.50 L of 0.700 M phosphoric acid reacts with 5.25 moles of sodium hydroxide. How many moles of hydrogen ions will completely neutralize the moles of hydroxide ions present in this amount of sodium hydroxide? a) 0.583 b) 1.75 c) 3.00 d) 15.75 e) 5.25

Answers

Answer:

5.25 moles of protons. Option e

Explanation:

Reaction between phosphoric acid and sodium hydroxide is neutralization.

We can also say, we have an acid base equilibrium right here:

H₃PO₄  +  3NaOH →  Na₃PO₄  +  3H₂O

Initially we have 5.25 moles of base.

We have data from the acid, to state its moles:

M = mol/L, so mol = M . L

mol = 1.75 moles of acid

If we think in the acid we know:

H₃PO₄  →  3H⁺  +  PO₄⁻³

We know that 1 mol of acid can give 3 moles of protons (hydrogen ions)

If we have 1.75 moles of acid, we may have

(1.75 . 3) /1 = 5.25 moles of protons

These moles will be neutralized by the 5.25 moles of base

H₃O⁺  +  OH⁻  ⇄  2H₂O     Kw

In a titration of a weak acid and a strong base, we have a basic pH

Question 9 of 25
How many hydrogen atoms are in a molecule of table sugar (C12H,2011)?
O A. 12
B. 45
C. 11
D. 22
SUBMIT

Answers

D.22

is my answer than welcome

C. A sample may contain any or all of the following ions Hg2 2, Ba 2, and Al 3. 1) No precipitate forms when an aqueous solution of NaCl was added to the sample solution. 2) No precipitate forms when an aqueous solution of Na2SO4 was added to the sample. 3) A precipitate forms when the sample solution was made basic with NaOH. Which ion or ions were present. Write the net ionic equation(s) for the the reaction (s).

Answers

Answer:

Al^3+

Explanation:

Solubility rules tell us what substances are soluble in water. Since NaCl was added and no precipitate was observed, the mercury II ion is absent.

Addition of Na2SO4 does not form a precipitate meaning that Ba^2+ is absent.

If a precipitate is formed when NaOH is added, the the ionic reaction is as follows;

Al^3+(aq) + 3 OH^-(aq) ------> Al(OH)3(s)

Identify the highest energy molecular process that occurs when a molecule absorbs a microwave photon.

Answers

The question is incomplete, the complete question is;

Identify the highest energy molecular process that occurs when a molecule absorbs a microwave photon

a) electronic excitation

b) bond breakage

c) molecular vibration

d) molecular rotation

Answer:

molecular rotation

Explanation:

Microwaves are part of the electromagnetic spectrum. They are lower energy, lower frequency radiation.

When molecules absorb infrared radiation, they transition between the rotational states of the molecule.

Hence, the highest energy molecular process that occurs when a molecule absorbs a microwave photon is molecular rotation.

Identify each of the following half-reactions as either an oxidation half-reaction or a reduction half-reaction.

half-reaction identification
Cu+(aq)--->Cu2+(aq) + e- _________
I2(s) + 2e--->2I-(aq) _________

Answers

Answer:

Cu+(aq)--->Cu2+(aq) + e- : oxidation

reason: there is loss of electrons.

I2(s) + 2e--->2I-(aq) : reduction

reason: There is reduction of electrons.

Balance the following chemical equation.

CCl4 -> ___ C+ ___ Cl2

Answers

Answer:

Explanation:

CCl4 => C + 2Cl2

Choose the correct answer to make the statement true.

a. An exothermic reaction has a positive ΔH and absorbs heat from the surroundings.
b. An exothermic reaction feels warm to the touch. a positive ΔH and gives off heat to the surroundings.
c. An exothermic reaction feels warm to the touch. a negative ΔH and absorbs heat from the surroundings.
d. An exothermic reaction feels warm to the touch. a negative ΔH and gives off heat to the surroundings.
e. An exothermic reaction feels warm to the touch.

Answers

D.
The prefix “exo” indicates a release. “-thermic” indicates heat. Because there is a release of heat, the reaction gives off heat and is warm to the touch. ΔH is negative because there is a loss of heat energy.

How many atoms are in .45 moles of P4010

Answers

Answer:

5×6.02×1023

Explanation:

there are 5×6.02×1023 molecules of p4010 in 5mole. there are four P atoms in a single molecule of p4010

Identify the oxidation half-reaction for this reaction:
Fe(s) + 2HCl(aq) → FeCl2(aq) + H2(g)
A. Fe2+ + 2e → Fe(s)
O B. H2(g) → 2H+ + 2e
O C. Fe(s) → Fe2+ + 2e
O D. 2H+ + 2e → H2(9)

Answers

Answer:

Fe(s)->Fe2+2e-

Explanation:

A.p.e.x

The oxidation half-reaction for the given reaction is Fe(s) → Fe²⁺ + 2e⁻ Hence, Option (C) is correct

What is Oxidation reaction ?

Oxidation reaction is a chemical reaction which can be described as follows ;

Addition of oxygen Removal of hydrogen Loss of ElectronAddition of electronegative atomRemoval of Electropositive element

In the given reaction ;

Fe(s) + 2HCl(aq) → FeCl₂(aq) + H₂(g)

Fe at RHS got converted to Fe²⁺ state at LHS which shows the gain of electron by Fe with in the reaction.

Therefore,

The oxidation half-reaction for the given reaction is Fe(s) → Fe²⁺ + 2e⁻ Hence, Option (C) is correct

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Si enfriamos mercurio de 100C. Calcular la cantidad de calor que se debe restar sabiendo que la masa de mercurio es de 1800gr

Answers

Answer:

I do not speak Spanish.

Explanation:

repining of fruits is which type of change​

Answers

Answer:

irreversible.

I hope this will help you

Chemical Change. Hope it will help you

How many colors are there in a rainbow?

Answers

[tex]\boxed{\large{\bold{\blue{ANSWER~:) }}}}[/tex]

There are 7 colours in a rainbow

The colours of the rainbow are Red, Orange, Yellow, Green, Blue, Indigo and Violet.

Explanation:

there r seven colors in a rainbow.red, orange, yellow, green, blue, indigo and violet.

hope it helps.stay safe healthy and happy..

When solid Ni metal is put into an aqueous solution of Pb(NO3)2, solid Pb metal and a solution of Ni(NO3)2 result. Write the net ionic equation for the reaction.

Answers

Answer:

[tex]Pb^{2+}(aq)+Ni(s)\rightarrow Ni^{2+}(aq)+Pb(s)[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out firstly necessary for us to write the complete molecular equation as shown below:

[tex]Pb(NO_3)_2(aq)+Ni(s)\rightarrow Ni(NO_3)_2(aq)+Pb(s)[/tex]

Now, we can separate the nitrates in ions as they are aqueous to obtain:

[tex]Pb^{2+}(aq)+2(NO_3)^-(aq)+Ni(s)\rightarrow Ni^{2+}(aq)+2(NO_3)^-(aq)+Pb(s)[/tex]

And then, we cancel out the nitrate ions as the spectator ones, for us to obtain the net ionic equation:

[tex]Pb^{2+}(aq)+Ni(s)\rightarrow Ni^{2+}(aq)+Pb(s)[/tex]

Best regards!

what is the hybridisation of the central carbon in CH3C triple bonded to N​

Answers

Explanation:

the carbon would be sp3 hybridized, and it doesn't matter which carbon, since either of them have a full octet

sự sắp xếp nguyên tử trong vật chất

Answers

Answer:

sosksjsjjs

Explanation:

even i know how to type şïllily

CAN HF USED TO CLEAVE ETHERS EXPLAIN

Answers

Answer:

no

Explanation:

Fluoride is not nucleophilic (having the tendency to donate electrons) enough to allow for the use of HF to cleave ethers in protic media(protic solvents are polar liquid compounds that have dissociable hydrogen atoms). The rate of reaction is comparably low, so that heating of the reaction mixture is required.

why beta carbon hydrogen is easily replaceable but not alpha carbon hydrogen​

Answers

Answer:

Four common types of reactions involving carbonyl reactions: 1) nucleophilic addition; 2) nucleophilic acyl substitution; 3) alpha substitution; 4) carbonyl condensations. The first two were previously discussed and the second two involve the properties of the carbon directly adjacent to the carbonyls, α carbons.

Alpha-substitution reactions results in the replacement of an H attached to the alpha carbon with an electrophile.  

The nucleophile in these reactions are new and called enols and enolates.

Explanation:

The carbon in the carbonyl is the reference point and the alpha carbon is adjacent to the carbonyl carbon.  

Hydrogen atoms attached the these carbons denoted with Greek letters will have the same designation, so an alpha hydrogen is attached to an alpha carbon.  

Aldehyde hydrogens not given Greek leters.  

α hydrogens display unusual acidity, due to the resonance stabilization of the carbanion conjugate base, called an enolate.  

Tautomers are readily interconverted constitutional isomers, usually distinguished by a different location for an atom or a group, which is different than resonance.  

The tautomerization in this chapter focuses on the carbonyl group with alpha hydrogen, which undergo keto-enol tautomerism.  

Keto refers to the tautomer containing the carbonyl while enol implies a double bond and a hydroxyl group present in the tautomer.  

The keto-enol tautomerization equilibrium is dependent on stabilization factors of both the keto tautomer and the enol tautomer, though the keto form is typically favored for simple carbonyl compounds.  

The 1,3 arrangement of two carbonyl groups can work synergistically to stabilize the enol tautomer, increasing the amount present at equilibrium.

The positioning of the carbonyl groups in the 1,3 arrangement allows for the formation of a stabilizing intramolecular hydrogen bond between the hydroxyl group of the enol and the carbonyl oxygen as well as the alkene group of the enol tautomer is also conjugated with the carbonyl double bond which provides additional stabilization.

Aromaticity can also stabilize the enol tautomer over the keto tautomer.

Under neutral conditions, the tautomerization is slow, but both acid and base catalysts can be utilized to speed the reaction up.  

Biological enol forming reactions use isomerase enzymes to catalyze the shifting of a carbonyl group in sugar molecules, often converting between a ketose and an aldose in a process called carbonyl isomerization.

9. How can you separate sugar from a sugar solution contained in a glass without taste? Explain​

Answers

Answer:

See explanation

Explanation:

Sugar is a polar crystalline substance. The sugar crystal is capable of dissolving in water since it is polar.

When sugar dissolves in water, a sugar solution is formed. If I want to separate the sugar from the water in the solution, I have to boil the solution to a very high temperature.

When I do that, the water in the sugar solution is driven off and the pure sugar crystal is left behind.

examples s name of thosse food items we can store for a month?​

Answers

Answer:

1. Nuts

2. Canned meats and seafood

3. Dried grains

4. Dark chocolate

5. Protein powders

In the formation of 1.0 mole of the following crystalline solids from the gaseous ions, the most energy is released by

A) NaF
B) MgF₂
C) MgBr₂
D) AlF₃
E) AlBr₃

Answers

In the formation of 1.0 mole of the following crystalline solids from the gaseous ions, the most energy is released by AlF₃. Hence , Option (D) is correct

What factors affect the magnitude of energy of ionic crystalline solids ?

For an ionic compound, there are two main terms that this magnitude depends upon: ion size and ion charge.

Ion size: the smaller the ionic radii, the shorter the internuclear distance and, therefore, the closer the ions. This factor makes lattice enthalpy increase

Ion charge: the greater the charge on ions, the greater the attractive forces between them and, therefore, the larger the lattice enthalpy.

The lattice enthalpy of AlF₃ (5215 kJ/mol) is indeed greater than that of other given solids

Therefore , In the formation of 1.0 mole of the following crystalline solids from the gaseous ions, the most energy is released by AlF₃. Hence , Option (D) is correct

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2. 27.8 mL of an unknown were added to a 50.0-mL flask that weighs 464.7 g. The total mass of the flask and the liquid is 552.4 g. Calculate the density of the liquid in Lbs/ in3.

Answers

Answer:

[tex]d=4.24x10^{-4}\frac{lb}{in^3}[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out firstly necessary for us to set the equation for the calculation of density and mass divided by volume:

[tex]d=\frac{m}{V}[/tex]

Thus, we can find the mass of the unknown by subtracting the total mass of the liquid to the mass of the flask and the liquid:

[tex]m=552.4g-464.7g=87.7g[/tex]

So that we are now able to calculate the density in g/mL first:

[tex]d=\frac{87.7g}{27.8mL}=3.15g/mL[/tex]

Now, we proceed to the conversion to lb/in³ by using the following setup:

[tex]d=3.15\frac{g}{mL}*\frac{1lb}{453.6g}*\frac{1in^3}{16.3871mL}\\\\d=4.24x10^{-4}\frac{lb}{in^3}[/tex]

Regards!

A student attempts to separate 4.656 g of a sand/salt mixture just like you did in this lab. After carrying out the experiment, she recovers 2.775 g of sand and 0.852 g of salt.a. What was the percent composition of sand in the mixture according to the student's data? b. What was the percent recovery?

Answers

Answer:

Explanation:

a ) Total mixture = 4.656 g

Sand recovered = 2.775 g

percent composition of sand in the mixture

= (2.775 g / 4.656 g ) x 100

= 59.6 % .

b )

Total of sand and salt recovered = 2.775 g + .852 g = 3.627 g .

Total mixture = 4.656 g

percent recovery = (3.627 / 4.656 ) x 100

= 77.9 % .

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