The molarity (M) of an aqueous solution containing 22.5 g of sucrose (C12H22O11) in 35.5 mL of solution is ________.

Answer:

(a). Kindly check the explanation section.

(b). 24.5 M^-1 cm^-

(c). 0.0318 M

Explanation:

So, we are given the following data parameters which is going to aid us in solving the above Question;

The path length = 1cm, extinction coefficient = unknown, absorbance is unknown and concentration= 0.02mM.

Using the formula below; we can determine the extinction coefficient as;

Absorbance = extinction coefficient × concentration × path length. -----------(1).

(a). Since, the absorbance is not given in the Question it won't be possible to determine the value of the extinction coefficient. Thus, say the value of absorbance = A.

Then, extinction coefficient = A/ 0.02 × 1 cm.

(b). Making use of the formula above, the extinction coefficient can be Determine as follows;

extinction coefficient = 1.84/0.075 × 1 = 24.5 M^-1 cm^-1.

(c). The concentration can be Determine by also making use of the formula above and making the concentration the subject of the formula;

Concentration = absorbance/ extinction coefficient × path length.

Concentration = 0.78/24.5 M^-1 cm^-1. × 1cm = 0.0318 M.

Answer:

No

Explanation:

Nitrogen cannot be used to build the wire for a cell phone charger.

What are the elements used to build charger ?

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Answer:

C |||| It is not practical because it takes a huge amount of water to produce a new pair of jeans

Explanation:

If you're doing flvs then it's C.

Answer:

C!

Explanation:

i got it right on the test UwU

Answer:

52.8 g of O2.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

4Al + 3O2 —> 2Al2O3

From the balanced equation above,

4 moles of Al reacted with 3 moles of O2 to produce 2 moles of Al2O3

Next, we shall determine the number of mole of O2 needed to react with 2.2 moles of Al. This can be obtained as follow:

From the balanced equation above,

4 moles of Al reacted with 3 moles of O2.

Therefore, 2.2 moles of Al will react with = (2.2 × 3)/4 = 1.65 moles of O2.

Thus, 1.65 moles of O2 is needed for the reaction.

Finally, we shall determine the mass of O2 needed as shown below:

Mole of O2 = 1.65 moles

Molar mass of O2 = 2 × 16= 32 g/mol

Mass of O2 =?

Mole = mass/Molar mass

1.65 = mass of O2 /32

Cross multiply

Mass of O2 = 1.65 × 32

Mass of O2 = 52.8 g

Therefore, 52.8 g of O2 is needed for the reaction.

Answer:

Explain some of the uses of metals based on their properties.

Explanation:

Metals have a shiny or metallic luster and are good conductors of heat and electricity, they can be bent and pounded in various shapes, so they can be used on cars, coins, some pipes, keys, and and a flag.

Answer:

carbon-13

Explanation:

Answer:

32.5

Explanation:

2C⁴H¹⁰ needs 13 O²

5 C⁴H¹⁰ needs (13/2 × 5) O² = 32.5 O²

The answer is D, 32.5

(I took the test)

Answer: The force on the firefly

The unfortunate firefly hitting the bus does not change the velocity of the bus very much. Technically there is a change, but it's so very small and miniscule that it barely registers. To any casual observer not paying very close attention, they don't notice anything at all. So effectively the force on the firefly is a lot greater since the firefly got the worst end of the deal.

So in short, we look at the velocity of each object and see which velocity changed the most. In this case, the firefly's velocity changed from whatever speed it was flying to 0 when it stops flying all together. That's why the force is greater on the bug.

Answer:

residue is whatever remains after something else has been removed while filtrate is the liquid or solution that has passed through a filter, and which has been separated from the filtride.

filtration id done by placing a filter paper on the beaker or container then pour the filtride then let it settle and it will pass through and you will have the fitrate and residue

Explanation:

Answer:

option C is correct

Explanation:

Considering the ideal gas law and the definition of Avogadro's Number, the correct option is option a. The temperature of a sample of CH₄ gas (10.34 g) in a 50.0 L vessel at 1.33 atm is 984 °C.

In first place, you have to know that ideal gases are a simplification of real gases that is done to study them more easily.

It is considered to be formed by point particles, do not interact with each other and move randomly. It is also considered that the molecules of an ideal gas, in themselves, do not occupy any volume.

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T).

The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P×V = n×R×T

In this case, being the molar mass of CH₄ being 16 [tex]\frac{g}{mole}[/tex], that is, the mass present in one mole of an element or compound, the number of moles that 10.34 grams contains is calculated as:

[tex]10.34 g*\frac{1 mole}{16.04 g} = 0.645 moles[/tex]

So, you know:

Replacing:

1.33 atm × 50 L= 0.645 moles× 0.082 (atm×L)/ (mol×K) ×T

Solving:

T= [1.33 atm × 50 L] ÷ [0.645 moles× 0.082 (atm×L)/ (mol×K) ]

T≅ 1257 K

Being 273 K equivalent to 0 C, then:

T= 1257 K= 984 C

In summary, the correct option is option a. The temperature of a sample of CH₄ gas (10.34 g) in a 50.0 L vessel at 1.33 atm is 984 °C.

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Answer:

[tex]k=8.63x10^{-8}\frac{1}{M*s}[/tex]

Explanation:

Hello!

In this case, since the differential rate law of a second-order reaction is:

[tex]\frac{dC_A}{dt}=-kC_A^2[/tex]

Whereas A stands for NOCl and the corresponding integrated rate law is:

[tex]\frac{1}{C_A} =kt+\frac{1}{C_A_0}[/tex]

Thus, since we know the concentrations and the elapsed time, we compute the rate constant as shown below:

[tex]k=( \frac{1}{C_A}-\frac{1}{C_A_0} )/t\\\\k=( \frac{1}{0.83M}-\frac{1}{0.878M} )/763,200s\\\\k=8.63x10^{-8}\frac{1}{M*s}[/tex]

Best regards!

Answer:

(a) 6332.812 N (b) 1469.59  pounds.

Explanation:

Given that,

The surface area of the top of yu head is 25 cm x 25 cm

Area = 625 cm² = 0.0625 m²

Atmospheric pressure on it is 101325 Pa.

(a) Pressure acting on an object is given by force acting per unit area. It can be written as :

[tex]P=\dfrac{F}{A}\\\\F=PA\\\\F=101325\ Pa\times 0.0625\ m^2\\\\F=6332.812\ N[/tex]

6332.812 N of force is acting on your head at sea level.

(b) If area, A = 100 in²

We know that,

1 atm = 14.6959 pounds/in²

[tex]P=\dfrac{F}{A}\\\\F=PA\\\\F=14.6959 \ \text{Pounds}/in^2\times 100\ in^2\\\\F=1469.59\ \text{pounds}[/tex]

So, the force is 1469.59  pounds.

A. The force required to push your head at sea level is 6332.8125 N

B. The force in pound is 1469.59 lb

A. Determination of the force required at sea level.

Pressure (P) = 101325 Nm²

Area (A) = 25 cm × 25 cm = 0.25 m × 0.25 m = 0.0625 m²

P = F / A

Cross multiply

F = PA

F = 101325 × 0.0625

Therefore, the force required to push your head at sea level is 6332.8125 N

B. Determination of the force in pounds

Area (A) = 100 in²

Pressure (P) = 14.6959 pound / in²

F = PA

F = 14.6959 × 100

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Answer:

B) 1st C sp3 ; 2nd C sp3 ; O sp3

Explanation:

Both carbons in ethanol have an sp³ configuration of their orbitals, given that each carbon atom binds to four other atoms, as can be seen on the extended formula in the attachment.

In these carbons (1st and 2nd) one s orbital and three p orbitals are hybridized to form four hybrid orbitals sp³. These orbitals form four bonds and their bond angles are close to the tetrahedral value of 109.5º,as can be seen for methanol which also has an sp³ configuration.

The oxygen atom can also be described as having sp³ hybrid orbitals. Two of the four hybrid sp³ orbitals in oxygen are occupied by lone electron pairs, and two are used to form bonds with other atoms (in this case the oxygen is bonded to a carbon atom and a hydrogen atom).

Therefore, the three central atoms have an sp³ configuration.

Answer:

Titration is a technique to determine the concentration of an unknown solution. As illustrated in the titration setup above, a solution of known concentration (titrant) is used to determine the concentration of an unknown solution (titrand or analyte).

Typically, the titrant (the solution of known concentration) is added through a burette to a known volume of the analyte (the solution of unknown concentration) until the reaction is complete. Knowing the volume of titrant added allows us to determine the concentration of the unknown analyte. Often, an indicator is used to signal the end of the reaction, the endpoint. Titrant and analyte is a pair of acid and base. Acid-base titrations are monitored by the change of pH as titration progresses.

Let us be clear about some terminologies before we get into the discussion of titration curves.

Answer:

When there is no spark from the reaction

Explanation:

The reaction between the acid and the magnesium metal goes into completion when there is no more spark from the magnesium. Also, the magnesium would have totally disappeared into solution.

These physical observation marks the end of the reaction.

When Mg reacts with HCl hydrogen bubbles are released. The completion of the given reaction can be determined by the presence of bubbles above the HCl.

[tex]\bold {Mg+ 2HCl\rightarrow MgCl_2 + H_2}[/tex]

In this reaction, Mg reacts with 2 moles of HCl to form 1 mole of Magnesium chloride and release hydrogen gas.

During reaction,

Therefore, the completion of the given reaction can be determined by the presence of bubbles above the HCl.

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Answer:

From the numerical steps highlighted under explanation, the average atomic mass of bromine is 79.91 u

Explanation:

The steps to be taken will involve;

1) Find the number of isotopes of bromine.

2) Identify the atomic mass and relative abundance of each of the isotopes.

3) Multiply the atomic mass of each of the isotopes by their corresponding values relative abundance value.

4) Add the value in step 3 above to get the average atomic mass of bromine.

Now;

Bromine has 2 isotopes namely;

Isotope 1: Atomic mass = 78.92amu and a relative abundance of 50.69%.

Isotope 2: Atomic mass = 80.92amu and a relative abundance of 49.31%.

Using step 3 above, we have;

(78.92 × 50.69%)

And (80.92 × 49.31%)

Using step 4 above, we have;

(78.92 × 50.69%) + (80.92 × 49.31%) ≈ 79.91 u

The atomic mass of bromine has been calculated as 79.9062 amu.

The relative atomic mass of the element has been given as the mass of each isotope with respect to their abundance.

The atomic mass has been given as:

[tex]amu=\sum mass\;\times\;\% Abundance[/tex]

The available isotopes of Bromine have been:

Isotope 1 = 78.92 amu, and 0.5069 % abundance

Isotope 2= 80.92 amu, and 0.4931 % abundance

Submitting the values for the atomic mass unit of bromine ([tex]amu_{\text{Br}}[/tex]):

[tex]amu_{\text {Br}}= (78.92\;\times\;0.5069)\;+\;(80.92\;\times\;0.4931)\\amu_{\text{Br}}=40.0045\;+\;39.9016\\amu_{\text{Br}}=79.9062[/tex]

The atomic mass of bromine has been calculated as 79.9062 amu.

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Answer:

ABC

Explanation:

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