Discuss the any two applications of Beer’s law with suitable justification.

Answer:

Keep temperature constant and increase the pressure of the reaction. The rate of reaction increases.

Explanation:

First of all, the question is asking us to design an experiment to investigate the effect of pressure on the rate of reaction hence the pressure can not be held constant since it is the variable under investigation. This eliminates the first option.

Secondly, increasing the pressure of the reaction means that particles of the gas collide more frequently leading to a greater number of effective collisions and a consequent increase in the rate of reaction according to the collision theory.

Hence the answer above.

Answer:

2.35 × 10²² particles

Explanation:

To get the number of particles in a substance, we multiply the number of moles (n) of that substance by Avogadro's number (nA).

That is, we say;

number of atoms = n × 6.02 × 10²³

In this question, Neon gas is said to have 0.0391 moles. The number of atoms in that field.

number of particles = 0.0391 × 6.02 × 10²³

number of particles = 0.235 × 10²³

number of particles = 2.35 × 10²² particles

Answer:

If a standard iodine solution is used as a titrant for an oxidizable analyte, the technique is iodimetry. If an excess of iodide is used to reduce a chemical species while simultaneously forming iodine.

Iodine always used in a solution excess KI is given to aid in the solubilization of free iodine, which would be insoluble in clean water during normal circumstances.

Iodine is a kind of element which are mainly used in iodometry titration. It can be represented by I.

A solution would be a homogenous mixture of two components, usually a solute as well as a solvent.

Iodimetry would be a technique that uses a standard iodine solution as a titrant for such an oxidizable analyte. When an excessive amount of iodide is used to decrease a chemical while somehow producing iodine.

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Answer:

[tex]m_{Al_2O_3}=118.27gAl_2O_3[/tex]

Explanation:

Hello there!

In this case, if we consider the following chemical reaction, whereby Al2O3 is produced from Al and FeO:

[tex]3FeO+2Al\rightarrow 3Fe+Al_2O_3[/tex]

Thus, since there is 3:1 mole ratio of FeO to Al2O3, it turns out feasible for us to use their molar masses, 71.844 g/mol and 101.96 g/mol respectively, to obtain the grams of the latter as follows:

[tex]m_{Al_2O_3}=250.gFeO*\frac{1molFeO}{71.844gFeO}*\frac{1molAl_2O_3}{3molFeO} *\frac{101.96gAl_2O_3}{1molAl_2O_3}\\\\m_{Al_2O_3}=118.27gAl_2O_3[/tex]

Regards!

Answer: The equilibrium constant for the given chemical reaction is [tex][O_2]^3[/tex]

Explanation:

The equilibrium constant is defined as the ratio of the concentration of the products to the concentration of reactants each raised to the power of their stoichiometric coefficients.

The concentration of all the solids and liquids are considered to be 1 in the expression of equilibrium constant

For the given chemical equation:

[tex]2KClO_3(s)\rightleftharpoons 2KCl(s)+3O_2(aq)[/tex]

The expression of equilibrium constant follows:

[tex]K_{eq}=[O_2]^3[/tex]

Hence, the equilibrium constant for the given chemical reaction is [tex][O_2]^3[/tex]

Answer:

Explanation:

NaL

Difluorine - F2

Molar Mass Bond Polarity F₂ Fluorine Gas Fluorine

Products

Sodium Fluoride - NaF

Molar Mass Bond Polarity Oxidation State Floridine Sodium Monofluoride Naf

L2

Answer:

3.50 × 10⁻³ g/L

Explanation:

Step 1: Make an ICE chart for the solution of SrCO₃

"S" represents the molar solubility.

         SrCO₃(s) ⇄ Sr²⁺(aq) + CO₃²⁻(aq)

I                             0               0

C                           +S              +S

E                             S               S

The solubility product constant (Ksp) is:

Ksp = [Sr²⁺] [CO₃²⁻] = S²

S = √Ksp = √5.60 × 10⁻¹⁰ = 2.37 × 10⁻⁵ M

Step 2: Convert "S" to g/L

The molar mass of SrCO₃ is 147.63 g/mol.

2.37 × 10⁻⁵ mol/L × 147.63 g/mol = 3.50 × 10⁻³ g/L

Answer: A. 142.04 g/mol

Explanation:

Answer:

6.85 g

Explanation:

Step 1: Given data

Step 2: Calculate the moles of sucrose (solute) required

Molarity is equal to the moles of solute divided by the liters of solution.

M = moles of solute / liters of solution

moles of solute = M × liters of solution

moles of solute = 0.200 mol/L × 0.10000 L = 0.0200 mol

Step 3: Calculate the mass corresponding to 0.0200 moles of sucrose

The molar mass of sucrose is 342.3 g/mol.

0.0200 mol × 342.3 g/mol = 6.85 g

The amount of sucrose (in grams) must be weighed out for the given reaction is 6.846g.

Mass of any substance will be calculated by using the moles as:
n = W/M, where

W = required mass

M = molar mass

Given that, molarity of sucrose = 0.2 M

Volume of solution = 100mL = 0.1 L

Relation between moles and molarity is represented as:
M = n/V

On putting values on the above equation we get,

n = (0.2)(0.1) = 0.02 moles

We know that molar mass of sucrose (C₁₂H₂₂O₁₁) = 342.3 g/mole

Now we calculate the required mass by putting values on the first equation as:
W = (0.02)(342.3) = 6.846g

Hence, the required mass of sucrose is 6.846g.

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Answer:

Cu

Explanation:

Groups 3 - 12 (or groups IIA - IIB) of the periodic table contain transition elements. Transaction elements start from period four (4) of the periodic table. The phrase alludes to the fact that the d sublevel is filling at a lower main energy level than the s sublevel that came before it.

The transition elements' arrangement is inverted from the fill order, with the 4 s filled prior to the actual 3 d begins. The transition elements are commonly referred to as transition metals since they are all metals. They are less reactive than the metals in Groups 1 and 2 and have normal metallic characteristics.

From the options given Cu is the only transition metal in the fourth period on the periodic table.

Answer:

t(9.32% remaining) = 203 yrs (3 sig. figs.)

Explanation:

All radioactive decay follows a 1st order decay profile. This is defined by the expression ...

A =A₀e^-k·t

A = final activity = 9.32%

A₀ = initial activity = 100%

e = base of natural logs

k = rate constant = 0.693/t(1/2) = (0.693/30.2) yrs⁻¹ = 0.023 yrs⁻¹

t = time of decay = ln(A/A₀)/-k = ln(9.32%/100%)/-0.023 yrs⁻¹

 = 203.286637 yrs (calc. ans.)

 ≅ 203 yrs (3 sig. figs.)

Answer:

The empirical formula is, C4H4S

Explanation:

Number of moles of carbon = 1.119 g/ 44g/mol = 0.025 moles

Mass of Carbon= 0.025 moles × 12 g/ mole = 0.3 g

Number of moles of hydrogen = 0.229/18g/mol × 2 = 0.025 moles

Mass of hydrogen = 0.025 moles × 1 = 0.025 g

Number of moles of sulphur = 0.407g/ 64 g/mol = 0.0064 moles

Mass of sulphur= 0.0064 moles ×32 = 0.2 g

Now we obtain the mole ratios by dividing through by the lowest ratio.

C- 0.025 moles/ 0.0064 moles, H- 0.025 moles/ 0.0064 moles, S- 0.0064 moles/0.0064 moles

C4H4S

Explanation:

If there are 15 protons, 15 nuclear particles of unit positive charge, then

Z

=

15

. Now

Z

≡

the atomic number

, and you look at your copy of the Periodic Table, and you find that for

Z

=

15

, the element phosphorus is specified.

But we are not finished. Along with the 15 defining protons, there are also 16 neutrally charged, massive nuclear particles, 16 neutrons, and the protons and neutrons together determine the atomic mass. The isotope is thus

31

P

, which is almost 100% abundant, and an important nucleus for

NMR spectroscopy.

Answer: A volume of 59 mL of 0.220 M HBr solution is required to produce 0.0130 moles of HBr.

Explanation:

Given: Moles = 0.0130 mol

Molarity = 0.220 M

Molarity is the number of moles of solute present in liter of a solution.

[tex]Molarity = \frac{moles}{volume (in L)}[/tex]

Substitute the values into above formula as follows.

[tex]Molarity = \frac{moles}{volume (in L)}\\0.220 M = \frac{0.0130 mol}{Volume (in L)}\\Volume (in L) = 0.059 L[/tex]

As 1 L = 1000 mL

So, 0.059 L = 59 mL

Thus, we can conclude that a volume of 59 mL of 0.220 M HBr solution is required to produce 0.0130 moles of HBr.

Answer:

Mass of sample in mg = 15,285 mg

Explanation:

Given:

Volume of urine sample = 15 ml

Density of sample = 1.019 g/ml

FInd:

Mass of sample in mg

Computation:

Mass = density x volume

Mass of sample in mg = Volume of urine sample x Density of sample

Mass of sample in mg = 1.019 x 15

Mass of sample in mg = 15.285 gram

Mass of sample in mg = 15.285 x 1,000

Mass of sample in mg = 15,285 mg

Answer:

B. The movement of weathered materials by wind, water or ice.

Answer:

CH3CH2CH3

Explanation:

Dipole moment is the measure of the polarity of a chemical bond. It is the extent of charge separation in a molecule.

Dipole moment is the product of the magnitude of charge and the distance separating the charges from each other.

The molecule having the lowest dipole moment among the options is the molecule that has the least polarity. The least polar molecule among the options is CH3CH2CH3, it has no polar bonds in its structure.

We have that for the Question "Identify the compound with the lowest dipole moment. Identify the compound with the lowest dipole moment. CH3CH2CH3 CH3OCH3 CH3CHO CH3OH CH3CN "

it can be said that

 From the question we are given

CH3CH2CH3

CH3OCH3

CH3CHO

CH3OH

CH3CN

Generally alkanes have the lowest dipole moment, they have C-H bond which are non polar.

Therefore,

[tex]CH_3CH_2CH_3[/tex] have the lowest dipole moment

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Answer:

40.0L of SO2 are produced

Explanation:

To solve this question we need to find the moles of O2 using PV = nRT in order to find the moles. Thus, we can find the limiting reactant and the moles (And volume) of SO2 produced as follows:

Moles O2:

n = PV/RT

n = 1.20atm*55.0L / 0.082atmL/molK*358K

n = 2.25 moles of O2.

Clearly, limiting reactant is O2.

The moles of SO2 produced are:

2.25 moles of O2 * (8mol SO2 / 11mol O2) = 1.6351 moles SO2

Volume SO2:

V = nRT/P

V = 1.6351 moles SO2*0.082atmL/molK*358K / 1.20atm

V = 40.0L of SO2 are produced

Answer:

D

Explanation:

thermal energy must be transferred from the ice to it's surroundings

Thermal energy is the heat and energy present in the system. The thermal energy must be dissipated from ice to the surroundings to melt and produce liquid. Thus, option D is correct.

Thermal energy is present in any substance in the form of energy and heat that depends on kinetic energy, orderliness, randomness, temperature, etc. It is passed from a substance to the surrounding at a lesser temperature.

The ice can melt into a liquid when it loses its thermal energy and heat to the surrounding with increased kinetic energy and releases the temperature to the area with a temperature less than the ice.

Therefore, the heat from ice will get dissipated to the outer environment to melt.

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Answer:

Explanation:

Chemical change

Answer:

both drugs will be equally effective

Explanation:

A racemic mixture is a substance that contains a 50/50 mixture of the (+) and (-) enantiomers of an optically active compound.

It is important to note that even if a drug occurs as a racemic mixture, only the biologically active enantiomer will interact with the substrate on which the drug acts. The other enantiomer present is just irrelevant. Similarly, its presence does not affect the biological activity of the active enantiomer.

Hence, both Prilosec and Nexium are both equally effective.

Answer:

Answer:

Concentration: 0.185M HX

Ka = 9.836x10⁻⁶

pKa = 5.01

Explanation:

A weak acid, HX, reacts with NaOH as follows:

HX + NaOH → NaX + H2O

Where 1 mole of HX reacts with 1 mole of NaOH

To solve this question we need to find the moles of NaOH at equivalence point (Were moles HX = Moles NaOH).

18.50mL = 0.01850L * (0.20mol / L) = 0.00370 moles NaOH = Moles HX

In 20.0mL = 0.0200L =

0.00370 moles HX / 0.0200L = 0.185M HX

The equilibrium of HX is:

HX(aq) ⇄ H⁺(aq) + X⁻(aq)

And Ka is defined as:

Ka = [H⁺] [X⁻] / [HX]

Where [H⁺] = [X⁻] because comes from the same equilibrium

As pH = 2.87, [H+] = 10^-pH = 1.349x10⁻³M

Replacing:

Ka = [H⁺] [H⁺] / [HX]

Ka = [1.349x10⁻³M]² / [0.185M]

Ka = 9.836x10⁻⁶

pKa = -log Ka

Answer:

154 g

Explanation:

Step 1: Write the balanced decomposition equation

2 NaN₃(s) ⇒ 2 Na(s) + 3 N₂(g)

Step 2: Calculate the moles corresponding to 79.5 L of N₂ at STP

At STP, 1 mole of N₂ occupies 22.4 L.

79.5 L × 1 mol/22.4 L = 3.55 mol

Step 3: Calculate the number of moles of NaN₃ needed to form 3.55 moles of N₂

The molar ratio of NaN₃ to N₂ is 2:3. The moles of NaN₃ needed are 2/3 × 3.55 mol = 2.37 mol.

Step 4: Calculate the mass corresponding to 2.37 moles of NaN₃

The molar mass of NaN₃ is 65.01 g/mol.

2.37 mol × 65.01 g/mol = 154 g

Explanation:

An atom consists of two regions. The first is the tiny atomic nucleus, which is in the center of the atom and contains positively charged particles called protons and neutral, uncharged, particles called neutrons. ... Most atoms contain all three of these types of subatomic particles—protons, electrons, and neutrons.

Answer:

skeletal system

Explanation:

to create and fliter blood and provide frame-work to the human body and support

Answer: The mass of hydrogen formed when 26.98 g of aluminum reacts with excess hydrochloric acid according to the given balanced equation is 3.03 g.

Explanation:

The given balanced reaction equation is as follows.

[tex]2Al + 6HCl \rightarrow 2AlCl_{3} + 3H_{2}[/tex]

Here, the mole ration of Al and hydrogen produced is 2 : 3

As mass of aluminum is given as 26.98 g. So, moles of aluminum (molar mass = 26.98 g/mol) is as follows.

[tex]Moles = \frac{mass}{molar mass}\\= \frac{26.98 g}{26.98 g/mol}\\= 1 mol[/tex]

So, when 1 mole of Al reacted then 1.5 moles of hydrogen is produced as per the given mole ratio.

Therefore, mass of hydrogen formed is calculated as follows.

[tex]mass = moles \times molar mass\\= 1.5 mol \times 2.02 g/mol\\= 3.03 g[/tex]

Thus, we can conclude that the mass of hydrogen formed when 26.98 g of aluminum reacts with excess hydrochloric acid according to the given balanced equation is 3.03 g.

Answer:

it would be option C

Explanation:

Speed of light = 3×10^8m/s

Planck's constant = 6.626×10^-34 Js

Wavelength = 8 x 10^-9 m

Energy = [(3×10^8) * (6.626×10^-34)] / 8 x 10^-9

Energy = [19.878×10^(8-34)] / 8 x 10^-9

Energy = 2.48475 × 10^(-26+9)

Energy = 2.48×10^-17 J