What is the difference in height between the top surface of the glycerin and the top surface of the alcohol? Suppose that the density of glycerin is 1260 kg/m3 and the density of alcohol is 790 kg/m3.

Answer:

40mL of glycerol are needed to make a 20% v/v solution

Explanation:

This problem can be solved with a simple rule of three:

20%  v/v is a sort of concentration. In this case, 20 mL of solute are contained in 100 mL of solution.

Therefore, in 100 mL of solution you have 20 mL of solvent (glycerol)

In 200 mL, you would have,  (200 .20)/ 100 = 40 mL

Answer:

M₂ = 0.0745 M

Explanation:

In case of titration , the following formula can be used -

M₁V₁ = M₂V₂

where ,

M₁ = concentration of acid ,

V₁ = volume of acid ,

M₂ = concentration of base,

V₂ = volume of base .

from , the question ,

M₁ = 0.0952 M

V₁ = 38.73 mL

M₂ = ?

V₂ = 49.48 mL

Using the above formula , the molarity of ammonia , can be calculated as ,

M₁V₁ = M₂V₂  

0.0952 M * 38.73 mL = M₂* 49.48 mL

M₂ = 0.0745 M

Explanation:

(a)  Chemical reaction equation is given as follows.

           [tex]Ag^{+} + Cl^{-} \rightarrow AgCl(aq) \rightarrow K_{1}[/tex]

Also,  

         [tex]AgCl(s) \rightarrow Ag^{+} + Cl^{-} \rightarrow K_{3}[/tex]

Therefore, the net reaction equation is as follows.

         [tex]AgCl(s) \rightarrow AgCl(aq)[/tex]

Now, we will calculate the value of K for this reaction as follows.

          K = [tex]K_{1} \rightarrow K_{2}[/tex]

             = [tex]2.0 \times 10^{3} \times 1.8 \times 10^{-10}[/tex]

             = [tex]3.6 \times 10^{-7 }[/tex]

Hence, the value of K for the given reaction is  [tex]3.6 \times 10^{-7 }[/tex].

(b)  As the reaction  is given as follows.

              [tex]AgCl(s) \rightarrow AgCl(aq)[/tex]

Therefore, when excess of AgCl(s) is added then the amount of [AgCl(aq)] present in equilibrium is as follows.

                   K = [AgCl(aq)] = [tex]3.6 \times 10^{-7 }[/tex]

Thus, the value of [AgCl(aq)] in equilibrium with excess AgCl(s) is [tex]3.6 \times 10^{-7 }[/tex].

(a) The equilibrium constant K for the reaction[tex]\(\text{AgCl(s)} \rightleftharpoons \text{AgCl(aq)}\) is \( 3.6 \times 10^{-9} \).[/tex]

(b) The concentration of AgCl(aq) in equilibrium with excess AgCl(s) is [tex]\( 3.6 \times 10^{-9} \, \text{M} \).[/tex]

Part (a): Finding K for the reaction-

We are given the following equilibria and their constants:

1. [tex]\(\text{Ag}^+ + \text{Cl}^- \rightleftharpoons \text{AgCl(aq)} \quad K_1 = 20\)[/tex]

2. [tex]\(\text{AgCl(aq)} \rightleftharpoons \text{Ag}^+ + \text{Cl}^- \quad K_2 = 93\)[/tex]

3. [tex]\(\text{AgCl(s)} \rightleftharpoons \text{Ag}^+ + \text{Cl}^- \quad K_3 = 1.8 \times 10^{-10}\)[/tex]

We need to find K for the reaction:

[tex]\[ \text{AgCl(s)} \rightleftharpoons \text{AgCl(aq)} \][/tex]

We can see that we need to combine these equilibria in a way that gets us from AgCl(s) to AgCl(aq)

Let's write the reactions again:

1. [tex]\(\text{Ag}^+ + \text{Cl}^- \rightleftharpoons \text{AgCl(aq)} \quad K_1 = 20\)[/tex]

2. [tex]\(\text{AgCl(aq)} \rightleftharpoons \text{Ag}^+ + \text{Cl}^- \quad K_2 = 93\)[/tex]

3. [tex]\(\text{AgCl(s)} \rightleftharpoons \text{Ag}^+ + \text{Cl}^- \quad K_3 = 1.8 \times 10^{-10}\)[/tex]

The relationship between these equilibria can be described as follows:

- [tex]\(\text{AgCl(s)} \rightleftharpoons \text{Ag}^+ + \text{Cl}^- \quad (K_3)\)[/tex]

- [tex]\(\text{Ag}^+ + \text{Cl}^- \rightleftharpoons \text{AgCl(aq)} \quad (K_1)\)[/tex]

If we multiply K₃ by K₁, we will get the desired equilibrium:

[tex]\[ \text{AgCl(s)} \rightleftharpoons \text{Ag}^+ + \text{Cl}^- \quad (K_3) \][/tex]

[tex]\[ \text{Ag}^+ + \text{Cl}^- \rightleftharpoons \text{AgCl(aq)} \quad (K_1) \][/tex]

Multiplying these two equilibria together, the intermediate [tex]\(\text{Ag}^+\)[/tex] and [tex]\(\text{Cl}^-\)[/tex] ions cancel out, giving us:

[tex]\[ \text{AgCl(s)} \rightleftharpoons \text{AgCl(aq)} \][/tex]

The equilibrium constant for this reaction is:

[tex]\[ K = K_3 \times K_1 \][/tex]

Substituting the given values:

[tex]\[ K = (1.8 \times 10^{-10}) \times 20 \] \\[/tex]

[tex]\[ K = 3.6 \times 10^{-9} \][/tex]

So, the equilibrium constant K for the reaction [tex]\(\te{AgCl(s)} \rightleftharpoons \text{AgCl(aq)}\)[/tex] is [tex]\( 3.6 \times 10^{-9} \).[/tex]

Part (b): Finding AgCl(aq) in equilibrium with excess AgCl(s)

When AgCl(s) is in equilibrium with AgCl(aq), the equilibrium expression for this reaction is:

[tex]\[ K = [\text{AgCl(aq)}] \][/tex]

From Part (a), we found that:

[tex]\[ K = 3.6 \times 10^{-9} \][/tex]

Therefore, the concentration of [tex]\(\text{AgCl(aq)}\)[/tex] in equilibrium with excess [tex]\(\text{AgCl(s)}\)[/tex] is:

[tex]\[ [\text{AgCl(aq)}] = 3.6 \times 10^{-9} \, \text{M} \][/tex]

Answer:

Explanation:

A >1000 °C does not conduct electricity : covalent ( usually do not conduct electricity, they are formed by the sharing of electrons)

B 850 °C conducts electricity in the liquid state, but not in the solid state: Ionic ( ionic or electrovalent compounds are formed between atoms where one loses an electron while the other gains e.g NaCl, they conduct electricity when dissolved in a polar solvent because they dissociate into ions and have high melting and boiling points)

750 °C conducts electricity in the solid state : Metallic ( metals generally have delocalized electrons that enables them to conduct electron since they are no associated with bond and are therefore free to move)

D 150 °C does not conduct electricity : Molecular ( consist mainly of molecules; they do not have charge)

The bromination of trans-stilbene results in meso-1,2-dibromo-1,2-diphenylethane, an achiral molecule. In contrast, the bromination of cis-stilbene yields enantiomeric products, (1R,2S)-1,2-dibromo-1,2-diphenylethane and (1S,2R)-1,2-dibromo-1,2-diphenylethane.

The bromination of trans-stilbene and cis-stilbene leads to different products due to the arrangement of the double bonds in the starting materials.

1. **Trans-Stilbene:**

  - In trans-stilbene, the two phenyl rings are on opposite sides of the double bond. Bromination in the presence of bromine ([tex]\(Br_2\)[/tex]) or other brominating agents typically occurs with syn-addition across the double bond.

  - The product is meso-1,2-dibromo-1,2-diphenylethane. The meso compound has a plane of symmetry, resulting in an achiral molecule.

2. **Cis-Stilbene:**

  - In cis-stilbene, the two phenyl rings are on the same side of the double bond. Bromination of cis-stilbene can yield two enantiomeric products due to anti-addition across the double bond.

  - The products are (1R,2S)-1,2-dibromo-1,2-diphenylethane and (1S,2R)-1,2-dibromo-1,2-diphenylethane.

Answer:

The molecules are hybridized at different angles as shown in the explanation.

Explanation:

The hybridization and the geometries of the carbon atoms are as follows:

C1     → sp², trigonal planar, angle = 120⁰

C (II)→  sp³, tetrahedral, angle = 109.5⁰

C(III)→ sp², trigonal planar, 120⁰

C(IV)→ sp³, trigonal pyramidal, 109, 5⁰

Answer:

1. Zn is oxidize.

Cu is reduced.

2. Fe is oxidize.

H2 is reduced.

Explanation:

1. Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s)

From the equation,

Zn changes oxidation number from 0 to +2. Therefore Zn is oxidized.

Cu2+ changes oxidation number from +2 to 0. Therefore, Cu2+ is reduced

2. 3Fe(s)+4H2O(g)→Fe2O3(s)+4H2(g)

From the equation,

Fe changes oxidation number from 0 to +3. Therefore, Fe is oxidized.

H changes oxidation number from +1 to 0. Therefore, H is reduced

Answer:

The concentration  of methyl isonitrile will become 15% of the initial value after 10.31 hrs.

Explanation:

As the data the rate constant is not given in this description, However from observing the complete question  the rate constant is given as a rate constant of 5.11x10-5s-1 at 472k .

Now the ratio of two concentrations is given as

[tex]ln (\frac{C}{C_0})=-kt[/tex]

Here C/C_0 is the ratio of concentration which is given as 15% or 0.15.

k is the rate constant which is given as [tex]5.11 \times 10^{-5} \, s^{-1}[/tex]

So time t is given as

[tex]ln (\frac{C}{C_0})=-kt\\ln(0.15)=-5.11 \times 10^{-5} \times t\\t=\frac{ln(0.15)}{-5.11 \times 10^{-5} }\\t=37125.6 s\\t=37125.6/3600 \\t= 10.31 \, hrs[/tex]

So the concentration will become 15% of the initial value after 10.31 hrs.

Complete question:

The rearrangement of methyl isonitrile (CH3NC) to acetonitrile (CH3NC) is a first-order reaction and has a rate constant of 5.11x10-5s-1 at 472k . If the initial concentration of CH3NC is 3.00×10-2M. How many hours will it take for the concentration of methyl isonitrile to drop to 15.0 % of its initial value?

Answer:

The time taken for the concentration of methyl isonitrile to drop to 15.0 % of its initial value is 10.3 hours

Explanation:

The initial concentration of CH3NC is 3.00 X 10⁻²M =0.03M

The rate constant  K= 5.11 X 10⁻⁵s⁻¹

If the concentration of methyl isonitrile to drop to 15.0 %;

The new concentration of methyl isonitrile becomes 0.15 X 0.03 = 0.0045 M

The time taken to drop to 0.0045 M, can be calculated as follows:

[tex]t = -ln[\frac{(CH_3NC)}{(CH_3NC)_0}]/K[/tex]

[tex]t = (-ln[\frac{0.0045}{0.03}]/5.11 X 10^{-5})X(\frac{1 min}{60 s}) = 618.8 mins[/tex]

→ 618.8 mins X 1hr/60mins = 10.3 hours

Therefore, the time taken for the concentration of methyl isonitrile to drop to 15.0 % of its initial value is 10.3 hours

Answer:

a. alkyne

b. alkane

c. alkyne

d. alkene

Explanation:

The general formula for each class of compound is given below

Alkane: [tex]C_nH_{2n+2}[/tex]

Alkene: [tex]C_nH_{2n}[/tex]

Alkyne: [tex]C_nH_{2n-2}[/tex] (assuming single multiple bonds)

Now let us classify according to the above formulas:

a. It has two hydrogen atoms less than the two times of carbon atoms hence, it's alkyne

b. It has two hydrogen atoms more than the two times of carbon atoms hence, it's alkane

c. It has two hydrogen atoms less than the two times of carbon atoms hence, it's alkyne

d. It has hydrogen atoms two times of carbon atoms hence, it's alkene

Answer:Based on the molecular formula, the following compounds belongs to the following groups:

a. C6H10 ( alkyne)

b. C4H10 ( alkane)

c. C8H14 (alkyne)

d. C8H16 (alkene)

Explanation: Hydrocarbon is an organic chemical compound that contains Hydrogen and carbons. There are three main types of Hydrocarbons which includes:

- Saturated Hydrocarbons.( They are composed entirely of single bonds and are saturated with hydrogen.). They are the alkane.

- Unsaturated Hydrocarbons: They are composed of either a double or triple bonds. The double bonds are the alkene and the triple bonds are the alkynes.

- Aromatic Hydrocarbons: they contain an aromatic ring. Example is the Benzene.

I hope this helps. Thanks.

Answer:

The mass of CaCO3 reacted  is 5.00 grams

Explanation:

Step 1 :Data given

Before the reaction, the container, including the CaCO3, had a mass of 24.20 g

After the reaction the container with product had a mass of only 22.00 g because the CO2 gas produced did not remain in the container.

Molar mass of CO2 = 44.01 g/mol

Molar mass CaCO3 = 100.09 g/mol

Step 2: The balanced equation

CaCO3 → CaO + CO2

Step 3: Calculate mass of CO2

Mass of CO2 = 24.20 grams - 22.00 grams

Mass of CO2 = 2.20 grams

Step 4: Calculate moles CO2

Moles CO2 = mass CO2 / molar mass CO2

Moles CO2 = 2.20 grams / 44.01 g/mol

Moles CO2 = 0.0500 moles

Step 5: Calculate moles CaCO3

For 1 mol CaCO3 we'll have 1 mol CaO and 1 mol CO2

For 0.0500 moles CO2 we need 0.0500 moles CaCO3

Step 6: Calculate mass CaCO3

Mass CaCO3 = moles CaCO3 * molar mass CaCO3

Mass CaCO3 = 0.0500 moles  * 100.09 g/mol

Mass CaCO3 = 5.00 grams

The mass of CaCO3 reacted  is 5.00 grams

Final answer:

2.20 grams of calcium carbonate (CaCO3) reacted, as calculated by the difference in mass before and after the decomposition reaction, which released carbon dioxide gas from the container.

Explanation:

The initial mass of the reaction container with calcium carbonate (CaCO3) was 24.20 g, while the final mass after the reaction was 22.00 g. The mass of CaCO3 that reacted can be found by subtracting the final mass from the initial mass, because the only mass lost would be the carbon dioxide (CO2) gas that escaped from the container.

Mass of CaCO3 that reacted = Initial mass - Final mass
= 24.20 g - 22.00 g
= 2.20 g

Therefore, 2.20 grams of calcium carbonate reacted, decomposing into calcium oxide (CaO) and releasing carbon dioxide gas according to the reaction CaCO3(s) → CaO(s) + CO2(g).

Answer:

(a) ν = 3.1 × 10¹³ s⁻¹

(b) λ = 3.467 μm

Explanation:

We can solve both problems using the following expression.

c = λ × ν

where,

c: speed of light

λ: wavelength

ν: frequency

(a)

c = λ × ν

ν = c / λ

ν = (3.000 × 10⁸ m/s) / (9.6 × 10⁻⁶ m)

ν = 3.1 × 10¹³ s⁻¹

(b)

c = λ × ν

λ = c / ν

λ = (3.000 × 10⁸ m/s) / (8.652 × 10¹³ s⁻¹)

λ = 3.467 × 10⁻⁶ m

λ = 3.467 × 10⁻⁶ m (10⁶ μm/ 1 m)

λ = 3.467 μm

Answer:

a) F⁻ < Cl⁻ < Br⁻

b) Mg²⁺ < Na⁺ < F⁻

c) Cr³⁺ < Cr²⁺

Explanation:

The ions in (a) part of the question belong to a halogen group. F, Cl, and Br are present in periods 2, 3, and 4 respectively. As we move down the group the size of atoms increases hence their ions will be in the same order. (ion from top to bottom of group 7)

The ions in (b) part of the question are isoelectronic. The relative size of such species can be estimated by the charge on their nucleus. Lower the nucleus charge greater will be the size of the ion.

Nuclear charge of Mg²⁺ = no. of protons = 12

Nuclear charge of Na⁺   = no. of protons = 11

Nuclear charge of F⁻     = no. of protons =  9

The ions in (c) part are the two oxidized states of chromium. In such cases, higher the number of nuclear charge smaller will be the ion.

Answer:

(a) ml = 0, ±1, ±2

(b) ml = 0

(c) ml = 0, ±1, ±2, ±3, ±4

Explanation:

The rules for electron quantum numbers are:

1. Shell number, 1 ≤ n

2. Subshell number, 0 ≤ l ≤ n − 1

3. Orbital energy shift, -l ≤ ml ≤ l

4. Spin, either -1/2 or +1/2

So in our exercise,

(a) l = 2; equivalent with with sublevel d

-l ≤ ml ≤ l, ml = 0, ±1, ±2, equivalent with dxy, dxz, dyz, dx2-y2, dz2

(b) n = 1;

n = 1, only 01 level

l = 0, equivalent with sublevel s

ml = 0

(c) n = 4, l = 3.

l = 3, equivalent with sublevel f

ml = 0, ±1, ±2, ±3, ±4

Answer:

Explanation:

Calcium is the element of the group 2 and period 4 which means that the valence electronic configuration is [tex]1s^22s^22p^63s^23p^64s^2[/tex] or [tex][Ar]4s^2[/tex].

Chlorine is the element of the group 17 and period 3 which means that the valence electronic configuration is [tex]1s^22s^22p^63s^23p^5[/tex] or [tex][Ne]3s^23p^5[/tex].

Thus, calcium losses 2 electrons to 2 atoms of chlorine and these 2 atoms of chlorine accepts each electron to form ionic bond. This is done in order that the octet of the atoms are complete and they become stable.

Thus, the formula of calcium chloride is [tex]CaCl_2[/tex].

Hence, in [tex]CaCl_2[/tex], Calcium exits in [tex]Ca^{2+}[/tex] form which has electronic configuration of [tex]1s^22s^22p^63s^23p^6[/tex] or [tex][Ar][/tex]

The ground-state electron configuration for a calcium ion (Ca²+) is 1s²2s²2p63s²3p6, the same configuration as the noble gas Argon (Ar). This is because a calcium ion loses its two 4s valence electrons when it forms a cation.

The calcium ion, denoted as Ca²+, has lost its valence electrons to form a cation. In an uncharged, ground state calcium atom, it contains 20 electrons with configuration as 1s²2s²2p63s²3p64s². When it loses its two 4s valence electrons to become a cation, its electron configuration changes to 1s²2s²2p63s²3p6 which is similar to the noble gas Argon (Ar).

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Answer:

Electromagnetic waves are usually defined as those waves that are generated due to the vibrations between an electric as well as a magnetic field. Here, the component comprising the electric field and the component comprising the magnetic field vibrates perpendicular to each other, and both are in phase. Some of the examples of this type of wave are microwaves, infrared, ultra-violet, visible light, X-rays and many more.

The microwave and ultraviolet radiations are two electromagnetic waves that have similar characteristics, travel at a similar speed of about 300,000 km per second.

They differ from one another in many ways. It is because the microwaves have a higher wavelength, low frequency, and low energy. On the other hand, ultraviolet radiations have a low wavelength, high frequency, and high energy.

Final answer:

The genome of an organism contains all its genetic material necessary for life and reproduction. Genome size varies widely across species and can inform us about genetic diversity and evolutionary history. The Human Genome Project and similar research have deepened our understanding of genomes and their role in growth, development, and evolution.

Explanation:

The genome of an organism includes all of its genetic material, which contains the necessary information for sustaining life and reproduction. As the complexity of organisms increases, generally, so does the size of their genomes. However, genome size doesn't correlate directly to the number of genes an organism has.

For instance, humans have about 3.5 pg of DNA in their genome, which translates to roughly 3.4 billion base pairs, despite not having the largest genome when compared to certain plants or even other animals. Intriguingly, a significant portion of the genome consists of non-coding DNA that doesn't seem to have a direct function in gene encoding.

Genome size differs widely across species, with eukaryotes often having multiple chromosomes and bacteria generally having smaller genomes. The genome size can provide information on an organism's genetic diversity and evolutionary history. The role of the genome is vital in the growth and development of organisms, dictating the specific instructions for these processes.

Additionally, the sequencing of genomes, as in the Human Genome Project, has provided valuable insights into genetic variations between and within species, fostering a greater understanding of evolutionary biology and potential medical advancements.

Answer: Molarity of this solution is 0.88 M

Explanation:

Molarity : It is defined as the number of moles of solute present in one liter of solution.

Formula used :

[tex]Molarity=\frac{n\times 1000}{V_s}[/tex]

where,

Given : 3.0 g of hydrogen peroxide is present in 100 g of solution.

n = moles of solute = [tex]\frac{\text {given mass}}{\text {molar mass}}=\frac{3.0g}{34g/mol}=0.088mol[/tex]

[tex]V_s[/tex] = volume of solution in ml= [tex]\frac{\text {mass of solutuion}}{\text {density of solution}}=\frac{100g}{1.01g/ml}=99.0ml[/tex]

Now put all the given values in the formula of molarity, we get

[tex]Molarity=\frac{0.088moles\times 1000}{99.0ml}=0.88mole/L[/tex]

Thus molarity of this solution is 0.88 M

Answer:

The formula is Be(NO3)2*3H2O. The number of waters in the hydrate is 3

Explanation:

Step 1: Data given

Mass of the hydrate sample = 3.41 grams

Mass after heating = 2.43 grams

Step 2: Calculate mass of water

After heating, all the water is gone. So the mass of water can be calculated by

Mass of hydrate before heating - mass after heating

Mass of water = 3.41 -2.43 = 0.98 grams

Step 2 : Calculate moles H2O

Moles H2O = mass H2O / molar mass H2O

Moles H2O = 0.98 grams / 18.02 g/mol

Moles H2O = 0.054 moles

Step 3: Calculate moles Be(NO3)2

Moles Be(NO3)2 = 2.43 grams / 133.02

Moles Be(NO3)2 = 0.0183 moles

Step 4: Calculate molecules water

Molecules H2O = 0.054 moles / 0.0183 moles

Molecules H2O = 3.0

The formula is Be(NO3)2*3H2O. The number of waters in the hydrate is 3

The number of waters of hydration in the hydrate is 3.

Based on the given information,  

• The formula of the beryllium nitrate hydrate is Be(NO3)2⋅xH2O.  

• The weight of the sample is 3.41 grams.

• The weight of the sample after heating of the sample is 2.43 grams {weight of Be(NO3)2}.  

Now, the mass of xH2O will be,  

= 3.41 g - 2.43 g  

= 0.98 grams

The moles of Be(NO3)2 will be calculated as,  

Mass of Be(NO3)2/Molecular weight of Be(NO3)2 = 2.43 g/133 = 0.01827 moles

The moles of xH2O = 0.98/18 = 0.05444 moles

Now the value of x will be,  

= 0.05444/0.01827  

= 3

Thus, the number of waters of hydration in the hydrate is 3.

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Answer:

0.76

Explanation:

N = No(0.5)^t/t1/2

N/No = (0.5)^t/t1/2

t = 2300 years, t1/2 = 5730 years

N/No = (0.5)^2300/5730 = 0.5^0.401 = 0.76

Answer:

a) Thymol represents a minor concentration.

b) Of the ingredients listed, the ones that qualify as being contained in trace amounts are the ones with concentrations less than 0.01%, 100ppm or 0.1mg/L. And in this question, none of the given concentrations is less than 0.01%. Maybe one of the other constituents whose concentrations weren't given is in trace amounts.

c) The mouthwash brought for analysis is the sample.

d) Thymol is the analyte.

e) The matrix is every component of the mouthwash sample apart from the analyte constituent, thymol.

Menthol, eucalyptol, methyl salicylate, caramel, water, alcohol in addition to sodium benzoate, color, poloxamer 407 and benzoic acid are all member substances of the matrix.

f) The blank can contain every constituent apart from the analyte constituent, thymol.

Explanation:

a) The composition range for major, minor and trace components are given thus.

1-100% Major

0.01-1% Minor

1 ppb to 100ppm Trace

<1ppb ultra trace

In Chemistry terms,

•Major constituents: Substances in concentrations over 1mg/L

•Minor constituents: Substances in concentrations between 1mg/L and 0.1 mg/L

•Trace constituents: Substances in concentrations under 0.1 mg/L

Hence, thymol with a concentration of 0.064% represents a minor constituent.

b) Of the ingredients listed, the ones that qualify as being contained in trace amounts are the ones with concentrations less than 0.01%, 100ppm or 0.1mg/L. And in this question, none of the given concentrations is less than 0.01%. Maybe one of the other constituents whose concentrations weren't given is in trace amounts.

c) In chemical analysis or Analytical chemistry, a sample is a portion of material selected from a larger quantity of material.

Therefore, for this question, the mouthwash is the sample.

d) In Analytical Chemistry, an analyte or analyte component is a substance or chemical constituent that is of interest in an analytical procedure.

For this question, the analyte is thymol; the constituent the analyser is interested in.

e) In chemical analysis, matrix refers to the components of a sample other than the analyte of interest.

Therefore, the matrix is every component of the mouthwash sample apart from the analyte constituent, thymol.

Menthol, eucalyptol, methyl salicylate, caramel, water, alcohol in addition to sodium benzoate, color, poloxamer 407 and benzoic acid are all member substances of the matrix.

f) A blank solution is a solution containing little to no analyte of interest, so, the blank can contain every constituent apart from thymol.

Hope this helps!

Answer:

[tex]7.3\times 10^2\ days[/tex]

Explanation:

Given that:

Half life = [tex]9.8\times 10^3[/tex] days

[tex]t_{1/2}=\frac{\ln2}{k}[/tex]

Where, k is rate constant

So,  

[tex]k=\frac{\ln2}{t_{1/2}}[/tex]

[tex]k=\frac{\ln2}{9.8\times 10^3}\ days^{-1}[/tex]

The rate constant, k = 0.00007 days⁻¹

Using integrated rate law for first order kinetics as:

[tex][A_t]=[A_0]e^{-kt}[/tex]

Where,  

[tex][A_t][/tex] is the concentration at time t

[tex][A_0][/tex] is the initial concentration

Given:

5 % is lost which means that 0.05 of [tex][A_0][/tex] is decomposed. So,

[tex]\frac {[A_t]}{[A_0]}[/tex] = 1 - 0.05 = 0.95

t = ?

[tex]\frac {[A_t]}{[A_0]}=e^{-k\times t}[/tex]

[tex]0.95=e^{-0.00007\times t}[/tex]

t = 732.76 days = [tex]7.3\times 10^2\ days[/tex]

It will take approximately [tex]\( 7.25 \times 10^2 \)[/tex] days for benzoyl peroxide to lose 5% of its potency when refrigerated.

To determine the time, it takes for benzoyl peroxide to lose 5% of its potency, we can use the first-order reaction kinetics formula:

[tex]\[ N(t) = N_0 \times e^{-kt} \][/tex]

where:

[tex]\( N(t) \)[/tex] is the amount of substance remaining after time [tex]\( t \)[/tex],

[tex]\( N_0 \)[/tex] is the initial amount of substance,

[tex]\( k \)[/tex] is the rate constant,

[tex]\( t \)[/tex] is the time.

The half-life [tex]\( t_{1/2} \)[/tex] is related to the rate constant [tex]\( k \)[/tex] by the equation:

[tex]\[ t_{1/2} = \frac{\ln(2)}{k} \][/tex]

Given the half-life [tex]\( t_{1/2} = 9.8 \times 10^3 \)[/tex] days, we can solve for [tex]\( k \)[/tex]:

[tex]\[ k = \frac{\ln(2)}{t_{1/2}} = \frac{\ln(2)}{9.8 \times 10^3} \][/tex]

Now, we want to find the time [tex]\( t \)[/tex] when 95% of the substance remains, so [tex]\( N(t) = 0.95N_0 \)[/tex]. Plugging this into the first-order reaction formula:

[tex]\[ 0.95N_0 = N_0 \times e^{-kt} \][/tex]

[tex]\[ 0.95 = e^{-kt} \][/tex]

Taking the natural logarithm of both sides:

[tex]\[ \ln(0.95) = -kt \][/tex]

Solving for [tex]\( t \)[/tex]:

[tex]\[ t = -\frac{\ln(0.95)}{k} \][/tex]

Substituting [tex]\( k \)[/tex] with the expression involving the half-life:

[tex]\[ t = -\frac{\ln(0.95)}{\ln(2)/t_{1/2}} \][/tex]

[tex]\[ t = -\frac{\ln(0.95) \cdot t_{1/2}}{\ln(2)} \][/tex]

[tex]\[ t = -\frac{\ln(0.95) \cdot 9.8 \times 10^3}{\ln(2)} \][/tex]

[tex]\[ t \approx -\frac{\ln(0.95) \cdot 9.8 \times 10^3}{0.693} \][/tex]

[tex]\[ t \approx -\frac{-0.051293 \cdot 9.8 \times 10^3}{0.693} \][/tex]

[tex]\[ t \approx \frac{0.5027 \times 10^3}{0.693} \][/tex]

[tex]\[ t \approx 7.25 \times 10^2 \][/tex]

Answer:

The volume of NaOH required is 80mL

pH of the resulting solution is 7 since reaction produces a normal salt

Explanation:

First, we generate a balanced equation for the reaction as shown below:

NaOH + HCl —> NaCl + H2O

From the equation,

nA = 1

nB = 1

From the question,

Ma = 2.0 mol/L

Va = 20 mL

Mb = 0.5 mol/L

Vb =?

MaVa / MbVb = nA /nB

2x20 / 0.5 x Vb = 1

0.5 x Vb = 2 x 20

Divide both side by 0.5

Vb = (2 x 20)/0.5

Vb = 80mL

The volume of NaOH required is 80mL

Since the reaction involves a strong acid and a strong base, a normal salt solution will be produced which is neutral to litmus paper. So since the salt solution is neutral to litmus paper then the pH of the resulting solution is 7

The pH of a solution prepared by mixing HCl and NaOH, we need to determine the concentration of the resulting solution. Since HCl and NaOH react in a 1:1 ratio to form water and salt (NaCl), we can use the concept of neutralization to find the resulting concentration.

Moles of HCl = volume (in L) × molarity = 0.020 L × 0.30 mol/L = 0.006 mol

Moles of NaOH = volume (in L) × molarity = 0.015 L × 0.60 mol/L = 0.009 mol

Since HCl and NaOH react in a 1:1 ratio, the limiting reagent is HCl. This means that all of the HCl will react with NaOH, and we will have an excess of NaOH. Therefore, the number of moles of HCl remaining will be zero.

Total volume = 20.0 mL + 15.0 mL = 35.0 mL = 0.035 L

Concentration of resulting solution = (Moles of HCl + Moles of NaOH) / Total volume

= (0.006 mol + 0.009 mol) / 0.035 L

= 0.429 mol/L

pH = -log[H+]

pH = -log(0.429)

≈ 0.366

Therefore, the pH of the solution prepared by mixing 20.0 mL of 0.30 M HCl with 15.0 mL of 0.60 M NaOH is approximately 0.366.

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Explanation:

Electron affinity of 7th group elements whose electron affinity of their anionic forms ( EA2) are higher than the electron affinity of their neutral form(EA1). This is because their anionic forms are more stable than their neutral form. If the reaction is endothermic ,change in the energy is negative this means that electron affinity is positive. And vice versa for negative electron affinity.

The EA2 of oxygen is positive because energy is required to add an electron to a small space that is already dense with negative charge. The EA2 of sulfur is also positive.

The electron affinity refers to the energy evolved when one mole of electrons is added to an atom. We know that the energy required to add the second electron to oxygen is positive rather than negative implying that the process is endothermic.

This is because, energy is required to add an electron to a small space that is already dense with negative charge. The EA2 of sulfur is also positive.

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Answer:HCl(aq) + NaOH (aq) -> NaCl (aq) + H2O(l)

Explanation:

The reaction of HCl and NaOH is a neutralization reaction. When an acid and a base react salt and water is produced.

HCl(aq) + NaOH (aq) ----> NaCl (aq) + H2O(l)

Because the reactants and products are ionic compounds, they exist as ions in a solution.

H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) -------> Na+(aq) + Cl-(aq) + H2O(l)

The aim of a neutralization reaction is the formation of water. Na+(aq) + Cl-(aq) remain the same on both the reactants side and the products side, the net ionic reaction will be

H+(aq) + OH-(aq) -------> H2O(l)

The reaction of HCl and NaOH forms water in a neutralization reaction. The net ionic equation, H⁺ + OH⁻ → H₂O, emphasizes the essential combination of ions to produce water.

The reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH) is a neutralization reaction, resulting in the formation of water and salt. In molecular form, the equation is written as:

HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

Considering the ionic nature of these compounds in solution, they exist as ions. The complete ionic equation includes the dissociation of each compound into its constituent ions:

H⁺(aq) + Cl⁻(aq) + Na⁺(aq) + OH⁻(aq) → Na⁺(aq) + Cl⁻(aq) + H₂O(l)

Sodium cations (Na⁺) and chloride anions (Cl⁻) remain unchanged on both sides, acting as spectator ions. Eliminating these spectator ions yields the net ionic equation:

H⁺(aq) + OH⁻(aq) → H₂O(l)

This simplified representation emphasizes the essential components of the neutralization reaction, focusing on the combination of hydrogen ions (H⁺) from the acid and hydroxide ions (OH⁻) from the base to form water. The net ionic equation succinctly captures the core chemical transformation occurring in the neutralization process.

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Answer:

C

Explanation:

The aqueous solution C has three ions where as b 2 ions and A no Ionization

Answer:

F

Explanation:

simple subtration

In the manufacture of nitric acid by the oxidation of ammonia, the first product is nitric oxide. The nitric oxide is then oxidized to nitrogen dioxide. The standard reaction enthalpy for the reaction is −114.1 kJ/mol rxn.

A thermodynamic system's enthalpy is the sum of its internal energy and the product of its pressure and volume. It is a constant-pressure state function that is used in many measurements in chemical, biological, and physical systems.

Enthalpy is significant because it tells us how much heat is present in a system (energy). Heat is important because it allows us to generate valuable work. An enthalpy shift indicates how much enthalpy was lost or gained during a chemical reaction, with enthalpy referring to the heat energy of the system.

During chemical reactions, atom bonds can dissolve, reform, or both in order to absorb or release energy. The heat absorbed or emitted by a device under constant pressure is referred to as enthalpy.

Thus, option F is correct.

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Answer: The molecular weight of unknown non-electrolyte is 61.75 g/mol

Explanation:

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

[tex]\Delta T_f=\text{Freezing point of benzophenone}-\text{Freezing point of solution}[/tex]

To calculate the depression in freezing point, we use the equation:

[tex]\Delta T_f=iK_fm[/tex]

or,

[tex]\text{Freezing point of benzophenone}-\text{Freezing point of solution}=iK_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}[/tex]where,

i = Vant hoff factor = 1 (for non-electrolyte)

[tex]K_f[/tex] = molal freezing point depression constant = 9.80°C/m

[tex]m_{solute}[/tex] = Given mass of unknown non-electrolyte = 1.0223 g

[tex]M_{solute}[/tex] = Molar mass of unknown non-electrolyte = ?

[tex]W_{solvent}[/tex] = Mass of solvent (benzophenone) = 10.2685 g

Putting values in above equation, we get:

[tex]47.5-31.7=1\times 9.80\times \frac{1.0223\times 1000}{M_{solute}\times 10.2685}\\\\M_{solute}=\frac{1\times 9.80\times 1.0223\times 1000}{15.8\times 10.2685}=61.75g/mol[/tex]

Hence, the molecular weight of unknown non-electrolyte is 61.75 g/mol

The molecular weight of the unknown nonelectrolyte dissolved in benzophenone can be calculated using the formula for freezing point depression and the information provided in the question. After performing the necessary calculations, the molecular weight of the unknown compound is found to be 331 g/mol.

To calculate the molecular weight of the unknown compound, we use the formula for freezing point depression:

ΔT = iKfm

Where ΔT is the change in temperature, i is the van't Hoff factor which is 1 for nonelectrolytes, Kf is the cryoscopic constant for benzophenone, which, in this case, is 5.12°C kg/mol, and m is the molality of the solution. The change in temperature is the difference between the freezing points of pure benzophenone and the solution: 47.5°C - 31.7°C = 15.8°C.

The molality can be calculated by rearranging the formula:

m = ΔT / (iKf) = 15.8 / (1 * 5.12) = 3.086 mol/kg

Since we are given grams and want to convert to kilograms:

1.0223g of unknown compound / molar mass (unknown) = 3.086 mol/kg

Rearrange to find the molar mass (molecular weight) of the compound:

molar mass (unknown) = 1.0223g / 3.086 kg = 0.331 kg/mol = 331 g/mol

Hence the molecular weight of the unknown nonelectrolyte dissolved in benzophenone is 331 g/mol.

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Answer:

0.0634 M.

Explanation:

Equation of the neutralisation reaction:

CH3COOH + NaOH--> CH3COONa + H2O

Number of moles = molar concentration * volume

= 0.1002 * 0.01581

= 0.00158 mol.

By stoichiometry,

1 mole of acetic acid reacts with 1 mole of NaOH

Number of moles of acetic acid = 0.00158 mol.

Concentration in 2nd dilution = moles/ volume

= 0.00158/0.25

= 0.00634 M

At 25 ml,

Concentration =

C1 * V1 = C2 * V2

= (0.00634 * 0.25)/0.025

= 0.0634 M.

Given:

Neutralization reaction's equation:

Now,

The number of moles will be:

= [tex]Molar \ concentration\times Volume[/tex]

= [tex]0.1002\times 0.01581[/tex]

= [tex]0.00158 \ mol[/tex]

and,

The concentration of second dilution will be:

= [tex]\frac{Moles}{Volume}[/tex]

= [tex]\frac{0.00158}{0.25}[/tex]

= [tex]0.00634 \ M[/tex]

hence,

The concentration at 25 mL will be:

→ [tex]C_1\times V_1 = C_2\times V_2[/tex]

or,

→         [tex]C_2 = \frac{C_1\times V_1}{V_2}[/tex]

                 [tex]= \frac{0.00634\times 0.25}{0.025}[/tex]

                 [tex]= 0.0634 \ M[/tex]

Thus the above response is right.  

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Answer: The number of seconds in a solar year is [tex]3.2\times 10^7s[/tex]

Explanation:

We are given some conversion factors:

1 minute = 60 seconds

1 hour = 60 minutes

1 solar day = 24 hours

1 solar year = 365.24 solar days

Calculating the number of seconds in 1 solar year by using the conversion factors, we get:

[tex]\Rightarrow (\frac{60s}{1min})\times (\frac{60min}{1hr})\times (\frac{24hr}{1\text{solar day}})\times (\frac{365.24\text{solar days}}{1\text{ solar year}})\\\\\Rightarrow (\frac{31556736s}{1\text{ solar year}})[/tex]

There are 31556736 seconds in 1 solar year.

Hence, the number of seconds in a solar year is [tex]3.2\times 10^7s[/tex]

Answer:

238,485 Joules

Explanation:

The amount of energy required is a summation of heat of fusion, capacity and vaporization.

Q = mLf + mC∆T + mLv = m(Lf + C∆T + Lv)

m (mass of water) = 75 g

Lf (specific latent heat of fusion of water) = 336 J/g

C (specific heat capacity of water) = 4.2 J/g°C

∆T = T2 - T1 = 119 - (-20) = 119+20 = 139°C

Lv (specific latent heat of vaporization of water) = 2,260 J/g

Q = 75(336 + 4.2×139 + 2260) = 75(336 + 583.8 + 2260) = 75(3179.8) = 238,485 J