Diet cola drinks have a pH of about 3.0, while milk has a pHof
about 7.0. How many times greater is theH3O+
concentration in diet cola than inmilk?
A. 2.3 times higher in diet cola than in milk.
B. 400 times higher in diet cola than in milk.
C. 0.43 times higher in diet cola than in milk.
D. 1,000 times higher in diet cola than in milk.
E. 10,000 times higher in diet cola than in milk.

Explanation:

The given data is as follows.

         Initial concentration = 0.15 mg/ml,        

        Final concentration = 0.1 mg/ml, (as [tex]\frac{0.1 microgram}{1 microliter} \times \frac{10^{-3}}{1 microgram} \times \frac{1 microliter}{10^{-3}ml}[/tex])

        Final volume = [tex]100 microliter \times \frac{10^{-3} ml}{1 microliter}[/tex] = 0.1 ml

According to the dilution formula we get the following.

              [tex]C_{i} \times V_{i} = C_{f} \times V_{f}[/tex]

or,                        [tex]V_{i}[/tex] = [tex]\frac{C_{f} \times V_{f}}{C_{i}}[/tex]

Putting the given values into the above formula we get the following.

                   [tex]V_{i}[/tex] = [tex]\frac{C_{f} \times V_{f}}{C_{i}}[/tex]

                               = [tex]\frac{0.1 mg/ml \times 0.1 ml}{0.15 mg/ml}[/tex]

                               = 0.0667 ml

                               = [tex]6.67 \times 10^{-2} ml \times \frac{1 microliter}{10^{-3} ml}[/tex]                                

                              = 66.7 microliter

This means that volume of protein stock which is required is 66.7 ml. Hence, calculate the volume of water required as follows.

                  Volume of water required = Total volume - volume of protein stock

                                                     = (100 - 66.7)  microliter

                                                     = 33.3 microliter

Thus, we can conclude that we need 33.3 microliter of water and 66.7 microliter of protein.

Answer : The chemical formula of a compound is, [tex]Ga_2O_3[/tex]

Solution :  Given,

Mass of gallium = 1.25 g

Mass of gallium oxide = 1.68 g

Mass of oxygen = Mass of gallium oxide - Mass of gallium

Mass of oxygen = 1.68 - 1.25 = 0.43 g

Molar mass of Ga = 69.72 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of Ga = [tex]\frac{\text{ given mass of Ga}}{\text{ molar mass of Ga}}= \frac{1.25g}{69.72g/mole}=0.0179moles[/tex]

Moles of O = [tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{0.43g}{16g/mole}=0.027moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Ga = [tex]\frac{0.0179}{0.0179}=1[/tex]

For O = [tex]\frac{0.027}{0.0179}=1.5[/tex]

The ratio of Ga : O = 1 : 1.5

To make in whole number we multiple ratio by 2, we get:

The ratio of Ga : O = 2 : 3

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = [tex]Ga_2O_3[/tex]

The empirical formula weight = 2(69.72) + 3(16) = 187.44 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

[tex]n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}[/tex]

[tex]n=\frac{187.44}{187.44}=1[/tex]

Molecular formula = [tex](Ga_2O_3)_n=(Ga_2O_3)_1=Ga_2O_3[/tex]

Therefore, the chemical of the compound is, [tex]Ga_2O_3[/tex]

The empirical formula of gallium oxide, formed by the reaction of gallium with excess oxygen, is Ga₂O₃.

To determine the chemical formula of gallium oxide formed, follow these steps:

            Mass of O = 1.68 g (mass of Ga oxide) - 1.25 g (mass of Ga) = 0.43 g

           Moles of Ga = 1.25 g / 69.72 g/mol (molar mass of Ga) = 0.0179 moles

           Moles of O = 0.43 g / 16.00 g/mol (molar mass of O) = 0.0269 moles

             Ratio of Ga to O = 0.0179 : 0.0269 ≈ 2:3

The given molar mass of the gallium oxide (187.44 g/mol) confirms the empirical formula Ga₂O₃.

Answer:

Hydrogen bonded materials

Explanation:

Metals have metallic bonds, polymers usually have covalent bonds and ceramic materials have covalent and ionic bonds; on the other hand, hydrogen bonds are a type of bonding characterized for have Van der Waals interactions, a type of intermolecular interactions, so the correct answer is Van der Waals interactions.

Answer: Distillation is a process in which we use the boiling point, condensation of the substances to separate them from each other, it is a physical separation not the chemical because physical property of substance is used in it.

Now, the distinction in between a laboratory distillation and an industrial distillation is that, the distillation process in laboratory distillation works in batches, while in the industrial distillation it occurs in continuous manner, industrial distillation is the large scale distillation.

Final answer:

The main difference between laboratory scale distillation and industrial distillation lies in the scale and complexity of the process.

Explanation:

The main difference between laboratory scale distillation and industrial distillation lies in the scale and complexity of the process. Laboratory scale distillation is typically used for smaller quantities and is carried out using simple apparatus. It is commonly used for research, quality control, and small-scale production.

On the other hand, industrial distillation is conducted on a much larger scale and involves more complex equipment. It is used in industries such as oil refining and chemical manufacturing, where large quantities of materials need to be processed. Industrial distillation may include fractionating columns and other specialized equipment to separate components from mixtures.

Answer:

a) When the 4 tanks operate in parallel the retention time is 1.26 hours.  

b) If the tanks are in series, the retention time would be 0.31 hours

Explanation:

The plant has 4 tanks, each tank has a volume V = 600 [tex]m_3[/tex]. The total flow to the plant is Ft = 12 MGD (Millions of gallons per day)

When we use the tanks in parallel, it means that the total flow will be divided in the total number of tanks. F1 will be the flow of each tank.

Firstly, we should convert the MGD to [tex]m_3 /day[/tex]. In that sense, we can calculate the retention time using the tank volume in [tex]m_3[/tex].

[tex]Ft =12 \frac{MG}{day}  (\frac{1*10^6 gall}{1MG} ) (\frac{3.78 L}{1 gall} ) (\frac{1 dm^3}{1L} ) (\frac{1 m^3}{10^3 dm^3} ) = 45360 \frac{m^3}{day}[/tex]

After that, we should divide the total flow by four, because we have four tanks.

[tex]F1 = Ft/4 =(45360 \frac{m^3}{d} )/4 = 11 340 \frac{m^3}{d}[/tex]

To calculate the retention time we divide the total volume V by the flow of each tank F1.

[tex]t1 =\frac{V1}{F1} = \frac{600 m^3}{11340  \frac{m^3}{d} }  = 0.0529 day\\t1 = 0.0529 d (\frac{24 h}{1d} ) = 1.26 h[/tex]

After converting t1 to hours we found that the retention time when the four reactors are in parallel is 1.26 hours.

b)

If the four reactors were working in series, the entire flow goes first through one tank, then the second and so on. It means the total flow will be the flow of each tank.

In that order of ideas, the flow for reactors in series will be F2, and will have the same value of F0.

F2 = F0

To calculate the retention time t2 we divide the total volume V by the flow of each tank F2.

[tex]t2 =\frac{V1}{F2} = \frac{600 m^3}{45360  \frac{m^3}{d} }  = 0.01 day\\t2 = 0.01  d (\frac{24 h}{1d} ) = 0.31 h[/tex]

After converting t2 to hours we found that the retention time when the four reactors are in series is 0.31 hours.

a) Retention time in each settling tank for parallel operation is approximately 317.0 hours.

b) Retention time in each settling tank for series operation is also approximately 317.0 hours.

To find the retention time in each settling tank, we'll use the formula:

[tex]\[ \text{Retention Time} = \frac{\text{Volume of Tank}}{\text{Flow Rate}} \][/tex]

Where:

- Volume of Tank is the volume of each settling tank.

- Flow Rate is the total flow rate.

We'll first convert the flow rate to cubic meters per day (m\(^3\)/day) to match the units of the tank volume:

[tex]\[ 1 \, \text{MGD} = 1 \, \text{million gallons} \times \frac{3.785 \times 10^3 \, \text{liters}}{1 \, \text{m}^3} \times \frac{1 \, \text{day}}{24 \, \text{hours}} \]\[ = 1 \, \text{MGD} \times \frac{3.785 \times 10^3 \, \text{liters}}{1 \, \text{m}^3} \times \frac{1 \, \text{day}}{24 \, \text{hours}} \times \frac{10^{-6} \, \text{m}^3}{1 \, \text{liter}} \]\[ = \frac{3.785 \times 10^3}{24 \times 10^6} \, \text{m}^3/\text{hour} \]\[ = 0.1577 \, \text{m}^3/\text{hour} \][/tex]

Now we can calculate the retention time for both scenarios:

a) For parallel operation:

[tex]\[ \text{Retention Time (Parallel)} = \frac{V_{\text{tank}}}{Q} = \frac{600 \, \text{m}^3}{1.8924 \, \text{m}^3/\text{hour}} \]\[ \text{Retention Time (Parallel)} \approx 317.0 \, \text{hours} \][/tex]

b) For series operation:

In series, the total flow rate remains the same, but the volume is effectively multiplied by the number of tanks in series. So, the retention time for each tank remains the same as in the parallel case:

[tex]\[ \text{Retention Time (Series)} = \frac{V_{\text{tank}}}{Q} = \frac{600 \, \text{m}^3}{1.8924 \, \text{m}^3/\text{hour}} \]\[ \text{Retention Time (Series)} \approx 317.0 \, \text{hours} \][/tex]

Therefore, the retention time in each settling tank is approximately 317.0 hours for both parallel and series operation.

Answer:

Option d, Amount left = 16 μg

Explanation:

Half life = 88 days

Initial concentration = 128 μg

[tex]No.\ of\ half\ life=\frac{Total\ days}{t_{1/2}}[/tex]

[tex]No.\ of\ half\ life=\frac{264}{88} = 3[/tex]

Amount of material left after n half life = [tex]\frac{Amount\ of\ starting\ material}{2^n}[/tex]

Where, n = No. of half life

Amount of material left = [tex]\frac{128}{2^3}[/tex]

                                       = [tex]\frac{128}{8} = 16[/tex]μg

SO, option d is correct

Answer : The correct option is, (b) 0.087

Explanation :

The formula used for relative saturation is:

[tex]\text{Relative saturation}=\frac{P_A}{P_A^o}[/tex]

where,

[tex]P_A[/tex] = partial pressure of ethyl acetate

[tex]P_A^o[/tex] = vapor pressure of ethyl acetate

Given:

Relative saturation = 50 % = 0.5

Vapor pressure of ethyl acetate = 16 kPa

Now put all the given values in the above formula, we get:

[tex]0.5=\frac{P_A}{16kPa}[/tex]

[tex]P_A=8kPa[/tex]

Now we have to calculate the molar saturation.

The formula used for molar saturation is:

[tex]\text{Molar saturation}=\frac{P_{vapor}}{P_{\text{vapor free}}}[/tex]

and,

P(vapor free) = Total pressure - Vapor pressure

P(vapor) = [tex]P_A[/tex] = 8 kPa

So,

P(vapor free) = 100 kPa - 8 kPa = 92 kPa

The molar saturation will be:

[tex]\text{Molar saturation}=\frac{P_{vapor}}{P_{\text{vapor free}}}[/tex]

[tex]\text{Molar saturation}=\frac{8kPa}{92kPa}=0.087[/tex]

Therefore, the molar saturation is 0.087

Answer:

133g

Explanation:

If one litre of ocean water contains 35.06g of salt, all we have to to to get the mass of salt in 3.79 L of ocean water is multiply 35.06g/L by 3.79 L.

35.06g/L × 3.79 L = 133g

The amount of salt in 3.79 L of ocean water can be found by multiplying the volume of the water by the constant ratio of 35.06 g of salt per liter. This gives 132.8774 grams, which when we round to 4 significant figures is equivalent to 132.9 g.

To find the amount of salt in 3.79 L of ocean water, we use the given ratio of salt to water: 1 liter of ocean water contains 35.06 grams of salt. So to find the amount of salt in any given volume of ocean water, we just multiply the volume (in liters) by this constant ratio of 35.06 grams of salt per liter.

Applying this to the volume of 3.79 L we get: 3.79 L * 35.06 g/L = 132.8774 grams. Since we should express our answer using the correct number of significant figures (4 in this case, because there are 4 significant figures in the 3.79 L volume measurement), the answer is rounded to 132.9 g.

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Answer:

c)The Boltzmann distribution is not valid at very low temperatures.

Explanation:

Option cis correct because the Boltzman distribution can be applied when the temperature is high enough or the particle density is low enough to render quantum effects negligible.

The Boltzman distribution in statistics describe the distribution of particles over different energy states when there is thermal equilibrium (that is why option b is INCORRECT).

Answer:

The alkaline earth metals are barium and calcium.

Explanation:

Alkaline earth metals is the denomination given to the six elements that occupy Group 2 in the Periodic Table, therefore they have an electron configuration that ends in the orbital s².

From the list: barium and calcium are alkaline earth metals. On the other hand, lithium and sodium occupy Group 1, and aluminum occupies Group 3.

Among the listed elements, barium and calcium are the alkaline earth metals, which belong to Group 2 of the periodic table.

The question concerns the identification of alkaline earth metals, which are elements in Group 2 of the periodic table. From the given options, the alkaline earth metals are barium and calcium. These elements are known for being shiny, silvery-white, and somewhat reactive. They exhibit distinctive properties such as higher ionization energies than alkali metals and the ability to lose electrons to form compounds with a +2 oxidation state.

Answer: The concentration of cow's milk after 5 years is 10691 Bq/L

Explanation:

All the radioactive reactions follow first order kinetics.

The equation used to calculate rate constant from given half life for first order kinetics:

[tex]t_{1/2}=\frac{0.693}{k}[/tex]

We are given:

[tex]t_{1/2}=30yrs[/tex]

Putting values in above equation, we get:

[tex]k=\frac{0.693}{30}=0.0231yr^{-1}[/tex]

The equation used to calculate time period follows:

[tex]N=N_o\times e^{-k\times t}[/tex]

where,

[tex]N_o[/tex] = initial concentration of Cow's milk = 12000 Bq/L

N = Concentration of cow's milk after 5 years = ?

t = time = 5 years

k = rate constant = [tex]0.0231yr^{-1}[/tex]

Putting values in above equation, we get:

[tex]N=12000Bq/L\times e^{-(0.0231yr^{-1}\times 5yr)}\\\\N_o=10691Bq/L[/tex]

Hence, the concentration of cow's milk after 5 years is 10691 Bq/L

Answer:

Take account the molar mass of this compound (first of all u should know the formula, S2F10). As the molar mass is 254,1 g/mol you will know that in 1 mol, you have 254,1 g so make a rule of three to solve it. If we find 254,1 g  of S2F10 in 1 mol, 2,45 g of it are in 9,64 *10^-3 moles. That is the right number.

Explanation:

Final answer:

The number 0.70894 is written as 709 × 10⁻³ in Engineering Notation with 3 significant figures.

Explanation:

To write the number 0.70894 in Engineering Notation with 3 significant figures, we first identify the significant figures in the number. Since leading zeros are not significant, the significant figures are 708. We then adjust the number to have a multiple of three as its exponent for the base unit, which is standard in Engineering Notation. The number can be written as 709 (rounded up from 708.94 to maintain 3 significant figures) multiplied by 10 to the power of -3. Therefore, the number in Engineering Notation is 709 × 10⁻³.

Answer:

The rate constant of the reaction is [tex]1.0185\times 10^{-2} dm^3/ mol s[/tex].

[tex]1.31\times 10^3 s[/tex] is the half life of the reactant.

Explanation:

A+B → P

Integrated rate law for second order kinetics is given by:

[tex]\frac{1}{A}=kt+\frac{1}{A_0}[/tex]

A = concentration left after time t = [tex]0.02 mol /dm^3[/tex]

[tex]A_o[/tex] = Initial concentration = [tex]0.075 mol /dm^3[/tex]

t = 1 hour = 3600 seconds

[tex]\frac{1}{0.02 mol /dm^3}=k\times 3600 s+\frac{1}{0.075 mol /dm^3}[/tex]

[tex]k=\frac{1}{3600 s}\times (\frac{1}{0.02 mol /dm^3}-\frac{1}{0.075 mol /dm^3)}[/tex]

[tex]k = 1.0185\times 10^{-2} dm^3/ mol s[/tex]

Half life for second order kinetics is given by:

[tex]t_{\frac{1}{2}}=\frac{1}{k\times a_0}[/tex]

[tex]t_{\frac{1}{2}}=\frac{1}{1.0185\times 10^{-2} dm^3/ mol s\times 0.075 mol /dm^3}=1.31\times 10^3 s[/tex]

The rate constant of the reaction is [tex]1.0185\times 10^{-2} dm^3/ mol s[/tex].

[tex]1.31\times 10^3 s[/tex] is the half life of the reactant.

Explanation:

The given data is as follows.

           Volume = 1 L,    Concentration of Ca = 5 ppm or 5 mg/L

As 1 mg = 0.001 g so, 5 mg /L will be equal to 0.005 g/l. Molar mass of calcium is 40.078 g/mol.

Hence, calculate molarity of calcium as follows.

           Molarity of Ca = [tex]\frac{\text{given concentration}}{\text{molar mass}}[/tex]

                                  = [tex]\frac{0.005 g/l}{40.078 g/mol}[/tex]

         Molarity of Ca = [tex]1.25 \times 10^{-4} M[/tex]

Hence, molarity of [tex]CaCl_{2}[/tex] is [tex]1.25 \times 10^{-4} M[/tex]. Since, volume is same so, moles of calcium chloride will be [tex]1.25 \times 10^{-4} mol[/tex].

Thus, we can conclude that mass of [tex]CaCl_{2}[/tex] will be as follows.

             [tex]1.25 \times 10^{-4} \times 110.984[/tex]       (molar mass of [tex]CaCl_{2}[/tex] = 110.984 g/mol)

               = 0.0138 g

Thus, we can conclude that mass of [tex]CaCl_{2}[/tex] is 0.0138 g.

Final answer:

To convert 1.248×1010 g to different units, one must use specific conversion factors for each unit, resulting in 1.248×107 kg, 1.248×104 Mg or metric tons, and 1.248×1013 mg, all while maintaining four significant figures.

Explanation:

To convert 1.248×1010 g to various units, we'll use the relationships between grams, kilograms, milligrams, and metric tons.

It's important to keep significant figures in mind, maintaining four significant figures through each conversion to ensure precision.

Answer:

4= Si

5=P

3= Al

6= S

2= Mg

Explanation:

4 → Si

Silicon is part of the carbon group and has 14 electrons, spread over 3 shells:

⇒ on the first shell (K) 2 electrons

⇒on the second shell (L) 8 electrons

⇒ on the third shell (M) the remaining electrons : 14-8-2 = 4

5 → P

phosphorus  is part of the nitrogen group and has 15 electrons, spread over 3 shells:

⇒ on the first shell (K) 2 electrons

⇒on the second shell (L) 8 electrons

⇒ on the third shell (M) the remaining electrons : 15-8-2 = 5

3 → Al

Aluminium is part of the Boron group and has 13 electrons, spread over 3 shells:

⇒ on the first shell (K) 2 electrons

⇒on the second shell (L) 8 electrons

⇒ on the third shell (M) the remaining electrons : 13-8-2 = 3

6 → S

Sulfur is part of the oxygen group and has 16 electrons, spread over 3 shells:

⇒ on the first shell (K) 2 electrons

⇒on the second shell (L) 8 electrons

⇒ on the third shell (M) the remaining electrons : 16-8-2 = 6

2 → Mg

Magnesium is part of the Beryllium or alkaline earth metals group and has 12 electrons, spread over 3 shells:

⇒ on the first shell (K) 2 electrons

⇒on the second shell (L) 8 electrons

⇒ on the third shell (M) the remaining electrons : 12-8-2 = 2

The completed pairs of element with the number of valence electrons are:

Valence electrons are the electrons in the outermost shell of an atom that can participate in chemical bonding. The number of valence electrons an element has determines its chemical properties.

For example, sodium (Na) has 1 valence electron, which it can donate to other atoms to form bonds. This is why sodium is a reactive metal.

Sulfur (S) has 6 valence electrons, which it can share with other atoms to form bonds. This is why sulfur can form a variety of compounds with other elements.

The number of valence electrons an element has can also be determined by looking at its periodic table group. Elements in the same group have the same number of valence electrons.

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Answer:

19°C

Explanation:

When you use a psychrometric chart the thing that you need to know is the value of a least two variables of your system, then you need to intercept these variables and then you can find the other variables seeing the others axis. In your case, you know the dry temperature and the percentage of humidity, so as you see in the attach image your temperature is between 15°C and 20°C. The only choice in that range is  19°C.

Answer : 3.4 mL of this suspension should be given to an infant weighing 15 lb.

Explanation :

First we have to convert weight of infant from 'lb' to 'kg'.

Conversion used :

As, 1 lb = 0.454 kg

So, 15 lb = [tex]\frac{15lb}{1lb}\times 0.454kg=6.81kg[/tex]

Now we have to calculate the recommended dose.

As, 1 kg weight recommended dose 10 mg

So, 6.81 kg weight recommended dose [tex]\frac{6.81kg}{1kg}\times 10mg=68.1mg[/tex]

Now we have to calculate the milliliter of ibuprofen suspension.

As, 100 mg dose present in 5.0 mL

So, 68.1 mg dose present in [tex]\frac{68.1mg}{100mg}\times 5.0mL=3.4mL[/tex]

Therefore, 3.4 mL of this suspension should be given to an infant weighing 15 lb.

Answer: The number of atoms of plutonium present in given number of moles are [tex]3.7\times 10^{24}[/tex]

Explanation:

We are given:

Number of moles of plutonium metal = 6.1 moles

According to mole concept:

1 mole of an element contains [tex]6.022\times 10^{23}[/tex] number of atoms.

So, 6.1 moles of plutonium will contain = [tex]6.1\times 6.022\times 10^{23}=3.7\times 10^{24}[/tex] number of plutonium atoms.

Hence, the number of atoms of plutonium present in given number of moles are [tex]3.7\times 10^{24}[/tex]

Final answer:

To determine the number of plutonium atoms in 6.1 moles, multiply 6.1 by Avogadro's number (6.022 x 10^23 atoms/mole), yielding 3.673 x 10^24 atoms of plutonium in scientific notation.

Explanation:

The question asks how many plutonium atoms are present in 6.1 moles of plutonium metal. To find this, we use Avogadro's number, which is 6.022 × 1023 atoms/mole. This means there are 6.022 × 1023 atoms of any element per mole of that element.

To calculate the total number of atoms in 6.1 moles of plutonium, you multiply the number of moles by Avogadro's number:

Number of atoms = 6.1 moles × 6.022 × 1023 atoms/mole

This calculation gives you the answer in scientific notation: 3.673 × 1024 atoms of Pu.

Answer:

Explanation:

To calculate pH you need to use Henderson-Hasselbalch formula:

pH = pka + log₁₀ [tex]\frac{[A^-]}{[HA]}[/tex]

Where HA is the acid concentration and A⁻ is the conjugate base concentration.

The equilibrium of acetic acid is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺ pka: 4,75

Where CH₃COOH is the acid and CH₃COO⁻ is the conjugate base.

Thus, Henderson-Hasselbalch formula for acetic acid equilibrium is:

pH = 4,75 + log₁₀ [tex]\frac{[CH_{3}COO^-]}{[CH_{3}COOH]}[/tex]

a) The pH is:

pH = 4,75 + log₁₀ [tex]\frac{[2 mol]}{[2 mol]}[/tex]

pH = 4,75

b) The pH is:

pH = 4,75 + log₁₀ [tex]\frac{[2 mol]}{[1mol]}[/tex]

pH = 5,05

I hope it helps!

Answer:

x = 130 Ms⁻¹

Explanation:

The equation is:

x = (2.4 x 10³ M⁻²s⁻¹)(0.38 M)³

The expression raised to the power of 3 is simplified:

x = (2.4 x 10³ M⁻²s⁻¹)(0.054872 M³)

The two values are multiplied together and units canceled:

x = 130 Ms⁻¹

Answer:

x = 131.69Ms⁻¹

Explanation:

x = (2.4×10³M⁻²s⁻¹) (0.38M)³

Upon expanding, we have;

x = (2.4 × 10³M⁻²s⁻¹) (0.38M) * (0.38M) * (0.38M)

x = (2.4 × 10³M⁻²s⁻¹) (0.054872 M³)

x = 2.4 * 0.054872 * 10³ * M⁻²s⁻¹ * M³

x = 0.13169 * 10³ * M⁻²s⁻¹ * M³

x = 131.69 * M⁽⁻²⁺³ ⁾* s⁻¹

x = 131.69 * Ms⁻¹

x = 131.69Ms⁻¹

Explanation:

The least number of significant figures in any number of the problem determines the number of significant figures in the answer.

The least precise number present after the decimal point determines the number of significant figures in the answer.

a) 12.3 mm × 15.853 mm :

[tex] 12.3 mm\times 15.853 mm :194.9919 mm^2\approx 195 mm^2[/tex]

Three significant figures.

b) 72.98 · 15.830 ml

[tex]72.98\times 15.830 ml=1,155.2734 ml\approx 1,155 mL[/tex]

Three significant figures.

c) 12.3 mm + 15.853 mm

[tex]12.3 mm + 15.853 mm = 28.153 mm\approx 28.1 mm[/tex]

Three significant figures.

d)  172.3 cm - 15.853 cm

[tex] 172.3 cm - 15.853 cm= 156.447 cm\approx 156.4 cm[/tex]

Four significant figures.

Final answer:

The student's Mathematics question involves calculating with significant figures. When multiplying or dividing values, the result is rounded to the number of significant figures in the least precise value, while addition and subtraction are rounded to match the least number of decimal places.

Explanation:

The subject of this question is Mathematics, specifically related to significant figures in calculations, which is a concept typically covered in High School mathematics or physics courses.

Report the answer with the correct number of significant figures:

(a) To multiply 12.3 mm by 15.853 mm, we should round our answer to the same number of significant figures as the number with the fewest significant figures. In this case, 12.3 has three significant figures. Therefore, the result should also be reported with three significant figures.

(b) When multiplying 72.98 by 15.830 ml, the number with the fewest significant figures is 72.98, which has four significant figures. The answer should be reported with four significant figures.

(c) When adding 12.3 mm to 15.853 mm, we look for the least precise value in terms of decimal places, which is 12.3 mm (it is precise to the tenths place). Therefore, our result should also be rounded to the tenths place.

(d) In the subtraction 172.3 cm - 15.853 cm - 12.5 cm, since 12.5 cm has the least number of decimal places (one decimal place), our final result should also be rounded to one decimal place.

When it comes to significant figures, the rule for multiplication and division is to match the result to the number with the fewest significant figures. For addition and subtraction, we round the result to the least number of decimal places in the values used.

Explanation:

Aldose is monosaccharide sugar in which the carbon backbone chain has carbonyl group on endmost carbon atom which corresponds to an aldehyde, and the hydroxyl groups are connected to all other carbon atoms.

Example - Glucose

Ketone is functional group with structure RC(=O)R', where the groups, R and R' can be variety of the carbon-containing substituents.

Example, Ethylmethylketone

Final answer:

The simplest aldose is glyceraldehyde, and the simplest ketone is acetone. These classifications are based on the presence of a carbonyl group, with aldoses having the group at the end of the chain and ketones within the chain. This naming follows IUPAC nomenclature rules with the suffixes -al for aldehydes and -one for ketones.

Explanation:

The simplest aldose is glyceraldehyde, which is a three-carbon aldehyde with the molecular formula C₃H₆O₃. It contains a carbonyl group (CHO) at the end of the carbon chain, making it an aldehyde. The simplest ketone is acetone (propanone), which has the molecular formula C₃H₆O. It contains a carbonyl group (CO) within the carbon chain, which classifies it as a ketone.

Both aldehydes and ketones contain a carbonyl group, a functional group with a carbon-oxygen double bond. The nomenclature rules for naming these compounds involve the use of the suffixes -al for aldehydes and -one for ketones. For instance, the name glyceraldehyde is derived from glycerol (the alcohol form), with the ending changed to -al to indicate it's an aldehyde, while the name acetone is derived from the root word for two carbon groups (acet-) with the ending -one to signify it's a ketone.

Answer:

The result was that H₂SO₄ is a strong electrolyte, NH₃ is a weak electrolyte and sugar is a non-electrolyte.

Explanation:

The experiment will show that the H₂SO₄ solution will light the lightbulb brightly, the NH₃ solution will light it weakly and the sugar solution will not light it at all.

This happens because to light the lightbulb we need free charges moving in the solution. Therefore, the substance that produce more ions will be the best electrolyte.

H₂SO₄ → 2H⁺ + SO₄²⁻

NH₃ + H₂O → NH₄⁺ + OH⁻

Sugar will not generate ions

Explanation:

It is given that 44.45 g of hydrocarbon gas produces 2649 kJ or [tex]2649 \times 10^{3} J[/tex] of heat.

Also here, 1 lb = 453.592 g.

Therefore, amount of energy released by 453.592 g of hydrocarbon gas will be calculated as follows.

              [tex]\frac{2649 \times 10^{3} J \times 453.592 g}{44.45 g}[/tex]

             = [tex]27.03 \times 10^{6} J[/tex]

It is known that 1 J = [tex]2.778 \times 10^{-7} Kwh[/tex].

Hence, [tex]27.03 \times 10^{6} J[/tex] = [tex]27.03 \times 10^{6} J \times 2.778 \times 10^{-7} Kwh/J[/tex]

                              = 7.508 Kwh

Thus, we can conclude that the combustion of exactly one pound of this hydrocarbon gas produce 7.508 Kwh energy.

Answer:

London dispersion forces and dipole-dipole interactions

Explanation:

Intermolecular forces are the interactive forces that are present between the molecules. These intermolecular forces can be attractive or repulsive.

Hydrogen bromide is a diatomic covalent polar molecule, having chemical formula: HBr. Since, hydrogen bromide is a polar molecule, it has a permanent dipole.

Therefore, the intermolecular forces present between the hydrogen bromide molecules are dipole-dipole interactions and the london dispersion forces.