Which statement is true about obligate anaerobes? View Available Hint(s) Which statement is true about obligate anaerobes? They obtain energy by oxidizing ferrous ions. They will use O2 if it is present, but may obtain energy by fermentation if needed. They use O2 for cellular respiration and cannot grow without it. They are poisoned by O2. They live exclusively by cellular respiration or by anaerobic respiration.

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:

49.95 g of HCl

Explanation:

Let's formulate the chemical equation involved in the process:

Ca(OH)2 + 2 HCl → CaCl2 + 2 H2O

This means that we need 1 mole of Calcium hydroxide to neutralize 2 moles of hydrochloric acid. From this, we calculate the quantity of HCl moles that would be neutralized by 0.685 moles of Ca(OH)2

1 mole Ca(OH)2 ---- 2 moles HCl

0.685 moles Ca(OH)2 ---- x = 1.37 moles HCl

Now that we know the quantity of HCl moles that would react, let's calculate the quantity of grams this moles represent:

1 mole of HCl ---- 36.46094 g

1.37 moles ------ x = 49.95 g of HCl

Answer:

The value of y = 5.1478

Explanation:

The linear equation is an equation obtained when a linear polynomial is equated to zero. When the solution obtained on solving the equation is substituted in the equation in place of the unknown, the equation gets satisfied.

The given equation: 5.3 x 10- (y)(2y) = 0

⇒ 53 - 2y² = 0

⇒ 2y² = 53

⇒ y² = 53 ÷ 2 = 26.5

⇒ y = √26.5 = 5.1478

Explanation:

Inexhaustible resource -

It is the resource which never get reduced or deplete on constant usage , they are mainly the type of naturally occurring resources , hence they reappear in nature again .

The example of inexhaustible resources are - the wind energy , tides , solar energy and the geothermal energy .

Hence , the solar energy is also a form of inexhaustible resource .

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:

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:

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:

Molar mass of unknown solute is 679 g/mol

Explanation:

Let us assume that the solute is a non-electrolyte.

For a solution with non-electrolyte solute remains dissolved in it -

Depression in freezing point of solution, [tex]\Delta T_{f}=K_{f}.m[/tex]

where, m is molality of solute in solution and [tex]K_{f}[/tex] is cryogenoscopic constant of solvent.

Here [tex]\Delta T_{f}=(-22.9^{0}\textrm{C})-(-28.7^{0}\textrm{C})=5.8^{0}\textrm{C}[/tex]

If molar mass of unknown solute is M g/mol then-

                 [tex]m=\frac{\frac{2.2}{M}}{0.0167}mol/kg[/tex]

So, [tex]5.8^{0}\textrm{C}=29.9^{0}\textrm{C}/(mol/kg)\times \frac{\frac{2.2g}{M}}{0.0167}mol/kg[/tex]

so, M = 679 g/mol

Answer:

Tube 1: 10⁻¹, tube 2: 10⁻², tube 3: 10⁻³, tube 4: 10⁻⁴, tube 5: 10⁻⁵.

Explanation:

Serial dilutions are dilutions that the concentration decreases by the same quantity in each successive step. It means that the undiluted will be used for the first step, then the first will be used for the second, and successively. So, for a 10-fold, the concentration must decrease 1/10 in each step, it means that the dilution will be 1/10 in the first one (because it's 1 in tube 0).

In tube 1, the dilution is 1/10 = 0.1 = 10⁻¹;

In tube 2, the dilution will decrease more 1/10, so it will be 1/100x10 = 1/100 = 0.01 = 10⁻²;

In tube 3, it will be 1/1000x1/10 = 1/1000 = 0.001 = 10⁻³

In tube 4, it will be 10⁻⁴, and

In tube 5, it will be 10⁻⁵.

21.077 × 10²¹ silver atoms are there in 3.86 g of silver.

Number of atoms = Number of moles × Avogadro's Number

Avogadro's number is the number of particles in one mole of substance. 6.022 × 10²³ is known as Avogadro's constant / Avogadro's number.

Avogadro's Number = 6.022 × 10²³

To calculate the number of moles of a substance we have to divide the given mass/weight of the substance by the molar mass of the substance.

Number of moles = [tex]\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

                             = [tex]\frac{3.86\ g}{107.87\ g/mol}[/tex]

                             = 0.035 mol

Now put the value of number of moles in above formula, we get

Number of atoms = Number of moles × Avogadro's Number

                             = 0.035 × 6.022 × 10²³

                             = 0.21077  × 10²³

                             = 21.077 × 10²¹

Thus, from the above conclusion we can say that 21.077 × 10²¹ silver atoms are there in 3.86 g of silver.

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Answer: Option (B) is the correct answer.

Explanation:

Whenever there is less concentration of solute particles in a solvent then it means less number of solute particles are available. As a result, there will occur less number of collisions between the solvent and solute particles.

It means that there will be a decrease in rate of reaction.

But if there is more concentration of solute particles then it shows more number of solute particles are available for reaction. As a result, more number of collisions will take place between the particles of solute and solvent.

Hence, then there will occur an increase in rate of reaction.

Thus, we can conclude that a lower concentration of dissolved particles decrease the reaction rate because when there are less dissolved particles, less collisions take place.

Answer:

B

Explanation:

Answer: Option (d) is the correct answer.

Explanation:

The given data is as follows.

Tube diameter d = 10 mm = 0.01 m

Velocity of glycerol, v = 0.5 m/s

Density of glycerol ([tex]\rho[/tex]) = 1240 kg/m3

Dynamic viscosity of glycerol ([tex]\mu[/tex]) = 0.0813 pa.s

Reynolds number (Re) = [tex]\rho \times velocity \times \frac{density}{\mu}[/tex]

                                     = [tex]1240 \times 0.5 \times \frac{0.01}{0.0813}[/tex]

                                     = 76.26

Therefore, according to Reynolds number we can say that flow is laminar.

                     Lt = [tex]0.05 \times Re \times d[/tex]

                         = [tex]0.05 \times 76.26 \times 0.01[/tex]

                         = [tex]0.03813 m[/tex]

As it is known that 1 m = 1000 mm. Hence, in 0.03813 m will be equal to [tex]0.03813 m \times \frac{1000 mm}{1 m}[/tex]

                         = 38.13 mm

Thus, we can conclude that the transition length of glycerol is 38.13 mm.

Explanation:

- Simple Distillation: its a separation method that can be used when the two or more liquids in the mix have at least 50 degrees of difference between their boiling points.

-Azeotropic distillation: is a technique to break an azeotrope (constant boiling point mixtures), that can't be separated by simple distillation, by adding another component to generate a new azeotrope (between one initial component and the new one added) with lower boiling point.

-Extractive distillation: is a process to separate mixtures with close boiling points by adding a miscible, high boiling or none volatile solvent to increase the relative volatility of the liquids in the mix, this increases the separation factor. It differences from the azeotropic method because it doesn't form an azeotrope.

-Liquid-liquid extraction: is a method to separate compounds based on their relative solubilities in two different immiscible liquids.

After describing all the methods we can conclude that all of them are methods to separate substances based on their physical properties, this is their similarity. The difference between this method is the property it uses to separate (solubility in the case of extraction and boiling point in the case of destinations), the cases in which they bare used (when the liquids difference in boiling points is bigger [simple] or close [attractive and azeotropic]) and the formation of azeotropes (present in azeotropic and absent in extractive).

I hope you find this information useful and interesting! Good luck!

Answer:

c. 1.4 x 10²³ oxygen atoms

Explanation:

The number of oxygen atoms in one molecule of CH₃COOH is 2.

Avogadro's constant relates the number of molecules in one mole:

6.022 × 10²³ mol⁻¹

Thus, the number of oxygen atoms in one molecule of acetic acid can be converted to the number of oxygen atoms in one mole of acetic acid:

(2 oxygen atoms / molecule)(6.022 × 10²³ molecule / mol) = 1.204 x 10²⁴ atoms per mole

Finally, the number of oxygen atoms in 0.12 moles of acetic acid are calculated:

(1.204 x 10²⁴ atoms / mol)(0.12 mol) = 1.4 x 10²³ atoms/mol

Answer:

0.01348 M (mol L-1)

Explanation:

Water can be autoionized:

H2O + H2O ⇄ H3O+ + OH-

This a reversible reaction and then it has a unique constant Kw that at 25ºC it takes the value of 10^-14

[tex]10^{-14} =[H_{3} O+][OH-][/tex]

pH is the concentration of [H3O+] ions in a logaritmic scale:

[tex]pH = -log([H_{3}O+][/tex]

Then you can solve for the [OH-]

[tex][OH-]=(Kw)/(10^{-pH})=(10^{-14})/(10^{-12.03})=0.01348 M[/tex]

Hope it helps!

Answer:

[tex]\boxed{\text{115.2 torr}}[/tex]

Explanation:

Let’s call water Component 1 and lactose Component 2.

According to Raoult’s Law,  

[tex]p_{1} = \chi_{1}p_{1}^{\circ} \text{ and}\\p_{2} = \chi_{2}p_{2}^{\circ}[/tex]

where

p₁ and p₂ are the vapour pressures of the components above the solution

χ₁ and χ₂ are the mole fractions of the components

p₁° and p₂° are the vapour pressures of the pure components.

Data:

m₁ = 110.0 g;    p₁° = 118.0 torr

m₂ = 50.00 g; p₂° =    0    torr

1. Calculate the moles of each component

[tex]n_{1} = \text{110.0 g} \times \dfrac{\text{1 mol}}{\text{18.02 g}} = \text{6.104 mol}\\\\n_{2} = \text{50.00 g} \times \dfrac{\text{1 mol}}{\text{342.3 g}} = \text{0.1461 mol}[/tex]

2. Calculate the mole fraction of each component

[tex]\begin{array}{rcl}\chi_{2} & = & \dfrac{n_{2}}{n_{1} + n_{2}}\\\\&= & \dfrac{0.1461}{6.104 + 0.1461}\\\\& = &\dfrac{0.1461}{6.250}\\\\& = & \mathbf{0.023 37}\\\chi_{1}& = &1 - \chi_{2}\\& = &1 - 0.023 37\\& = & \mathbf{0.976 63}\\\end{array}[/tex]

3. Calculate the vapour pressure of the mixture

[tex]p_{1} = 0.97663 \times \text{118.0 torr} = \text{ 115.2 torr}\\p_{2} = 0\\p_{\text{tot}} = p_{1} + p_{2} = \text{115.2 torr + 0} = \textbf{115.2 torr}\\\text{The partial pressure of water above the solution is $\boxed{\textbf{115.2 torr}}$}[/tex]

To find the partial pressure of water above the solution, calculate the mole fraction of water in the solution using the given masses of lactose and water. Multiply the mole fraction of water by the vapor pressure of pure water at 55 °C to find the partial pressure of water above the solution.

To find the partial pressure of water above the solution, we need to calculate the mole fraction of water in the solution using the given masses of lactose and water. First, find the moles of lactose by dividing its mass by its molecular weight. Next, find the moles of water by dividing its mass by its molecular weight. Then, calculate the mole fraction of water by dividing the moles of water by the total moles in the solution. Finally, multiply the mole fraction of water by the vapor pressure of pure water at 55 °C to find the partial pressure of water above the solution.

Moles of lactose = 50.00 g / 342.3 g/mol = 0.146 mol

Moles of water = 110.0 g / 18.015 g/mol = 6.105 mol

Mole fraction of water = moles of water / (moles of lactose + moles of water) = 6.105 mol / (0.146 mol + 6.105 mol) ≈ 0.976

Partial pressure of water = mole fraction of water * vapor pressure of pure water at 55°C = 0.976 * 118.0 torr ≈ 115.2 torr

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Answer:

We need  2.933 L of 0.15 mg /mL of protein solution.

Explanation:

Concentration of given solution[tex]C_1 = 0.15 mg/mL[/tex]

1 mg = 0.001 g , 1 mL = 0.001 L

[tex]C_1=\frac{0.15\times 0.001 mg}{1\times 0.001 L}=0.15 g/L[/tex]

Molecular weight of protein = 22,000 Da =22,000 g/mol

Initial concentration in moles/liter:

[tex]C_1=\frac{0.15 g/L}{22,000 g/mol}=6.8182\times 10^{-6} mol/L[/tex]

Initial concentration in micromoles/mL :

1 L = 1000 mL

[tex]C_1=6.8182\times 10^{-6} mol/L=\frac{6.8182\times 10^{-6}\times 10^6 \mu mol}{1000 mL}=6.8182\times 10^{-3} \mu mole/ mL[/tex]

Initial concentration in micromoles/microLiter :

1 L = 1000,000 μL

[tex]C_1=6.8182\times 10^{-6} mol/L=\frac{6.8182\times 10^{-6}\times 10^6 \mu mol}{1000000 \mu L}=6.8182\times 10^{-6}\mu mol/\mu L[/tex]

Moles of protein required = 20 μmoles

n(Moles)=C(concentration) × V(Volume of solution)

[tex]20 \mu mol=6.8182\times 10^{-6}\mu mol/\mu L\times V[/tex]

[tex]V =\frac{20 \mu mol}{6.8182\times 10^{-6}\mu mol/\mu L}[/tex]

[tex]V=2.933\times 10^6 \mu L = 2.933 L[/tex]

We need  2.933 L of 0.15 mg /mL of protein solution.

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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Explanation:

The volumetric flow rate of water will be as follows.

       q = [tex]600 gpm \times \frac{0.000063 m^{3} s^{-1}}{1 gpm}[/tex]

         = 0.0378 [tex]m^{3}/s [/tex]

    Diameter = [tex]8 in \times \frac{0.0254 m}{1 in}[/tex]

                   = 0.2032 m

Relation between area and diameter is as follows.

           A = [tex]\frac{\pi}{4} \times D^{2}[/tex]

               = [tex]\frac{3.14}{4} \times (0.2032 m)^{2}[/tex]

               = 0.785 x 0.2032 x 0.2032

               = 0.0324 [tex]m^{2}[/tex]

Also,     q = A × V

or,         V = [tex]\frac{q}{A}[/tex]

                = [tex]\frac{0.0378 m^{3}/s}{0.0324 m^{2}}[/tex]

                = 1.166 m/s

As, viscosity of water = 1 cP = [tex]10^{-3}[/tex] Pa-s

Density of water = 1000 [tex]kg/m^{3}[/tex]

Therefore, we will calculate Reynolds number as follows.

 Reynolds number = [tex]\frac{D \times V \times density}{viscosity}[/tex]

                                 = [tex]\frac{0.2032 m \times 1.166 m/s  \times 1000}{10^{-3}}[/tex]  

                                = 236931.2

Hence, the flow will be turbulent in nature.

Thus, we can conclude that the Reynolds number is 236931.2 and flow is turbulent.

Answer:

The compound is [tex]Mn_{3}O_{4}[/tex]

Explanation:

The mass percentage of Mn is 72.1% and the mass percentage of O is 27.9%.

The mass percentage of a compound is given by:

[tex]percentage_{A}=\frac{n*MM_{A}}{MM_{C}} *100[/tex]

where:

n is its coefitient in the compund formula

MMa=Molar mass of the element A

MMc=Molar mass of the compound

So, we can figure out which compound is by dividing the percentage by its molar mass

Mn=72.1÷54.938045=1.31239

O=27.9÷15.9994=1.74382

Then, we divide each result by the smaller one (Mn)

Mn=1.31239÷1.31239=1

O=1.74382÷1.31239=1.3287

Each the realation of Mn:O is 1:1.3287

Then we multiply each result by 3:

Mn=1×3=3

O=1.3287×3=3.986≈4

Finally we figure out that the compound has 3 atoms of Mn and 4 atoms O. Result= [tex]Mn_{3}O_{4}[/tex]

Mn3O4 is sometimes used as a starting material in the production of soft ferrites e.g. manganese zinc ferrite, and lithium manganese oxide, used in lithium batteries.

The empirical formula of a compound with 72.1% Mn and 27.9% O is MnO.

The empirical formula of a compound can be determined based on its mass percentage composition of different elements. In this case, the compound is composed of 72.1% Mn (manganese) and 27.9% O (oxygen).

To find the empirical formula, we need to consider the relative number of atoms in the compound. Assuming we have 100g of the compound, we have 72.1g of Mn and 27.9g of O.

Using the molar masses of Mn (54.938 g/mol) and O (15.999 g/mol), we can calculate the number of moles of each element:

Now, we divide the number of moles by the smallest value to get the mole ratio:

Since we want whole-number ratios, we round the ratios to the nearest whole number:

Therefore, the empirical formula of the compound with 72.1% Mn and 27.9% O is MnO.

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:

0,542 g of Na₂HPO₄ and 0,741 g of NaH₂PO₄.

0,856 g of Tris-HCl and 0,553 g of Tris-base

Explanation:

It is possible to use Henderson–Hasselbalch equation to estimate pH in a buffer solution:

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

Where A⁻ is conjugate base and HA is conjugate acid

The equilibrium of phosphate buffer is:

H₂PO₄⁻ ⇄ HPO4²⁻ + H⁺    Kₐ₂ = 6,20x10⁻⁸; pka=7,21

Thus, Henderson–Hasselbalch equation for phosphate buffer is:

pH = 7,21 + log₁₀ [tex]\frac{HPO4^{2-} }{H2PO4^{-} }[/tex]

If desire pH is 7,0 you will obtain:

0,617 =  [tex]\frac{HPO4^{2-} }{H2PO4^{-} }[/tex] (1)

Then, if desire concentration of buffers is 0,10 M:

0,10 M = [HPO₄²⁻] + [H₂PO₄⁻] (2)

Replacing (1) in (2) you will obtain:

[H₂PO₄⁻] = 0,0618 M

And with this value:

[HPO₄²⁻] = 0,0382 M

As desire volume is 100mL -0,1L- the weight of both Na₂HPO₄ and NaH₂PO₄ is:

Na₂HPO₄ = 0,1 L× [tex]\frac{0,0382mol}{1L}[/tex]× [tex]\frac{141,96g}{1mol}[/tex] = 0,542 g of Na₂HPO₄

NaH₂PO₄ = 0,1 L× [tex]\frac{0,0618mol}{1L}[/tex]× [tex]\frac{119,96g}{1mol}[/tex] = 0,741 g of NaH₂PO₄

For tris buffer the equilibrium is:

Tris-base + H⁺ ⇄ Tris-H⁺ pka = 8,075

Henderson–Hasselbalch equation for tris buffer is:

pH = 8,075 + log₁₀ [tex]\frac{Tris-base }{Tris-H^{+} }[/tex]

If desire pH is 8,0 you will obtain:

0,841 =  [tex]\frac{Tris-base }{TrisH^{+} }[/tex] (3)

Then, if desire concentration of buffers is 0,10 M:

0,10 M = [Tris-base] + [Tris-H⁺] (4)

Replacing (3) in (4) you will obtain:

[Tris-HCl] = 0,0543 M

[Tris-base] = 0,0457 M

As desire volume is 100mL -0,1L- the weight of both Tris-base and Tris-HCl is:

Tris-base = 0,1 L× [tex]\frac{0,0457mol}{1L}[/tex]× [tex]\frac{121,1g}{1mol}[/tex] = 0,553 g of Tris-base

Tris-HCl = 0,1 L× [tex]\frac{0,0543mol}{1L}[/tex]× [tex]\frac{157,6g}{1mol}[/tex] = 0,856 g of Tris-HCl

I hope it helps!

To make a 0.10 M solution of the desired buffers, you need to calculate the weight of the buffer needed based on the desired pH and molar mass of the buffer. For the phosphate buffer, you would need to weigh out 1.4196 grams of Na2HPO4 and NaH2PO4. For the Tris buffer, you would need to weigh out 1.211 grams of Tris base.

In order to calculate the weight of the buffer needed to make a 0.10 M solution, you need to use the formula:

Weight of buffer = (desired concentration) * (volume of solution) * (molar mass of buffer)

For the phosphate buffer, with a desired pH of 7.0, you can use the compounds Na₂HPO₄ and Na₂HPO₄. The weight of Na₂HPO₄ needed would be 0.10 mol/L * 0.1 L * 141.96 g/mol, which is 1.4196 grams. The weight of Na₂HPO₄ needed would be the same.

For the Tris buffer, with a desired pH of 8.0, you would use Tris base. The weight of the Tris base needed would be 0.10 mol/L * 0.1 L * 121.1 g/mol, which is 1.211 grams.

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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.

Answer: The mass of air is 1.18 g

Explanation:

To calculate the number of moles, we use the equation given by ideal gas equation:

[tex]PV=nRT[/tex]

Or,

[tex]PV=\frac{m}{M}RT[/tex]

where,

P = pressure of the gas = 150 psia = 10.2 atm   (Conversion factor:  1 psia = 0.068 atm)

V = Volume of gas = [tex]10in^3=0.164L[/tex]    (Conversion factor:  [tex]1in^3=0.0164L[/tex] )

m = mass of air = ?

M = Average molar mass of air = 28.97 g/mol

R = Gas constant = [tex]0.0820\text{ L atm }mol^{-1}K^{-1}[/tex]

T = temperature of the gas = [tex]440^oF=499.817K[/tex]  (Conversion factor: [tex](T(K)-273.15)=(T(^oF)-32)\times \frac{5}{9}[/tex]  )

Putting values in above equation, we get:

[tex]10.2atm\times 0.164L=\frac{m}{28.97g/mol}\times 0.0820\text{ L atm }mol^{-1}K^{-1}\times 499.817K\\\\m=1.18g[/tex]

Hence, the mass of air is 1.18 g

Answer : The 3D representation of [tex]CH_3CHCHI[/tex] are shown below.

Explanation :

The configuration of the geometrical isomers is designated by two system which are, Cis-trans system and E-Z system.

The rules for E-Z system are :

The atoms or groups attached to each olefinic carbon are given priority as per the sequence rule.

If the higher priority groups are present on same sides across the double bond, the geometrical isomer is said to have Z-configuration or 'cis'.

If the higher priority groups are present on opposite sides across the double bond, the geometrical isomer is said to have E-configuration or 'trans'.

In the given compound, higher priority groups are [tex]CH_3[/tex] and [tex]I[/tex].

When [tex]CH_3[/tex] and [tex]I[/tex] are present on same sides across the double bond, the geometrical isomer is said to have 'cis'.

When [tex]CH_3[/tex] and [tex]I[/tex] are present on opposite sides across the double bond, the geometrical isomer is said to have 'trans'.

The 3D representation is shown by dash and wedge bonds in which dash shows that the molecules are below the plane and wedge shows that the molecules are above the plane.

Answer:

The glycosylation reaction or glycoside formation is an organic reaction in which the hemiacetal group of cyclists ketoses or aldoses turns into acetals, named glycosides. Reaction in the attached picture.

Explanation:

Carbohydrates can be found in an open-chain form or a cyclic form. For the second one, the carbonyl group of the aldehyde could react with the alcohol group of the molecule to form the cycle. As shown in the attached picture, the alcohol group of this cyclic form could react with an alcohol (like methanol) in acidic conditions to form an acetal. These compounds are stable at neutral and acidic conditions, but they hydrolyze at basic conditions. This reaction produces both acetals anomers (α and β) because the attack of the nucleophile (alcohol) could be from both sides. However, the most stable anomer will predominate.

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:

the water concentration at equilibrium is

⇒ [ H2O(g) ] = 0.0510 mol/L

Explanation:

∴ Kc = ( [ CO(g) ] * [ H2 ]³ ) / ( [ CH4(g) ] * [ H2O(g) ] ) = 0,30

⇒ [ CO(g) ] = 0.206 mol / 0.778 L = 0.2648 mol/L

⇒ [ H2(g) ] = 0.187 mol / 0.778 L = 0.2404 mol/L

⇒ [ CH4(g) ] = 0.187 mol / 0.778 L = 0.2404 mol/L

replacing in Kc:

⇒ ((0.2648) * (0.2404)³) / ([ H2O(g) ] * 0.2404 ) = 0.30

⇒ 0.0721 [ H2O(g) ] = 3.679 E-3

⇒ [ H2O(g) ] = 0.0510 mol/L

Answer:

The concentration is 50,8 % w/v and radio strengths = 1,96.

Explanation:

Phenobarbital sodium is a medication that could treat insomnia, for example.

2,0 M of Phenobarbital sodium means 2 moles in 1L.

The concentration units in this case are %w/v that means 1g in 100 mL and ratio strengths that means  1g in r mL. Thus, 2 moles must be converted in grams with molar weight -254 g/mole- and liters to mililiters -1 L are 1000mL-. So:

2 moles × [tex]\frac{254 g}{1 mole}[/tex]= 508 g of Phenobarbital sodium.

1 L ×[tex]\frac{1000 mL}{ 1 L}[/tex] = 1000 mL of solution

Thus, % w/v is:

[tex]\frac{508 g}{1000 mL}[/tex] × 100 = 50,8 % w/v

And radio strengths:

[tex]\frac{1000 mL}{508 g}[/tex]  = 1,96. Thus, you have 1 g in 1,96 mL

I hope it helps!