If one mole of a substance has a mass of 56.0 g, what is the mass of 11 nanomoles of the substance? Express your answer in nanograms using the correct number of significant figures. Do not enter your answer using scientific notation.

Answer:

a) [tex]0.63\frac{kg}{m^{3}}[/tex]

Explanation:

Density is given by the expression [tex]d=\frac{m}{V}[/tex], where m is the mass of the substance and V is the volume occupied by the substance.

As the problem says that the liquid is enclosed in a cylinder, you should find the volume of that cylinder that will be the same volume of the liquid, so:

For a cylinder the volume is given by V=[tex]2\pi r^{2}h[/tex]

Replacing the values given, we have:

[tex]V=2\pi (1m)^{2}(5m)[/tex]

[tex]V=15.708m^{3}[/tex]

Replacing the values of m and V in the equation of density, we have:

[tex]d=\frac{10kg}{15.708m^{3}}[/tex]

[tex]d=0.63\frac{kg}{m^{3}}[/tex]

Answer:

v = 2.512 E-5 m³/mol

Explanation:

∴ P = 80 bar → V = 0.001384 m³/Kg......sat. liq water table

∴ P = 85 bar → V = 0.0014013 m³/Kg

⇒ P = 83 bar → V = ?

specific volume ( V ):

⇒ V = 0.001384 + (( 83 - 80 ) / ( 85 - 80 ))*( 0.0014013 - 0.001384 )

⇒ V = 0.00139438 m³/Kg

molar volume ( v ):

∴ Mw water = 18.01528 g/mol

⇒ v = 0.00139438 m³/Kg * ( Kg/1000g ) * ( 18.01528 g/mol )

⇒ v = 2.512 E-5 m³/mol

Answer:

The correct option is: b. pH 6.4-8.0

Explanation:

Phenol red is a weak acid that is used as a pH indicator and exists in the form of stable red crystals.

The color of the phenol red solution changes from yellow to red when the change in pH is observed. The color of phenol red transitions from yellow to red when the pH is 6.8 - 8.2 or 6.4 - 8.0

Above the pH of 8.2, the phenol red solution turns a bright pink in color.

Answer:

v=3.92*10¹⁷ m/s, and l=2.93*10⁻³⁰m

Explanation:

The velocity of the electrons in the x-ray tube is related to the mass of electrons and the energy applied, the equation is E=0.5*m*v², where E is the energy in kV, m is the mass of the electrons (9.11x10-31kg), and v the velocity. Substituting, 70000 = 0.5*(9.11x10-31kg)*v², and rearranging the terms, sqrt((70000)/(0.5)(9.11x10-31)) = v = 3.92*10¹⁷ m/s. Then the energy is related to the constant h (6.62607015×10−34 Js), the speed of light 299792458 m/s, and "l" that is the wavelength. So the equation is l=(h*c)/E, and l = 2.93*10⁻³⁰m.

Answer:

2486 mmHg

Explanation:

Gay-Lussac's Law states the pressure varies directly with temperature when volume remains constant:

P₁/T₁ = P₂/T₂

Where P₁ and T₁ are initial pressure and temperature and P₂ and T₂ are final pressure and temperatue.

The problem says initial pressure is 721 mmHg, initial temperature is 25°C and final temperature is 755°C. The question is final pressure.

°C must be converted to absolute temperature (K), thus:

25°C + 273,15 = 298,15 K

755°C + 273,15 = 1028,15 K

Thus, pressure P₂ is:

(T₂·P₁) / T₁ = P₂

1028,15K · 721mmHg / 298,15 K =  2486 mmHg

I hope it helps!

Answer:

make thermometers smaller using mercury

Explanation:

Daniel Fahrenheit invented first accurate thermometer which used mercury instead of the alcohol and the water mixtures. In laboratory, he used this invention of him to develop first temperature scale which was enough precise to become the worldwide standard.

The key to the Fahrenheit's thermometer was that the mercury is able to rise and fall within tube without sticking to sides. It was ideal substance for the reading temperatures since mercury expanded at more constant rate than the alcohol and is also able to be read the temperature at much higher and also lower temperatures.

Answer:

185.5156 g

Explanation:

The reaction of copper sulfate with zinc is shown below as:

[tex]CuSO_4+Zn\rightarrow ZnSO_4+Cu[/tex]

Given that :

Amount of copper sulfate = 454 g

Molar mass of copper sulfate = 160 g/mol

The formula for the calculation of moles is shown below:

[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]

Thus, moles are:

[tex]moles= \frac{454\ g}{160\ g/mol}[/tex]

[tex]moles\ of\ copper\ sulfate= 2.8375\ mol[/tex]

From the reaction,  

1 mole of copper sulfate reacts with 1 mole of zinc

Thus,

2.8375 moles of copper sulfate reacts with 2.8375 moles of zinc

Moles of Zinc that should react = 2.8375 moles

Mass of zinc= moles×Molar mass

Molar mass of zinc = 65.38 g/mol

Mass of zinc = 2.8375 ×65.38 g = 185.5156 g

Final answer:

185.595 grams of zinc would react with 454 g of copper sulfate, based on the one-to-one mole ratio between copper sulfate and zinc and the molar mass of zinc (65.38 g/mol).

Explanation:

The student is asking how many grams of zinc would react with 454 g of copper sulfate, which has a molar mass of 160 g/mol. According to the provided reaction, CuSO4 (aq) + Zn(s) → Cu(s) + ZnSO4 (aq), there is a one-to-one mole ratio between copper sulfate and zinc. Given the molar mass of copper sulfate (160 g/mol), we first find the number of moles of copper sulfate in 454 g.

Number of moles of CuSO4 = (454 g) / (160 g/mol) = 2.8375 mol

Since the mole ratio between copper sulfate and zinc is 1:1, an equal number of moles of zinc will react with the copper sulfate. We then need to find the molar mass of zinc to convert moles of zinc to grams. Zinc has a molar mass of approximately 65.38 g/mol.

Mass of zinc = Number of moles of Zn × molar mass of Zn = 2.8375 mol × 65.38 g/mol = 185.595 g

Therefore, 185.595 grams of zinc would react with 454 g of copper sulfate.

Answer:

400 cans of soda would be lethal.

Explanation:

In a can of soda, there is (2.13 mg/oz * 12 oz) 25 mg caffeine.

25 mg * (1g / 1000 mg) = 0.025 g

If in a can of soda there is 0.025 g of caffeine, a lethal dose of caffeine will be ingested after drinking (10.0 g * (1 can / 0.025 g)) 400 cans of soda.

Answer:

B) Increase in boiling point and decrease in vapor pressure

Explanation:

Vapor pressure is inversely related  to the Boiling point , as

higher the boiling point, lower the vapor pressure. and

Lower the boiling point, higher the vapor pressure.

Hydrogen bonding.

The electrostatic attraction between Hydrogen , bonded to electronegative atom like F, O, N and the more electronegative atom is called as Hydrogen bonding.

For example -

In alcohols, - OH group has Hydrogen that is bonded to more electronegative atom O.

As ,  

Extra energy is required to break Hydrogen bonds.

because the substance which exhibits Hydrogen bonding have lower vapor pressure than that of the substance with out Hydrogen bonding.

Hence , the substance with Hydrogen bonding , has higher boiling point,.

Hence , the correct option is  Increase in boiling point and decrease in vapor pressure .

Answer:

m= 1.84 m

M= 1.79 M

mole fraction (X) =

Xsolute= 0.032

Xsolvent = 0.967

Explanation:

1. Find the grams of FeCl3 in the solution: when we have a mass % we assume that there is 100 g of solution so 24% means 24 g of FeCl3 in the solution. The rest 76 g are water.

2. For molality we have the formula m= moles of solute / Kg solvent

so first we pass the grams of FeCl3 to moles of FeCl3:

24 g of FeCl3x(1 mol FeCl3/162.2 g FeCl3) = 0.14 moles FeCl3

If we had 76 g of water we convert it to Kg:

76 g water x(1 Kg of water/1000 g of water) = 0.076 Kg of water

now we divide m = 0.14 moles FeCl3/0.076 Kg of water

m= 1.84 m

3. For molarity we have the formula M= moles of solute /L of solution

the moles we already have 0.14 moles FeCl3

the (L) of solution we need to use the density of the solution to find the volume value. For this purpose we have: 100 g of solution and the density d= 1.280 g/mL

The density formula is d = (m) mass/(V) volume if we clear the unknown value that is the volume we have that (V) volume = m/d

so V = 100 g / 1.280 g/mL = 78.12 mL = 0.078 L

We replace the values in the M formula

M= 0.14 moles of FeCl3/0.078 L

M= 1.79 M

3. Finally the mole fraction (x)  has the formula

X(solute) = moles of solute /moles of solution

X(solvent) moles of solvent /moles of solution

X(solute) + X(solvent) = 1

we need to find the moles of the solvent and we add the moles of the solute like this we have the moles of the solution:

76 g of water x(1 mol of water /18 g of water) = 4.2 moles of water

moles of solution = 0.14 moles of FeCl3 + 4.2 moles of water = 4.34 moles of solution

X(solute) = 0.14 moles of FeCl3/4.34 moles of solution = 0.032

1 - X(solute) = 1 - 0.032 = 0.967

The average atomic weight of Hafnium is calculated by multiplying the atomic weight of each isotope by its relative abundance and summing the products. The result is an average atomic weight of 178.433 amu.

To calculate the average atomic weight of Hafnium (Hf), we multiply the atomic weight of each isotope by its natural abundance, expressed as a fraction, and then sum the results. Here are the calculations:

Adding these products together we get the average atomic weight of Hafnium:

(0.0016 × 173.940) + (0.0526 × 175.941) + (0.1860 × 176.943) + (0.2728 × 177.944) + (0.1362 × 178.946) + (0.3508 × 179.947) = 0.278304 + 9.255366 + 32.922978 + 48.523072 + 24.373172 + 63.079956 = 178.432848 amu

The average atomic weight of Hafnium truncated to three decimal places is 178.433 amu.

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Answer:

e. 18

Explanation:

A neutral P atom has an atomic number of 15, which means there are 15 protons in the atom. In order to be neutral, the P atom must also have 15 electrons.

The P³⁻ anion has 3 electrons more than the neutral P atom since it has a charge of -3.

Thus, the total number of electrons are 15 + 3 = 18 electrons.

Answer:

The temperature difference across the stainless steel is 18°C

Explanation:

The heat flows through the plastic layer by convection and then through the steel layer by conduction and then through the plastic layer on the other side.

The heat flux q/A can be expressed as:

[tex]q/A = h*\Delta T[/tex]

It can also be expressed as

[tex]q/A=h*(T_A-T_B) = h_1(T_A-T_1)=k/\Delta x*(T_1-T_2)=h_2(T_2-T_B)[/tex]

being Δx/k*(T1-T2) the conductive heat flux through the steel.

If we want to know the temperatur difference across the stainless steel (T1-T2) we can write:

[tex](k/\Delta x)*(T_1-T_2)=h*(T_A-T_B)\\\\(T_1-T_2)=(\Delta x/k)*h*(T_A-T_B)\\\\(T_1-T_2)=(0.04/16)*120*60=18 \, ^{\circ} C[/tex]

Answer:

The final temperature of the entire water sample after all the ice is melted, is 12,9°C. We should realize that if there is no loss of heat in our system, the sum of lost or gained heat is 0.  It is logical to say that the temperature has decreased because the ice gave the water "heat" and cooled it

Explanation:

This is the begin:

Q1 = Q which is gained from the ice to be melted

Q2 = Q which is lost from the water to melt the ice

Q1 + Q2 = 0

We are informed that the ice is at 0 ° so we have to start calculating how many J, do we need to melt it completely. If the ice had been at a lower temperature, it should be brought to 0 ° with the formula

Q = mass. specific heat. (ΔT)

and then make the change of state by the latent heat of fusion.

The heat of fusion for water at 0 °C is approximately 334 joules per gram.

So Q = Hf . mass

Q1 = 334 J/g . 8.32 g = 2778,88 J

For water we should use this:

Q = mass. specific heat. (ΔT)

Q2 = 55g . 4180 J/kg. K (Tfinal -T initial)

Q2 = 55g . 4180 J/kg. K (Tfinal -T initial)

(notice we have kg, so we have to convert 55 g, to kg, 0,055kg so units can be cancelled)

Q2 = 0,055kg . 4180 J/kg. K (Tfinal (The unknown) -25°)

T° should be in K for the units of Specific heat but it is the same. The difference is the same, in K either in °C

25°C = 298K

Q2 = 0,055kg . 4180 J/kg. K (Tfinal -298K)

Now the end

Q1 + Q2 = 0

334 J/g . 8.32 g + 0,055kg . 4180 J/kg. K (Tfinal -298K)

2778,88 J + 229,9 J/K (Tfinal - 298 K) = 0

2778,88 J + 229,9 J/K x Tfinal - 68510,2 J = 0

229,9 J/K x Tfinal = 68510,2 J - 2778,88 J

229,9 J/K x Tfinal = 65731,4 J

Tfinal = 65731,4 J / 229,9 K/J

Tfinal = 285,9 K

Tfinal = 285,9 K - 273K = 12,9 °C

The final temperature of the water after an 8.32 g ice cube at 0.00 °C is placed into 55 g of water at 25 °C and all the ice melts is 12.959 °C, using the conservation of energy and assuming no heat loss to the surroundings.

When an 8.32 g ice cube at 0.00 °C is placed into 55 g of water at 25 °C, we can find the final temperature after all the ice is melted by applying the principle of conservation of energy. The heat gained by the ice melting must be equal to the heat lost by the water cooling down. We will assume that no heat is lost to the surroundings in this perfectly insulated system.

We need to calculate the heat required to melt the ice cube (Qmelt) using the heat of fusion of water (which is 79.9 cal/g or 334 J/g), and the heat lost by the water as it cools down (Qwater) using the specific heat of water (which is 4.184 J/g°C).

First, calculate the heat necessary to melt the ice:

Qmelt = mass of ice * heat of fusion

Qmelt = 8.32 g * 334 J/g

Qmelt = 2778.08 J

Next, set up the equation based on the conservation of energy, where the heat lost by the water (Qwater) equals the heat gained by the ice (Qmelt):

Qwater = mass of water * specific heat of water * temperature change

Qwater = Qmelt

55 g * 4.184 J/g°C * (25 °C - final temperature) = 2778.08 J

Solving for the final temperature:

230.62 J/°C * (25 °C - final temperature) = 2778.08 J

25 °C - final temperature = 2778.08 J / 230.62 J/°C

25 °C - final temperature = 12.041 °C

Final temperature = 25 °C - 12.041 °C

Final temperature = 12.959 °C

Therefore, the final temperature of the water after all the ice has melted is 12.959 °C (rounded to three significant figures).

Explanation:

The given data is as follows.

          [tex]\Delta H[/tex] = 286 kJ = [tex]286 kJ \times \frac{1000 J}{1 kJ}[/tex]

                            = 286000 J

 [tex]S_{H_{2}O} = 70 J/^{o}K[/tex],      [tex]S_{H_{2}} = 131 J/^{o}K[/tex]

 [tex]S_{O_{2}} = 205 J/^{o}K[/tex]

Hence, formula to calculate entropy change of the reaction is as follows.

          [tex]\Delta S_{rxn} = \sum \nu_{i}S_{i}_(products) - \sum \nu_{i}S_{i}_(reactants)[/tex]

                     = [tex][(\frac{1}{2} \times S_{O_{2}}) - (1 \times S_{H_{2}})] - [1 \times S_{H_{2}O}][/tex]

                    = [tex][(\frac{1}{2} \times 205) + (1 \times 131)] - [(1 \times 70)][/tex]

                    = 163.5 J/K

Therefore, formula to calculate electric work energy required is as follows.

             [tex]\Delta G_{rxn} = \Delta H_{rxn} - T \Delta S_{rxn}[/tex]

                            = [tex]286000 J - (163.5 J/K \times 298 K)[/tex]

                            = 237.277 kJ

Thus, we can conclude that the electrical work required for given situation is 237.277 kJ.

Final answer:

To find the electrical work required for the electrolysis of water to produce 1 mole of hydrogen, calculate the Gibbs Free Energy (ΔG) for the reaction. Using the given thermodynamic data, ΔG at 298 K is 236.4 kJ, representing the minimum electrical work needed.

Explanation:

The student has asked how to determine the electrical work required to produce one mole of hydrogen in the electrolysis of liquid water at 298 K and 1 atm. The question involves understanding the thermodynamics of the reaction: H2O(l) → H2(g) + 0.5O2(g), with given data points including ΔH and standard entropies (S°) for the reactants and products. To find the electrical work required, you first calculate the ΔG (Gibbs Free Energy) of the reaction using the formula ΔG = ΔH - TΔS. Knowing ΔG allows you to determine the maximum work that can be extracted from the reaction, which, in the case of electrolysis, corresponds to the minimum work required to drive the reaction in reverse.

ΔS for the reaction can be calculated using the given entropies: ΔS = ⅛∑S°(products) - ⅛∑S°(reactants) = (131 + 0.5×205) - 70 = 166.5 J/K. Therefore, ΔG at 298 K can be calculated as ΔG = 286,000 J - (298K × 166.5 J/K) = 286,000 J - 49,617 J = 236,383 J or 236.4 kJ. This value represents the minimum electrical work required to produce one mole of hydrogen gas via electrolysis of water under the specified conditions.

Answer:

The hydroxide ion concentration is 0.003981 M.

The hydronium ion concentration is[tex] 2.5119\times 10^{-12} M[/tex].

Explanation:

The pOH of an aqueous solution at 25°C =  2.40

[tex]pOH=-\log[OH^-][/tex]

[tex]2.40=-\log[OH^-][/tex]

[tex][OH^-]=0.003981 M[/tex]

The hydroxide ion concentration is 0.003981 M.

The pH of an aqueous solution at 25°C =  ?

The relationship between pH and pOH :

pH + pOH = 14

[tex]pH=14- pOh = 14-2.40 =11.6 [/tex]

[tex]pH=-\log[H_3O^+][/tex]

[tex]11.6=-\log[H_3O^+][/tex]

[tex][H_3O+]=2.5119\times 10^{-12} M[/tex]

The hydronium ion concentration is[tex] 2.5119\times 10^{-12} M[/tex].

Answer:

The  pH of the resulting solution is 12.52.

Explanation:

[tex]Molarity=\frac{n}{V}[/tex]

n = number of moles

V = volume of the solution in Liters

1)1 mol of HCl gives 1 mol of hydrogen ion.

[tex][HCl]=[H^+]=0.1 m[/tex]

Concentration of the hydrogen ion = 0.1 M

Volume of the solution = 10 mL = 0.010 L

[tex]0.1 M=\frac{n}{0.010L}[/tex]

Moles of hydrogen ions =  = 0.001 mol

2) 1 mol of NaOH gives 1 mol of hydroxide ion.

[tex][NaOH]=[OH^-]=0.2 M m[/tex]

Concentration of the Hydroxide ions = 0.2 M

Volume of the solution ,V'= 8 mL = 0.008 L

[tex]0.2=\frac{n'}{V'}[/tex]

Moles of hydroxide ions ,n ' = 0.0016

1 mol of HCl neutralizes 1 mol of NaOH ,then 0.001 mol of HCl will neutralize 0.001 mol NaOH.

So left over moles of hydroxide ions in the solution will effect the pH of the solution:

Left over moles of hydroxide ions in the solution = 0.0016 mol - 0.0010 mol = 0.0006 mol

Left over concentration of hydroxide ions:

[tex][OH^-]'=\frac{0.0006 mol}{0.010 L+0.008 L}=0.0333 mol/L[/tex]

[tex]pOH=-\log[OH^-]=-\log[0.03333 M]=1.48[/tex]

pH +pOH = 14

pH = 14 - 1.48 = 12.52

The  pH of the resulting solution is 12.52.

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Answer:

150 × 10⁻⁹ μL

Explanation:

Data provided in the question:

Molarilty of the stock solution, M₁ = 200 M

Final Volume of the solution, V₂ = 100 μL = 100 × 10⁻⁶ L

Final concentration, M₂ = 300 nM = 300 × 10⁻⁹ M

Now,

M₁V₁ = M₂V₂

where,

V₁ is volume of the stock solution

Thus,

200 × V₁ = 100 × 10⁻⁶ × 300 × 10⁻⁹

or

V₁ = 150 × 10⁻⁹ μL

Answer: Option (b) is the correct answer.

Explanation:

Reduction is defined as the process in which there occurs gain of hydrogen. Whereas oxidation is defined as the process in which there occurs loss of hydrogen.

As the given reaction is as follows.

      [tex]C_{2}H_{4} + H_{2} \rightarrow C_{2}H_{6}[/tex]

Since, hydrogen is being added in this chemical reaction. It means that reduction is taking place and carbon atom is reduced.

Thus, we can conclude that in the given reaction carbon atoms are reduced.

Answer:

Macroscopic scale

Explanation:

Subatomic scale is the scale at which atomic constituents, such as nucleus which contains protons and neutrons, and electrons, which orbit in  the elliptical paths around nucleus exists.

Miniscopic scale is a reference scale and is not a standard scale for measurement. Usually, this refers to minute objects.

Atomic scale is the scale which is at size of the atoms.

Macroscopic scale is length scale on which the objects or the phenomena are enough large to be visible with naked eye, without magnifying the optical instruments. It is the largest scale.

Microscopic scale is scale of the objects that require microscope to see them.

Answer : The freezing point of solution is 273.467 K

Explanation : Given,

Mass of glucose (solute) = 0.3 g

Mass of water (solvent) = 1000 g = 1 kg

Moles of fructose (solute) = 0.5 mol

Mass of water (solvent) = 1000 g = 1 kg

Molar mass of glucose = 180 g/mole

First we have to calculate the moles of glucose.

[tex]\text{Moles of glucose}=\frac{\text{Mass of glucose}}{\text{Molar mass of glucose}}=\frac{0.3g}{180g/mole}=0.00167mole[/tex]

Now we have to calculate the total moles after mixing.

[tex]\text{Total moles}=\text{Moles of glucose}+\text{Moles of fructose}[/tex]

[tex]\text{Total moles}=0.00167+0.5=0.502moles[/tex]

Now we have to calculate the molality.

[tex]\text{Molality}=\frac{\text{Total moles}}{\text{Mass of water(solvent in kg)}}[/tex]

[tex]\text{Molality}=\frac{0.502mole}{(1+1)kg}=0.251mole/kg[/tex]

Now we have to calculate the freezing point of solution.

As we know that the depression in freezing point is a colligative property that means it depends on the amount of solute.

Formula used :  

[tex]\Delta T_f=K_f\times m[/tex]

[tex]T^o_f-T_f=K_f\times m[/tex]

where,

[tex]\Delta T_f[/tex] = change in freezing point

[tex]T_f^o[/tex] = temperature of pure water = [tex]0^oC[/tex]

[tex]T_f[/tex] = temperature of solution = ?

[tex]K_f[/tex] = freezing point constant of water = [tex]1.86^oC/m[/tex]

m = molality = 0.251 mole/kg

Now put all the given values in this formula, we get

[tex]0^oC-T_f=1.86^oC/m\times 0.251mole/kg[/tex]

[tex]T_f=-0.467^oC=273.467K[/tex]

conversion used : [tex]K=273+^oC[/tex]

Therefore, the freezing point of solution is 273.467 K

Answer : The mass of sodium fluoride added should be 0.105 grams.

Explanation : Given,

The dissociation constant for HF = [tex]K_a=6.8\times 10^{-4}[/tex]

Concentration of HF (weak acid)= 0.0310 M

First we have to calculate the value of [tex]pK_a[/tex].

The expression used for the calculation of [tex]pK_a[/tex] is,

[tex]pK_a=-\log (K_a)[/tex]

Now put the value of [tex]K_a[/tex] in this expression, we get:

[tex]pK_a=-\log (6.8\times 10^{-4})[/tex]

[tex]pK_a=4-\log (6.8)[/tex]

[tex]pK_a=3.17[/tex]

Now we have to calculate the concentration of NaF.

Using Henderson Hesselbach equation :

[tex]pH=pK_a+\log \frac{[Salt]}{[Acid]}[/tex]

[tex]pH=pK_a+\log \frac{[NaF]}{[HF]}[/tex]

Now put all the given values in this expression, we get:

[tex]2.60=3.17+\log (\frac{[NaF]}{0.0310})[/tex]

[tex][NaF]=0.00834M[/tex]

Now we have to calculate the moles of NaF.

[tex]\text{Moles of NaF}=\text{Concentration of NaF}\times \text{Volume of solution}=0.00834M\times 0.300L=0.0025mole[/tex]

Now we have to calculate the mass of NaF.

[tex]\text{Mass of }NaF=\text{Moles of }NaF\times \text{Molar mass of }NaF=0.0025mole\times 42g/mole=0.105g[/tex]

Therefore, the mass of sodium fluoride added should be 0.105 grams.

Answer:

The rate is [tex] 4,5 \times 10^{-5}\frac{mole}{Ls}[/tex]

Explanation:

Stoichiometry

[tex]CH_{3}Cl+NaOH \rightarrow CH_{3}OH+NaCl [/tex]

Kinetics

[tex]-r_{A}=k \times [CH_{3}Cl] \times [NaOH] [/tex]

The rate constant K can be calculated by replacing with the initial data

[tex] 1 \times 10^{-4}\frac{mole}{Ls}=k \times [0,2M] \times [1,0M]  =5 \times 10^{-4}\frac{L}{mole s}[/tex]

Taking as a base of calculus 1L, when half of the [tex] CH_{3}Cl [/tex] is consumed the mixture is composed by

[tex] 0,1 mole CH_{3}Cl [/tex] (half is consumed)

[tex] 0,9 mole NaOH [/tex] (by stoicheometry)

[tex] 0,1 mole CH_{3}OH [/tex]  

[tex] 0,1 mole NaCl [/tex]

Then, the rate is

[tex]-r_{A}=5 \times 10^{-4} \frac{L}{mole s}\times 0,1M \times 0,9 M=4,5 \times 10^{-5}\frac{mole}{Ls}[/tex]

The reaction rate decreases because there’s a smaller concentration of reactives.

Answer:

Measuring with a ruler and using final volume minus initial volume

Explanation:

You can measure the volume of a geometric object by measuring its sides with a ruler and calculating the volume according to the corresponding formula for each object. For example, for a rectangular prism it would be

[tex]volume=length*width*height[/tex]

You can also measure the volume of an object by measuring how much water it displaces. To do this you have to fill a measuring cylinder with enough water for the object to be completely submerged and take note of the volume. Then, add the object and note again the volume of the water+object. The difference between both is the volume of the object.

[tex]Volume of the object= volume of water and object - volume of water[/tex]

The advantage of the second method is that it can be used for objects with irregular shapes as long as they do not float.

Answer: The given number in scientific notation is [tex]2.511\times 10^{-2}[/tex]

Explanation:

Scientific notation is defined as the notation in which a number is expressed in the decimal form. This means that the number is always written in the power of 10 form. The numerical digit lies between 0.1.... to 9.9.....

If the decimal is shifting to right side, the power of 10 is negative and if the decimal is shifting to left side, the power of 10 is positive.

We are given:

A number having value = 0.0251089

Converting this into scientific notation, we get:

[tex]\Rightarrow 0.0251089=2.511\times 10^{-2}[/tex]

Hence, the given number in scientific notation is [tex]2.511\times 10^{-2}[/tex]

Answer: The time required will be 19.18 years

Explanation:

All the radioactive reactions follows first order kinetics.

The equation used to calculate half life for first order kinetics:

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

We are given:

[tex]t_{1/2}=4.7\times 10^1yrs[/tex]

Putting values in above equation, we get:

[tex]k=\frac{0.693}{4.7\times 10^1yr}=0.015yr^{-1}[/tex]

Rate law expression for first order kinetics is given by the equation:

[tex]k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}[/tex]

where,  

k = rate constant  = [tex]0.015yr^{-1}[/tex]

t = time taken for decay process = ?

[tex][A_o][/tex] = initial amount of the reactant = 2 g

[A] = amount left after decay process =  (2 - 0.5) = 1.5 g

Putting values in above equation, we get:

[tex]0.015yr^{-1}=\frac{2.303}{t}\log\frac{2}{1.5}\\\\t=19.18yrs[/tex]

Hence, the time required will be 19.18 years

Explanation:

Except for the use in the chemistry laboratory , were it is used to synthesize many useful products, silver nitrate is also has biolofical and medical relevance.

Silver nitrate is commonly used for silver staining, for demonstrating the reticular fibers, the proteins and the nucleic acids. It is also used as stain in the scanning electron microscopy.

Silver Nitrate is also used for the bone ulcers as well as the burns and the acute wounds.

Answer:

A loud ordinary conversation following the supplied information in the question is about 4500 dB. But, in the official decibel system measure a loud conversation does not overcome 100 dB.

Explanation:

Using the supplied data of the exercise, we say that in a restaurant conversation the value is 45 dB. If we multiply this by 100 we will have a value for a laud ordinary conversation.

45×100 = 4500 dB.

but as I mentioned in the answer, in the official decibel system measure a loud conversation between 2 man reaches a maximal of 100 dB.