Question 3 (1 point) The extinction coefficient for copper sulfate in aqueous solution is 12 M-1.cm-1 at 800 nm. If the absorbance of the copper sulfate solution in 0.50 cm cuvette is 0.50 at 800 nm, the concentration of copper sulfate in this solution is: O3M O 83 mm 8.3 mm 12 M-cm 0.042 M

Answer:

The percent by mass of chromium(II) acetate in the solution is 4.42%.

Explanation:

Molality of the chromium(II) acetate solution = 0.260 m = 0.260 mol/kg

This means that in 1 kg of solution 0.260 moles of chromium(II) acetate are present.

1000 g = 1 kg

So, in 1000 grams of solution 0.260 moles of chromium(II) acetate are present.

Then in 100 grams of solution = [tex]\frac{0.260 mol}{1000}\times 100=0.0260 mol[/tex]

Mass of 0.0260 moles chromium(II) acetate:

= 0.0260 mol × 170 g/mol = 4.42 g

[tex](w/w)\%=\frac{\text{mass of solute in 100 gram solution}}{100}\times 100[/tex]

[tex]=\frac{4.42 g}{100 g}\times 100=4.42\%[/tex]

The percent by mass of chromium(II) acetate in the solution is 4.42%.

Answer:

a) [Tris0] : [Tris] = 1 : 100

b) Range = 7.1 to 9.1

Explanation:

a) Calculation of ratio of the basic and the acidic forms of tris

pH of a buffer is calculate using Henderson-Hasselbalch equation

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

Conjugate acid of Tris dissociated as

[tex]Tris \leftrightharpoons Tris0 + H^+[/tex]

For tris,

Salt or Basic form = tris0

Acid or Acidic form = Tris

pKa = 8.1

pH = 6.1

[tex]pH = pKa+log\frac{Tris0}{Tris}[/tex]

[tex]6.1 = 8.1+log\frac{Tris0}{Tris}[/tex]

[tex]log\frac{Tris0}{Tris} = -2[/tex]

[tex]\frac{Tris0}{Tris} = antilog (-2)[/tex]

[tex]\frac{Tris0}{Tris} = 10^{-2}[/tex]

[Tris0] : [Tris] = 1 : 100

b) Range of Tris

Range of any buffer is:

From (pKa -1) to (pKa+1)

So, range of Tris is:

From (8.1 - 1) to (8.1 +1)

or from 7.1 to 9.1

To solve the equation pH = 4 + log(x) and find the value of x when the hydrogen ion concentration is 0.01 M, rearrange the equation by subtracting 4 from both sides. Substitute the value of pH with its equivalent -log(0.01) and simplify the equation. The value of x is 0.0001.

To solve the equation pH = 4 + log(x) and find the value of x when the hydrogen ion concentration is 0.01 M, we need to rearrange the equation. Subtracting 4 from both sides, we get log(x) = pH - 4. Using the fact that pH = -log[H+], we substitute the value of pH with its equivalent -log(0.01). Simplifying the equation, we have log(x) = -log(0.01) - 4. Taking the antilog (10^x) of both sides, we get x = 10^(-log(0.01) - 4). Evaluating this expression gives us x = 0.0001.

Answer:

The final diameter of the balloon is 20.84 cm.

Explanation:

To answer this problem, because both the number of moles (because no air was lost or added to the balloon) and the pressure remain constant, we can use Charles' law, or the law of volumes:

V₁T₂=V₂T₁

Where V is Volume (in L) and T is temperature (in K). The subscripts 1 and 2 refer to different states of the gas: for this problem, let's say that ₁ refers to the state of the gas at 7:00 am and that ₂ refers to the state 6 hours later.

To calculate the volume occupied by the gas we use the formula for the volume of a sphere:

[tex]V=\frac{4}{3}*\pi  (\frac{d}{2})^{3}[/tex]

Thus the volume of the balloon at 7:00 am is 4188.79 cm³

Then we convert °F into K:

Now we use Charles' law to calculate V₂:

V₂=V₁T₂/T₁

V₂= [tex]\frac{4188.79cm^{3}*310.93K}{274.82 K}=4739.18cm^{3}[/tex]

Lastly we calculate the final diameter of the balloon:

[tex]4739.18cm^{3}=\frac{4}{3}*\pi  (\frac{d}{2})^{3}\\3554.38cm^{3}=\pi  (\frac{d}{2})^{3}\\1131.40cm^{3}=(\frac{d}{2})^{3}\\10.42cm=\frac{d}{2}\\20.84=d[/tex]

Answer: The moles of water produced are 1.54 moles.

Explanation:

To calculate the number of moles, we use the equation:

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

Given mass of ethane = 15.42 g

Molar mass of ethane = 30.07 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of ethane}=\frac{15.42g}{30.07g/mol}=0.513mol[/tex]

The chemical equation for the combustion of ethane follows:

[tex]2C_2H_6+5O_2\rightarrow 4CO2+6H_2O[/tex]

By Stoichiometry of the reaction:

2 moles of ethane produces 6 moles of water

So, 0.513 moles of ethane will produce = [tex]\frac{6}{2}\times 0.513=1.54mol[/tex] of water

Hence, the moles of water produced are 1.54 moles.

Answer:

Polysaccharides, also known as glycans, are the polymers of carbohydrate. These polymers are composed of the monosaccharide monomeric units that are joined together by glycosidic bonds.  

A polysaccharide molecule composed of the same type of monosaccharide units is known as homopolysaccharide or homoglycan.

Example: Cellulose, glycogen, and cellulose.

Whereas, a polysaccharide molecule composed of more than type of monosaccharide is known as heteropolysaccharides or heteroglycans.

Examples: Hyaluronic acid and heparin

The price of one cubic inch of mercury is approximately $0.9308, and the price of one pound of mercury is approximately $1.9079.

Using the above data, we must perform several calculations to determine the cost of one cubic inch of mercury and one pound of mercury. Here's a step-by-step guide to do it:

Given:

We will convert the mass of the flask of mercury to grams

1 pound (lb) = 453.59237 grams (g)

Mass of flask of mercury in grams = 76.00 lb * 453.59237 g/lb

Then, calculate the volume of mercury in cm³

Density of mercury = Mass / Volume

Volume of mercury in cm³ = Mass of flask of mercury in grams / Density of mercury

Then, we will calculate the volume of mercury in cubic inches

1 cm³ = 0.0610237 in³ (approximately)

Volume of mercury in cubic inches = Volume of mercury in cm³ * 0.0610237

Calculate the price of one cubic inch of mercury

Price of one cubic inch of mercury = Cost of flask of mercury / Volume of mercury in cubic inches

We will calculate the price of one pound of mercury

Price of one pound of mercury = Cost of flask of mercury / Mass of flask of mercury in pounds

We will perform the calculations:

Mass of flask of mercury in grams = 76.00 lb * 453.59237 g/lb = 34473.53 g

Volume of mercury in cm³ = 34473.53 g / 13.534 g/cm³ = 2549.915 cm³

Volume of mercury in cubic inches = 2549.915 cm³ * 0.0610237 in³/cm³ = 155.61 in³

Price of one cubic inch of mercury = $145.00 / 155.61 in³ ≈ $0.9308/in³

Price of one pound of mercury = $145.00 / 76.00 lb ≈ $1.9079/lb

Therefore, the price of one cubic inch of mercury is approximately $0.9308, and the price of one pound of mercury is approximately $1.9079.

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The price of one pound of mercury can be calculated by dividing the total cost of $145.00 by the total weight, which is 76 pounds, resulting in a price of $1.9079 per pound.

To find the price per pound of mercury, we first need to find the total mass of mercury in pounds. Given that a 76.00 pound flask of mercury costs $145.00, the price per pound can be calculated directly by dividing the total cost by the total weight in pounds.

Price per pound of mercury = Total cost / Total weight in pounds
Price per pound of mercury = $145.00 / 76.00 pounds
Price per pound of mercury = $1.9079 (rounded to four decimal places)

Note that we do not need the density of mercury for this specific calculation since the total weight and cost are directly given.

1.Concentration of acetic acid is 89.99 mM. Concentration of sodium acetate is 160.0 mM . 2.0.180 moles of acetic acid. 0.32 moles of sodium acetate. 3. Mass of acetic acid in 2 L solution is 10.81 g. Mass of sodium acetate in 2 L solution is 26.25 g.

The amount of matter in an item is indicated by its mass, which is a fundamental physical attribute of matter. It is a scalar quantity that is typically measured in kilogrammes (kg) or grammes (g). Mass is separate from weight, which is the gravitational force exerted on an object and can change depending on the strength of the gravitational field.

1.pH = pKa + log [Sodium acetate] / [Acetic acid]

5.0 = 4.75 + log [Sodium acetate] / [Acetic acid]

log [Sodium acetate] / [Acetic acid] = 5.0 - 4.75

                                                          = 0.25

[Sodium acetate] = 1.778 x [Acetic acid]

[Sodium acetate] + [Acetic acid] = 250 mM

1.778 x [Acetic acid] +  [Acetic acid] = 250 mM

2.778 x [Acetic acid] = 250 mM

[Acetic acid] = 250 mM  / 2.778

                     = 89.99 mM

[Sodium acetate] = 1.778 x [Acetic acid]

                              = 1.778 x 89.99 mM

                              = 160.0 mM

2.Concentration of acetic acid = 89.99 mM

                                                  = 0.08999 M  

Concentration of sodium acetate = 160.0 mM

                                                      = 0.16 M

2 x 0.08999 = 0.180 moles of acetic acid.

2 x 0.16 = 0.32 moles of sodium acetate.

3.Mass of acetic acid = no of moles x molar mass

                                   = 0.180 mol x 60.05 g /mol

                                   = 10.81 g

Mass of sodium acetate = no of moles x molar mass

                                       = 0.32 mol x 82.03 g /mol

                                       = 26.25 g

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Final answer:

Using the Henderson-Hasselbalch equation, the concentrations of acetic acid and sodium acetate in the buffer solution are calculated to be approximately 89 mM and 161 mM, respectively. When considering a 2 L volume of this buffer, there would be approximately 0.178 moles of acetic acid and 0.322 moles of sodium acetate, which translates to 10.69 grams of acetic acid and 26.25 grams of sodium acetate.

Explanation:

To calculate the concentrations of acetic acid and sodium acetate in the buffer, we use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])

Given:

Let's calculate [A-] and [HA] (where [A-] is the concentration of sodium acetate and [HA] is the concentration of acetic acid):
5.0 = 4.75 + log([A-]/[HA])
=> log([A-]/[HA]) = 0.25
=> [A-]/[HA] = 10^0.25
=> [A-]/[HA] ≈ 1.78

Since [A-] + [HA] = 250 mM, we can set up a system of equations:
[A-] = 1.78[HA]
[A-] + [HA] = 250 mM

Solving for [HA] and [A-] gives us:
[HA] ≈ 89 mM
[A-] ≈ 161 mM

Next, we calculate the amount of acetic acid and sodium acetate in 2 L of the buffer:
amount of acetic acid = [HA] × 2 L = 0.089 mol/L × 2 L = 0.178 mol
amount of sodium acetate = [A-] × 2 L = 0.161 mol/L × 2 L = 0.322 mol

Finally, to calculate the mass of acetic acid and sodium acetate in 2 L of the buffer, we use their molar masses:
mass of acetic acid = 0.178 mol × 60.05 g/mol = 10.69 g
mass of sodium acetate = 0.322 mol × 82.03 g/mol = 26.25 g

Answer: The mass of carbon dioxide produced is 27.1 grams.

Explanation:

Combustion reaction is defined as the reaction in which a hydrocarbon reacts with oxygen gas to produce carbon dioxide gas and water molecule.

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]      .....(1)

Given mass of butane = 8.9541 g

Molar mass of butane = 58.12 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of butane}=\frac{8.9541g}{58.12g/mol}=0.154mol[/tex]

The chemical equation for the combustion of butane follows:

[tex]2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O[/tex]

By Stoichiometry of the reaction:

2 moles of butane produces 8 moles of carbon dioxide

So, 0.154 moles of butane will produce = [tex]\frac{8}{2}\times 0.154=0.616mol[/tex] of carbon dioxide

Now, calculating the mass of carbon dioxide by using equation 1, we get:

Moles of carbon dioxide = 0.616 moles

Mass of carbon dioxide = 44 g/mol

Putting values in equation 1, we get:

[tex]0.616mol=\frac{\text{Mass of carbon dioxide}}{44g/mol}\\\\\text{Mass of carbon dioxide}=27.1g[/tex]

Hence, the mass of carbon dioxide produced is 27.1 grams.

Answer:

(a) 282 kJ

(b) 67.4 Calories

Explanation:

(a) The molar enthalpy, ΔH = −2802.5 kJ/mol, means that the heat produced by the reaction is 2802.5 kJ per mol of glucose.

We can multiply the enthalpy by the number of moles of glucose to get the heat produced by the metabolism. Grams of glucose will be converted to moles using the molar mass of glucose (180.156 g/mol):

(18.1 g)(mol/180.156g)(2802.5 kJ/mol) = 282 kJ

(b) Using the result we obtained above, kJ will be converted to Calories using the conversion factor of 4.184J = 1 cal. Calorie with a capital C is the same as a kilocalorie.

(282 kJ)(1 cal/4.184J) = 67.4 kcal = 67.4 Calories

Answer:

For a: The amount of heat produced for given amount of glucose is -283.05 kJ

For b: The amount of heat produced for given amount of glucose is -67648.9 Cal

Explanation:

To calculate the number of moles, we use the equation:

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

Given mass of glucose = 18.1 g

Molar mass of glucose = 180.16 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of glucose}=\frac{18.1g}{180.16g/mol}=0.101mol[/tex]

The given chemical reaction follows:

[tex]C_6H_{12}O_6(s)+6O_2(g)\rightarrow 6CO_2(g)+6H_2O(l);\Delta H=-2802.5kJ/mol[/tex]

By Stoichiometry of the reaction:

When 1 mole of glucose is reacted, the amount of heat released is 2802.5 kJ

So, when 0.101 moles of glucose is reacted, the amount of heat released is [tex]\frac{2802.5}{1}\times 0.101=283.05kJ[/tex]

Hence, the amount of heat produced for given amount of glucose is -283.05 kJ

To convert the heat produced in kilo joules to calories, we use the conversion factor:

1 kJ = 239 Cal

So, [tex]-283.05kJ\times (\frac{239Cal}{1kJ})=-67648.9Cal[/tex]

Hence, the amount of heat produced for given amount of glucose is -67648.9 Cal

Answer:

a) C Cu = 1588.6 mg/L

b) C Cu = 1588600 μg/L

c) this concentration exceeds the acute freshwater criteria:

∴ C Cu = 1588.6 mg/L >> 0.065 mg/L

Explanation:

∴ C Cu = 2.5 E-2 mol/L

∴Mw Cu = 63.546 g/mol

a) C Cu = 2.5 E-2 mol/L * 63.546 g/mol = 1.5886 g/L

⇒ C Cu = 1.5886 g/L * ( 1000 mg/g ) = 1588,6 mg/L

b) C Cu =  1588.6 mg/L * ( μg / 0.001 mg ) = 1588600 μg/L

c) C Cu = 65 ppb * ( ppm / 1000ppb ) = 0.065 ppm = 0.065 mg/L

∴ ppm ≡ mg/L

⇒ C Cu = 1588.6 mg/L >> 0.065 mg/L; this concentration exceeds the acute freshwater criteria

Answer:

F (3.98)> O (3.44)> Cl (3.16)> N (3.04)> Kr (3.00)> Br (2.96)> I (2.66)> Xe (2.6)> S(2.58)

Explanation:

Electronegativity is described as the tendency or ability of an element to attract electron density or bond pair of electrons towards itself. It is a chemical property of an element.  

In the periodic table, the least electronegative element is caesium (0.79) and the most electronegative element is fluorine (3.98).

List of 9 most electronegative elements are:

F (3.98)> O (3.44)> Cl (3.16)> N (3.04)> Kr (3.00)> Br (2.96)> I (2.66)> Xe (2.6)> S(2.58).

Answer:

5x10^3 L

Explanation:

Charle's Law states that the volume of a fixed amount of gas maintained at constant pressure is directly proportional to the absolute temperature of the gas, for a constant amount of gas we can write:

[tex]\frac{V1}{T1}=\frac{V2}{T2}[/tex]

As the pressure of the balloon doesn't change, we can use Charle's Law to solve the problem. Firs we change the given temperatures to absolute temperature units ( °K), using the following relations:

°K=273,15+°C

Therefore:

V1=4.6 x 10^3 L, T1=273,15+31=304,15°K

V2=?,                   T2=273,15+56=329,15°K

[tex]V2=\frac{V1T2}{T1}=\frac{4.6*10^3*329,15}{304,15}=4978,1L=5*10^3L[/tex]

The new volume of the balloon is 5x10^3 L.

The volume of a helium-filled balloon at 56°C under constant pressure can be calculated using Charles's Law by converting temperatures to Kelvins and using the volume-temperature direct proportionality relationship.

The subject of this question involves the relationship between volume and temperature of a gas under constant pressure, described by Charles's Law. This law states that the volume of an ideal gas is directly proportional to its temperature (in Kelvins) when pressure is held constant. To use Charles's Law for calculating the new volume of a balloon, first, convert the temperatures from Celsius to Kelvins (K = °C + 273.15). The initial temperature is 31°C, which is 304.15K, and the final temperature is 56°C, which is 329.15K.

Apply Charles's Law using the formula V1/T1 = V2/T2. Let V1 be the initial volume (4.6 x 10³ L), T1 be the initial temperature (304.15K), V2 be the final volume, and T2 be the final temperature (329.15K). Solving for V2 gives:

V2 = (V1 x T2) / T1 = (4.6 x 10³ L x 329.15K) / 304.15K

Calculating this will give us the new volume V2 which we express in scientific notation.

Answer:

a. The buffering range is between 2.74 and 4.74.

b. The ratio of the formate to the formic acid is 10.23.

Explanation:

a. For every buffer solution, the optimal effective range is pH = pKa ± 1. Outside this range, the buffer does not work properly.

For the formic acid, the pKa is 3.74, thus the optimal range is between 2.74 and 4.74.

b. The Henderson-Hasselbalch equation is a chemical expression used to calculate the pH of a buffer knowing the ratio of the acid to base, or to calculate the ratio knowing the pH. The expression is:

[tex]pH = pKa + Log \frac{[A^{-}]}{[HA]}[/tex]

where [A^{-}] is the concentration of the conjugate base and [HA] is the concentration of the acid.

For a formic acid/potassium formate solution that has a pH of 4.75 and pka of 3.74:

[tex]pH - pKa =4.75 - 3.74 = 1.01 = Log \frac{[A^{-}]}{[HA]}[/tex]

[tex]\frac{[A^{-}]}{[HA]} = 10^{1.01}  = 10.23[/tex]

Answer : The moles of oxygen gas lost are 2.88 mole

Explanation :

According to the Avogadro's Law, the volume is directly proportional to the number of moles of the gas at constant pressure and temperature.

[tex]V\propto n[/tex]   (At constant temperature and pressure)

or,

[tex]\frac{V_1}{V_2}=\frac{n_1}{n_2}[/tex]

where,

[tex]V_1[/tex] = initial volume of oxygen gas = 12.6 L

[tex]V_2[/tex] = final volume of oxygen gas = 12.1 L

[tex]n_1[/tex] = initial moles of oxygen gas = 3.0 mole

[tex]n_2[/tex] = final moles of oxygen gas = ?

Now put all the given values in the above formula, we get the final moles of oxygen gas.

[tex]\frac{12.6L}{12.1L}=\frac{3.0mole}{n_2}[/tex]

[tex]n_2=2.88mole[/tex]

Therefore, the moles of oxygen gas lost are 2.88 mole

Answer: 1.86 m

Explanation:

Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.

[tex]Molality=\frac{n\times 1000}{W_s}[/tex]

where,

n = moles of solute

[tex]W_s[/tex] = weight of solvent in g

Given : 15.0 grams of [tex]MgCl_2[/tex] is present in 100 g of solution

Moles of solute [tex]H_2O=\frac{\text {given mass}}{\text {Molar mass}}=\frac{15.0g}{95.3g/mol}=0.16[/tex]

mass of solution = 100 g

density of solution = 1.127g/ml

mass of solvent = mass of solution - mass of solute = 100 g - 15.0g = 85.0 g

[tex]Molality=\frac{0.16\times 1000}{85.0g}=1.86[/tex]

Thus molality of a 15.0% by mass solution of [tex]MgCl_2[/tex] is 1.86m.

Answer:

Density=7.86g/cm³

Explanation:

To solve this exercise, we only need to convert units. To do so, we can use a three-step process:

In this case, the units we're given are kg and cm³, and the units we must convert them into are g and cm³. Keeping in mind the conversion factor of 1000 g = 1 kg, we convert the mass of the plate:

[tex]11.0kg*\frac{1000g}{1kg}=11000 g[/tex]

Then we divide the mass by the volume, that is already in cubic centimeters:

[tex]Density=\frac{11000g}{1400cm^{3}}=7.86\frac{g}{cm^{3} }[/tex]

Answer:

27,397.23 L would be needed for a successful trip.

Explanation:

The problem gives us the density (ρ) of the fuel, by telling us that there are 803 g of fuel in 1 L, in which case:

ρ=[tex]\frac{mass}{Volume}=\frac{803g}{1L}  =803\frac{g}{L}[/tex]

The required mass of fuel is 2.2 * 10⁴ kg, we can convert this value into g:

2.2 * 10⁴ kg * [tex]\frac{1000g}{1kg}[/tex] = 2.2 * 10⁷ g

We calculate the required volume (V), using the mass and density:

[tex]803 g/L = \frac{2.2*10^{7}g }{V} \\V=\frac{2.2*10^{7}g }{803g/L}\\ V=27397.26 L[/tex]

Thus 27,397.23 L would be needed for a successful trip.

Answer:

The correct option is: a) yes

Explanation:

Chromium is a chemical element that belongs to the group 6 of the periodic table and a member of the d-block. It is a hard and brittle transition metal having atomic number 24.

The hydrated chlorides of chromium can be obtained by dissolving the chromium metal in sulphuric acid or hydrochloric acid.

Answer:

Ionic bonding: C

Covalent bonding: B

Metallic bonding: D

Pauli exclusion principle: A

Explanation:

All the electrons in 1 atom are characterized by a series of 4 numbers, known as quantum numbers. These numbers (n, l, ml, ms) describe the state of each electron (in which level, sublevel, orbital it is and its spin). For 2 electrons to coexist in the same atom they must differ in at least of these numbers. If they occupy the same level, sublevel and orbital, then they must have different (and opposite) spins. This is known as Pauli exclusion principle.

Also, to gain stability atoms can gain, lose or share electrons. In doing so they form bonds. There are 3 kinds of bonds:

Answer:

The pressure in the vessel is 13,3 atm.

Explanation:

The global reaction is:

2 HCl (aq)+ CaCO₃ (s) → CaCl₂(aq)+ H₂O(l)+ CO₂(g)

The increase in the pressure of the vessel after the reaction is by formation of a gas (CO₂). So we have to find the produced moles of this gas and apply the gas ideal law to find the pressure.

We have to find the limit reactant, to do so, we have to calculate the moles of each reactant in the reaction, the one that have the less moles will be the limit reactant:

HCl:

1,0L × (35/100) × (1000 cm³/1L) × (1,28 g/ 1cm³) × (1mol HCl/ 36,46 g) ÷ 2mol

(Concentration)      (L to cm³)         (cm³ to g)      (g to mol)  (moles of reaction)

moles of HCl= 6,14 mol

CaCo₃:

   1,0 kg     ×       (96/100)                ×   (1000 g/1kg) × (1 mol/100,09g)

(Limestone) (CaCo₃ in limestone)          (kg to g)            (g to mol)

moles of CaCo₃= 9,59 mol

So, reactant limit is HCl

This reaction have a yield of 97%. So, the CO₂ moles are:

6,14 mol × 97÷ = 5,96 mol CO₂

The ideal gas formula to obtain pressure is:

P = nRT/V

Where: n = 5,96mol; R= 0,082 atm×L/mol×K; T = 273,15 (until STP conditions) and V= 10,0 L

Replacing this values in the equation the pressure is

P = 13,3 atm

I hope it helps!

Answer:

Confusion is eliminated when the correct measurement is applied

Explanation:

A measurement standard is often described as the reference to which other measurements are judged or based on. Most times, measurement standard are used to determine how accurate an experimental measurement is.

A measurement standard may not eliminate confusion when the correct measurement is applied. If the scientist does not agree with the measurement standard being used, then problem will set in.

The statement 'A standard need not agree with a previously defined size' does not describe a measurement standard because consistency with previously defined sizes is crucial for accuracy. Measurement standards ensure comparability of data across various domains, driving advancements in both science and technology.

The statement that does not describe a measurement standard is 'A standard need not agree with a previously defined size.' Measurement standards must indeed be consistent with previously defined sizes to ensure accuracy and eliminate confusion. Measurement standards are crucial to avoid ambiguity and to make sure that experimental data from different labs can be compared accurately. By having unchanging standards, like for the size of a meter, we ensure that a unit of measure remains constant over time, which is indispensable for consistency in scientific experiments and technology development.

Furthermore, the correct measurement standards are essential for the growth of technology as they allow for precise and reliable data. The relationship between science and technology is symbiotic; they drive each other's advancement. As technology improves, new measurement techniques become possible, leading to more detailed observations and more accurate scientific experiments, which in turn can further advance technology.

Final answer:

To determine the volume of a 1:5000 w/v solution that can be made from 125 mL of a 0.2% w/v solution of benzalkonium chloride, we calculate the mass of the solute in the initial solution and divide it by the concentration of the desired dilute solution, giving us 1250 mL.

Explanation:

To calculate the target volume of a 1:5000 w/v solution from a 0.2% w/v solution of benzalkonium chloride, we first need to understand the concentration terms. A 0.2% w/v solution means 0.2 g of solute is present in every 100 mL of solution. For a 1:5000 w/v solution, there is 1 g of solute in 5000 mL of solution, or essentially y, 0.0002 g/mL (1 g / 5000 mL).

We then figure out how much benzalkonium chloride is in the initial 125 mL of the 0.2% solution. Since 0.2% w/v is equivalent to 0.2 g/100 mL, we have 0.2 g in 100 mL, hence in 125 mL, we'll have 0.25 g (since 125 mL is 1.25 times 100 mL).

To work out how much of the 1:5000 solution we can make with 0.25 g, we use the concentration of the 1:5000 solution, which is 0.0002 g/mL. Dividing the total available mass of solute (0.25 g) by the concentration (0.0002 g/mL), we get 1250 mL. Thus, the answer is a. 1250 mL.

Explanation:

Relation between molarity and molar mass is as follows.

             Molarity = [tex]\frac{mass}{\text{molar mass}} \times \frac{1000}{V (in ml)}[/tex]

It is given that mass is 1 mg/ml which is also equal to [tex]10^{-3}[/tex] g.

Molecular mass = 58 Da = 58 g/mol

Volume = 1 ml

Therefore, calculate molarity as follows.

         Molarity = [tex]\frac{mass}{\text{molar mass}} \times \frac{1000}{V (in ml)}[/tex]

                       = [tex]\frac{10^{-3}g}{58 g/mol} \times \frac{1000}{1 ml}[/tex]  

                       = 0.0172 molar

It is known that 1 molar equals 1000 millimolar.

So,               0.0172 molar = [tex]0.0172 molar \times \frac{1000 millimolar}{1 molar}[/tex]

                                          = 17.2 millimolar

Thus, we can conclude that the concentration of given solution is 17.2 millimolar.

Answer:

a) Analyte: lead. Sample: paint.

b) Analyte: nitrate. Sample: soil.

c) Analyte: citric acid. Sample: Lime

1) Lead: Analyte.

2) Paint chips: Sample.

3) Soil: Sample.

4) Nitrate: Analyte.

5) Lime wedge: Sample.

6) Citric acid: Analyte.

Explanation:

A sample is a portion of material selected from a larger quantity of material while an analyte is the chemical of the system that will be analysed.

Thus:

a) Analyte is lead while you must take a sample of paint to analyze this lead.

b) Analyte is the nitrate while sample must be soil.

c) Analyte is citric acid and lime is the sample

1) Lead: Analyte.

2) Paint chips: Sample.

3) Soil: Sample.

4) Nitrate: Analyte.

5) Lime wedge: Sample.

6) Citric acid: Analyte.

Final answer:

In scenarios (a) to (c), the samples are paint chips, soil, and lime wedge, respectively, and the analytes are lead, nitrate, and citric acid, respectively. Item identifications are lead and nitrate as analytes, and paint chips, soil, and lime wedge as samples.

Explanation:

In analytical chemistry, the analyte is the substance whose chemical content is being measured or analyzed, and the sample is the material or product that contains the analyte. Here are the identifications for each scenario:

To clarify, here are the identifications for each item listed:

Answer : The final temperature will be, 292 K

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

[tex]q_1=-q_2[/tex]

[tex]m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)[/tex]

where,

[tex]c_1[/tex] = specific heat of ice = [tex]2.05J/g.K[/tex]

[tex]c_2[/tex] = specific heat of beer = [tex]4.2J/g.K[/tex]

[tex]m_1[/tex] = mass of ice = 50 g

[tex]m_2[/tex] = mass of beer = 450 g

[tex]T_f[/tex] = final temperature = ?

[tex]T_1[/tex] = initial temperature of ice = [tex]0^oC=273+0=273K[/tex]

[tex]T_2[/tex] = initial temperature of beer = [tex]20^oC=273+20=293K[/tex]

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

[tex]50g\times 2.05J/g.K\times (T_f-273)K=-450g\times 4.2J/g.K\times (T_f-293)K[/tex]

[tex]T_f=291.971K\approx 292K[/tex]

Therefore, the final temperature will be, 292 K

Answer: The pH of resulting solution is 8.7

Explanation:

To calculate the number of moles for given molarity, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex]

Molarity of TRIS acid solution = 0.1 M

Volume of solution = 50 mL

Putting values in above equation, we get:

[tex]0.1M=\frac{\text{Moles of TRIS acid}\times 1000}{50mL}\\\\\text{Moles of TRIS acid}=0.005mol[/tex]

Molarity of TRIS base solution = 0.2 M

Volume of solution = 60 mL

Putting values in above equation, we get:

[tex]0.2M=\frac{\text{Moles of TRIS base}\times 1000}{60mL}\\\\\text{Moles of TRIS base}=0.012mol[/tex]

Volume of solution = 50 + 60 = 110 mL = 0.11 L    (Conversion factor:  1 L = 1000 mL)

[tex]pH=pK_a+\log(\frac{[salt]}{[acid]})[/tex]

[tex]pH=pK_a+\log(\frac{[\text{TRIS base}]}{[\text{TRIS acid}]})[/tex]

We are given:

[tex]pK_a[/tex] = negative logarithm of acid dissociation constant of TRIS acid = 8.3

[tex][\text{TRIS acid}]=\frac{0.005}{0.11}[/tex]

[tex][\text{TRIS base}]=\frac{0.012}{0.11}[/tex]

pH = ?

Putting values in above equation, we get:

[tex]pH=8.3+\log(\frac{0.012/0.11}{0.005/0.11})\\\\pH=8.7[/tex]

Hence, the pH of resulting solution is 8.7

Answer: 20.1 kg.

Explanation:

To calculate mass of a substance, we use the equation:

[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]

We are given:

Density of gasoline = [tex]0.737kg/L[/tex]

Volume of gasoline = [tex]7.2gallons=7.2\times 3.785L=27.252L[/tex]     (Conversion factor: [tex]1gal=3.785L[/tex] )

Putting values in above equation, we get:

[tex]0.737kg/L=\frac{\text{Mass of gasoline}}{27.252L}\\\\\text{Mass of gasoline}=20.084kg[/tex]

The rule apply for multiplication and division is :

The least precise number present after the decimal point determines the number of significant figures in the answer, thus the answer musty have three significant digits.

Hence, the mass added is 20.1 kg.

Answer:

Variability of data around the mean is best expressed through the standard deviation.

Explanation:

We have some statistical metrics, and we need to find the best measure of the variability of data around the mean.  

First variable is mode. Mode of a set of data values is the value that appears most often. It means the most frequent data.  

Second variable is Median. The median is a simple measure of central tendency. It is located around 50% of data range.  

Third variable is standard deviation. The standard deviation is found by taking the square root of the average of the squared deviations of the values subtracted from their average value. It means, it will show the dispersion of data around the average value.

According to the definition presents for the three variables, Standard deviation is the most appropriate measure of variability of data around the mean.  

Answer:

Correct answer is: Variability of data around the mean is best expressed through the standard deviation.

Explanation:

Got it correct on Plato Hope This helped :D

Answer:

The answer is 0.005097 mL.

Explanation:

Significant figures : The figures in a number which express the value of the magnitude of a quantity to a specific degree of an accuracy is known as significant digits.

Given :

[tex]x=\frac{(1.145 x 109 g/mol)\times (0.0035 mol)}{(8.57 x 10 g/mL)}[/tex]

[tex]x=\frac{0.4368 g}{8.57\times 10 g/mL}=0.005097053 mL[/tex]

[tex]x=0.005097053 mL\approx 0.005097 mL[/tex]