Answer: The amount of heat needed to melt the given amount of octane is 84.6 kJ
Explanation:
We know:
Boiling point of Octane = [tex]125.6^oC[/tex]
Few processes involved are:
(1): [tex]C_8H_{18} (s) (-57^oC, 219K) \rightleftharpoons C_8H_{18}(s) (-57^oC, 219K[/tex]
(2): [tex]C_8H_{18}(l) (-57^oC, 219K) \rightleftharpoons C_8H_{18}(l) (99.2^oC,372.2K)[/tex]
Calculating the heat absorbed for the process having same temperature:
[tex]q=n\times \Delta H_{(f)}[/tex] ......(i)
where,
q is the amount of heat absorbed, n is the moles of sample and [tex]\Delta H_{(f)}[/tex] is the enthalpy of fusion
Calculating the heat released for the process having different temperature:
[tex]q=n\times C_{l}\times (T_2-T_1)[/tex] ......(ii)
where,
n = moles of sample
[tex]C_{l}[/tex] = specific heat of liquid
[tex]T_2\text{ and }T_1[/tex] are final and initial temperatures respectively
The number of moles is defined as the ratio of the mass of a substance to its molar mass.
The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(3)
Given mass of octane = 160. g
Molar mass of octane = 114.23 g/mol
Plugging values in equation 3:
[tex]\text{Moles of octane }=\frac{160.g}{114.23g/mol}=1.40 mol[/tex]
For process 1:We are given:
[tex]n=1.40mol\\\Delta H_{fusion}=20.740 kJ/mol[/tex]
Putting values in equation (i), we get:
[tex]q_1=1.40mol\times 20.470kJ/mol\\\\q_1=28.658kJ[/tex]
For process 2:We are given:
[tex]n=1.40mol\\C=255.68J/mol^oC\\T_2=99.2^oC\\T_1=-57^oC[/tex]
Putting values in equation (ii), we get:
[tex]q_2=1.40mol\times 255.68J/mol^oC\times (99.2-(-57))\\\\q_2=55912.10J=55.912kJ[/tex]
Calculating the total amount of heat released:
[tex]Q=q_1+q_2[/tex]
[tex]Q=[(28.658)+(55.912)]kJ=84.6kJ[/tex]
Hence, the amount of heat needed to melt the given amount of octane is 84.6 kJ
Radioactive radon-222, found in many homes, is a potential health hazard. The half-life of radon-222 is 3.82 days. If you begin with 35.3 mg of this isotope, what mass remains after 10.1 days have passed
Answer:
5.65mg of the isotope remains
Explanation:
The radioactive decay follows the equation:
Ln[A] = -kt + ln[A]₀
Where [A] is amount of the isotope after time t, k is decay constant, and [A]₀ is initial amount of the isotope.
k = ln 2 / Half-life
k = ln 2 / 3.82 days
k = 0.18145days⁻¹
Replacing:
Ln[A] = -0.18145days⁻¹*10.1days + ln[35.3mg]
ln[A] = 1.7312
[A] = 5.65mg of the isotope remains
Question 8 (5 points)
(08.02 MC)
A 10 M concentrated stock solution of NaCl was used to prepare 5 liters of diluted 1 M solution. Which of the following statements is true about
the
process used to achieve this required dilution? (5 points)
O a
The volume of stock solution used was less than 0.4 liters.
Oь
The volume of stock solution used was more than 5 liters.
Ос
The volume of the solvent used was less than 0.4 liters.
Od
The volume of the solvent used was less than 5 liters.
Answer:
d . The volume of the solvent used was less than 5 liters.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to calculate the volume of the stock (initial) solution by using the following equation:
[tex]M_1V_1=M_2V_2[/tex]
Thus, we solve for, V1, which stands for the aforementioned volume of stock solution:
[tex]V_1=\frac{M_2V_2}{M_1}[/tex]
Then, we plug in to obtain:
[tex]V_1=\frac{5L*1M}{10M}\\\\V_1=0.5L[/tex]
Now, since the final volume was 5 L, we can infer that the volume of solvent is 4.5 L and that of the stock solution 0.5 L for a total of 5 L of diluted solution; therefore, the correct reasoning is d . The volume of the solvent used was less than 5 liters.
Regards!
Answer:
The volume of the solvent is less than 5
Explanation:
Choose the substance with the highest viscosity. Choose the substance with the highest viscosity. CF4 C7H16 C2H4I2 HOCH2CH2CH2CH2OH (CH3CH2)2CO
Answer:
C2H4I2
Explanation:
Viscosity of a fluid has to do with the internal friction between the internal layers of the fluid.
Molecular weight is found to be related to the viscosity of a fluid even though the relationship may not be strictly linear.
However, the greater the molecular weight of a substance, the greater the viscosity of the material.
Since C2H4I2 has the greatest molecular weight (281.86 g/mol), it is also expected to display the greatest viscosity among all the compounds listed in the question.
The reaction of bromine gas with chlorine gas, shown here, has a Keq value of 7.20 at 200°C. If a closed vessel was charged with the two reactants, each at an initial concentration of 0.200 M, but with no initial concentration of BrCl, what would be the equilibrium concentration of Br2, Cl2 and BrCl(g)?
Br2(g) + Cl2(g) ↔ 2BrCl(g) K = 7.20
Answer:
[tex][Cl_2]=[Br_2]=0.856M[/tex]
[tex][BrCl]=0.229M[/tex]
Explanation:
Hello there!
In this case, for this chemical equilibrium problem, it turns out possible for us to solve for the equilibrium concentrations by firstly setting up the equilibrium expression:
[tex]Keq=\frac{[BrCl]^2}{[Cl_2][Br_2]}[/tex]
Thus, by plugging in an ICE chart, in terms of x (reaction extent), we can write:
[tex]7.20=\frac{(2x)^2}{(0.200-x)^2}[/tex]
And could be solved for x as follows:
[tex]\sqrt{7.20} =\sqrt{\frac{(2x)^2}{(0.200-x)^2} } \\\\2.68=\frac{2x}{0.200-x} \\\\x=0.1146M[/tex]
Therefore, the equilibrium concentrations turn out to be:
[tex][Cl_2]=[Br_2]=0.200M-0.1146M=0.856M[/tex]
[tex][BrCl]=2*0.1146M=0.229M[/tex]
Regards!
For a phase change, H0 = 2 kJ/mol and A S0 = 0.017 kJ/(K•mol). What are
AG and the spontaneity of the phase change at 500 K?
Answer:
ΔG = -6.5kJ/mol at 500K
Explanation:
We can find ΔG of a reaction using ΔH, ΔS and absolute temperature with the equation:
ΔG = ΔH - TΔS
Computing the values in the problem:
ΔG = ?
ΔH = 2kJ/mol
T = 500K
And ΔS = 0.017kJ/(K•mol)
Replacing:
ΔG = 2kJ/mol - 500K*0.017kJ/(K•mol)
ΔG = 2kJ/mol - 8.5kJ/mol
ΔG = -6.5kJ/mol at 500KThe average molecular speed in a sample of Ar gas at a certain temperature is 213 m/s. The average molecular speed in a sample of Ne gas is ______ m/s at the same temperature.
Answer:
300m/s is the average molecular speed of Ne
Explanation:
Based on Graham's law, the ratio of speed of two gases under constant temperature is equal to the square root of the inverse of their molar masses. The equation is:
v1 / v2 = √m2 / √m1
Where v is the speed of the gas and m the molar mass of the gas
Assuming gas 1 is Argon and gas 2 is Neon:
v1 = 213m/s
v2 = ?
m2 = 20.18g/mol
m1 = 39.948g/mol
213m/s / v2 = √20.18g/mol / √39.948g/mol
v2 = 213m/s / 0.71074
v2 = 300m/s is the average molecular speed of Ne
HELP WITH MY 2 QUESTION
1-What type of packaging is used for milk?
2-How do the physical and chemical properties (material, reactivity, shape, hardness, color)
of each packaging type help create new and better product?
Answer:
1. Milk comes in a range of sizes and forms, and is packaged in a range of materials. Glass bottles, plastic coated paper board, blow mold nonreturnable polyethylene containers, plastic pouches, and returnable rigid polycarbonate containers are among the commercial containers available.
Dont know the answer to the second one sorry :(
In this lab, you will be making solutions of potassium permanganate (KMnO4), which has a formula weight of 158.04 g/mole. Remember to show your calculations and include tne Correct unnits in your answers
a) How many grams of KMnO4 would you need to make 1 L of a 2M solution?
b) How many grams of KMnO4 would you need to make 350 mL of a 0.75 M solution?
c) How many grams of KMnO4 would you need to make 80 mL of a 0.01 M solution?
Answer:
A. Mass of KMnO₄ = 316.08 g
B. Mass of KMnO₄ = 41.49 g
C. Mass of KMnO₄ = 0.13 g.
Explanation:
A. Determination of the mass of KMnO₄
We'll begin by determining the number of mole of KMnO₄ in the solution. This can be obtained as follow:
Volume = 1 L
Molarity = 2 M
Mole of KMnO₄ =?
Mole = Molarity × Volume
Mole of KMnO₄ = 2 × 1
Mole of KMnO₄ = 2 moles
Finally, we shall determine the mass of KMnO₄. This can be obtained as follow:
Mole of KMnO₄ = 2 moles
Molar mass of KMnO₄ = 158.04 g/mole
Mass of KMnO₄ =?
Mass = mole × molar mass
Mass of KMnO₄ = 2 × 158.04
Mass of KMnO₄ = 316.08 g
B. Determination of the mass of KMnO₄
We'll begin by determining the number of mole of KMnO₄ in the solution. This can be obtained as follow:
Volume = 350 mL = 350/1000 = 0.35 L
Molarity = 0.75 M
Mole of KMnO₄ =?
Mole = Molarity × Volume
Mole of KMnO₄ = 0.75 × 0.35
Mole of KMnO₄ = 0.2625 mole
Finally, we shall determine the mass of KMnO₄. This can be obtained as follow:
Mole of KMnO₄ = 0.2625 mole
Molar mass of KMnO₄ = 158.04 g/mole
Mass of KMnO₄ =?
Mass = mole × molar mass
Mass of KMnO₄ = 0.2625 × 158.04
Mass of KMnO₄ = 41.49 g
C. Determination of the mass of KMnO₄
We'll begin by determining the number of mole of KMnO₄ in the solution. This can be obtained as follow:
Volume = 80 mL = 80/1000 = 0.08 L
Molarity = 0.01 M
Mole of KMnO₄ =?
Mole = Molarity × Volume
Mole of KMnO₄ = 0.01 × 0.08
Mole of KMnO₄ = 0.0008 mole
Finally, we shall determine the mass of KMnO₄. This can be obtained as follow:
Mole of KMnO₄ = 0.0008 mole
Molar mass of KMnO₄ = 158.04 g/mole
Mass of KMnO₄ =?
Mass = mole × molar mass
Mass of KMnO₄ = 0.0008 × 158.04
Mass of KMnO₄ = 0.13 g
A typical dollar bill is 15.50 cm by 6.50 cm.
Calculate the surface area in square meters, square centimeters and square nanometers
Answer:
0.010075 m²100.75 cm²1.0075x10¹⁶ nm²Explanation:
As the measurements are given to us in centimeters, let's start by calculating the surface area in square centimeters:
Area = 15.50 cm * 6.50 cm = 100.75 cm²Now we convert 100.75 cm² to m², as follows:
100.75 cm² * [tex](\frac{1m}{100cm}) ^2[/tex] = 0.010075 m²Finally we convert 0.010075 m² to nm², as follows:
0.010075 m² * [tex](\frac{1nm}{1x10^{-9}m}) ^2[/tex] = 1.0075x10¹⁶ nm²Which of the following compounds is composed of Al,Si,O and H
A.Epsom salt
B.Limestone
C.Clay
D.Urea
Explanation:
the answer is clay as seen from the picture
in
What are common features in
relation to electron arrangement
across a period and down a group
Igor s elements on the periodic
table.
Answer:
The order of sub-energy levels ( s » p » d » f )
The number of orbitals ( s » 2 ), ( p » 6), ( d» 10), ( f » 14 )
For an aqueous solution of sucrose (C12H22O11), determine:
the number of moles of sucrose in 1.50 L of this solution
Answer:WHAT GRADE WORK IS THIS ?
Explanation:
Which is most likely the reason the U.S government made such an investment
Answer:
The U.S government made such an investment because they're most likely to reduce harmful emissions.
What is leukemia
A
B
C
D
C. Cancer that affects the blood
Hope I could help!
NO2 (nitrogen dioxide) is a Greenhouse Gas that can be produced in car engines. The average diesel truck produces 80g of NO2 for every 1000 kilometers it drives. How many moles of NO2 are in 80g of NO2?
Answer:
[tex]1.74molNO_2[/tex]
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to perform mole-mass relationships by using the molar mass of the involved substance, in this case NO2, which is 46.0 g/mol; then we just set up a conversion factor like the one shown below:
[tex]80g*\frac{1mol}{46g}\\\\1.74molNO_2[/tex]
Regards!
With the temperature held constant at 300 K, use the Select mass slider to place weights on the lid. Record the pressure and volume of the gas for each added mass.
Added mass on the lid Total mass Pressure Volume
(Lid + added mass)
0 kg 10 kg
10 kg 20 kg
20 kg 30 kg
30 kg 40 kg
Solution :
When the temperature is held constant at 300 K, we can use the Select mass slider in order to place the weights on the lid. And we record the pressure and also the volume of the gas for each of the added mass.
Added mass on the lid Total mass Pressure Volume
0 kg 10 kg 98.1 [tex]N/m^2[/tex] [tex]2.54 \ m^3[/tex]
10 kg 20 kg [tex]196.2 \ N/ m^2[/tex] [tex]1.27 \ m^3[/tex]
20 kg 30 kg [tex]294.3 \ N/ m^2[/tex] [tex]0.85 \ m^3[/tex]
30 kg 40 kg [tex]392.4 \ N/ m^2[/tex] [tex]0.43 \ m^3[/tex]
As the pressure increases at a constant temperature, the volume of the gas decreases.
Thus we can see that the pressure is inversely proportional to the volume.
Rank the solutions below in order of increasing acidity. (Drag and drop into the appropriate area)
0.01 M CH3COOH
0.1 M NaOH
0.01 M H2SO4
3 M NH3
0.1 M HCl
Answer:
0.1 M NaOH, 3 M NH3, 0.01 M CH3COOH, 0.01 M H2SO4, 0.1 M HCl
Explanation:
Strong acids are more acids than weak acids. In the same way, strong bases are more basic than weak bases that are in the same concentration.
Then, the more concentrated acid or base will be more acidic or basic.
CH3COOH. Weak acid
NaOH. Strong base
H2SO4. Strong acid
NH3. Weak base.
HCl. Strong acid
The less acid (More basic):
0.1 M NaOH, 3 M NH3, 0.01 M CH3COOH, 0.01 M H2SO4, 0.1 M HClStrong base, weak base, weak acid, diluted strong acid, undiluted strong acid
g The activation energy for diffusion depends on the strength of atomic bonds. Higher strength bonds result in higher activation energies. How do you predict the activation energy for self diffusion will be related to melting temperature
Answer:
Yes, the activation energy for self diffusion is related to melting temperature.
Explanation:
The activation energy for self diffusion is related to melting temperature because bonds result due to activation energies and we know that bonds are break down when the temperature is increases and as a result the solid substance melts and change into liquid state. When the bonds between atoms are broken down with the use of heat energy the state of matter changes so we can say that activation energy is related to temperature.
A 29.00 mL sample of an H2SO4 solution of unknown concentration is titrated with a 0.1122 M KOH solution. A volume of 43.22 mL of KOH was required to reach the equivalence point.
What is the concentration of the unknown H2SO4 solution? Express your answer using four significant figures.
Answer: The concentration of the unknown [tex]H_{2}SO_{4}[/tex] solution is 0.167 M.
Explanation:
Given: [tex]V_{1}[/tex] = 29.00 mL, [tex]M_{1}[/tex] = ?
[tex]V_{2}[/tex] = 43.22 mL, [tex]M_{2}[/tex] = 0.1122 M
Formula used to calculate concentration of the unknown [tex]H_{2}SO_{4}[/tex] solution is as follows.
[tex]M_{1}V_{1} = M_{2}V_{2}[/tex]
Substitute the values into above formula as follows.
[tex]M_{1}V_{1} = M_{2}V_{2}\\M_{1} \times 29.00 mL = 0.1122 M \times 43.22 mL\\M_{1} = 0.167 M[/tex]
Thus, we can conclude that concentration of the unknown [tex]H_{2}SO_{4}[/tex] solution is 0.167 M.
Look at the equation CO2 + H2 = CO + H20.
CO is a ________
reactant and a product
reactant
product
Answer:
CO is considered as a product.
Explanation:
A general chemical equation for a combination reaction follows:
To write a chemical equation, we must follow some of the rules:
The reactants must be written on the left side of the direction arrow.
A '+' sign is written between the reactants, when more than one reactants are present.
An arrow is added after all the reactants are written in the direction where reaction is taking place. Here, the reaction is taking place in forward direction.
The products must be written on the right side of the direction arrow.
A '+' sign is written between the products, when more than one products are present.
For the given chemical equation:
are the reactants in the reaction and are the products in the reaction.
Hence, CO is considered as a product.
what is the similarities of freezing of water and electrolysis of water
Answer:
Electrolysis of water is the process of using electricity to decompose water into oxygen and hydrogen gas. Hydrogen gas released in this way can be used as hydrogen fuel, or remixed with the oxygen to create oxyhydrogen gas, which is used in welding and other applications.Ordinarily, the freezing point of water and melting point is 0 °C or 32 °F. The temperature may be lower if supercooling occurs or if there are impurities present in the water which could cause freezing point depression to occur. Under certain conditions, water may remain a liquid as cold as -40 to -42°
Explanation:
Use the drop-down menus to classify each of the following as an addition, substitution, elimination, or
condensation reaction.
CH3CHO+H2O → CH3OCH3
CH,CICH CI + Zn → C2H4 + ZnCl2
CH3CH3Br + OH – CH3CH3OH + Br
2CH2COOH
>>
(CH3CO)20 + H20
Answer:
CH3CHO+H2O → CH3OCH3 - addition
CH,CICH CI + Zn → C2H4 + ZnCl2 - elimination
CH3CH3Br + OH – CH3CH3OH + Br - substitution
2CH2COOH >>(CH3CO)20 + H20 - condensation
Explanation:
An addition reaction is a reaction in which a specie is added across the double bond as we can see in CH3CHO+H2O → CH3OCH3.
In an elimination reaction, a small molecule is lost from a saturated compound to form the corresponding unsaturated compound as in CH,CICH CI + Zn → C2H4 + ZnCl2
In a substitution reaction, a chemical moiety replaces another in a molecule as in; CH3CH3Br + OH – CH3CH3OH + Br .
A condensation reaction is in which two molecules are joined together to form a bigger molecule as in; 2CH2COOH >>(CH3CO)20 + H20.
Answer:
addition
elimination
substitution
condensation
For the equilibrium that exists in an aqueous solution of nitrous acid (HNO2, a weak acid), the equilibrium constant expression is: _________
a. K = [ H+] [NO2-] / [HNO2]
b. K = [ H+] [N] [O]2 / [HNO2]
c. K = [ H+] [NO2-] / [HNO2]
d. K = [H+]2 [NO2-] / [HNO2]
e. None of these
Answer: For the equilibrium that exists in an aqueous solution of nitrous acid (HNO2, a weak acid), the equilibrium constant expression is K = [ H+] [NO2-] / [HNO2].
Explanation:
An expression that depicts the ratio of products and reactants raised to the power of their coefficients at equilibrium is called equilibrium constant.
An equilibrium constant is denoted by the symbol 'K'.
For example, the dissociation of nitrous acid in aqueous solution is as follows.
[tex]HNO_{2} \rightleftharpoons H^{+} + NO^{-}_{2}[/tex]
Hence, its expression for equilibrium constant is as follows.
[tex]K = \frac{[H^{+}][NO^{-}_{2}]}{[HNO_{2}]}[/tex]
Thus, we can conclude that for the equilibrium that exists in an aqueous solution of nitrous acid (HNO2, a weak acid), the equilibrium constant expression is K = [ H+] [NO2-] / [HNO2].
The reduction of carbon dioxide by hydrogen gas takes place at 420°C to produce water vapor and carbon monoxide. The equation for
this reaction at equilibrium is shown below.
H2(g) + CO2(g) = H2O(g) + CO(g)
Which of the following changes in concentration occur when more water vapor is added to the system under equilibrium conditions?
O A [Hz] decreases, [CO] decreases, [CO] increases
OB (H) increases, (CO2) increases, [CO] decreases
OC [Hz, increases, [CO2, increases, [CO] increases
D. [H2, decreases, [CO] decreases, [CO] decreases
A solution made by dissolving 9.81 g of a nonvolatile, nonelectrolyte in 90.0 g of water boiled at 100.37 °C at 760 mm Hg. What is the molar mass of the substance? [kp = 0.51 °c/m]
Answer:
151 g/mol
Explanation:
In order to solve this problem we need to keep in mind the formula for the boiling point elevation:
ΔT = Kb * m * iWhere:
ΔT is the temperature difference between the boiling point of the solution and that of pure water. 100.37 °C - 100.00 °C = 0.37 °C.m is the molarity of the solutioni is the van't Hoff factor. As the solute is a nonelectrolyte, the factor is 1.Input the data and calculate m:
0.37 °C = 0.51 °C/m * m * 1 m =0.72 mWe now can calculate the number of moles of the substance, using the definition of molarity:
molarity = moles of solute / kg of solventIn this case kg of solvent = 90.0 g / 1000 = 0.090 kg
0.72 m = moles / 0.090 kgmoles = 0.065 molFinally we calculate the molar mass, using the number of moles and the mass:
9.81 g / 0.065 mol = 151 g/molFor the reaction...
H2 (g) + Cl2 (g) <=> 2HCl (g) + heat
If the pressure in the system is increased, the equilibrium position will _______.
a)shift to the left
b)shift to the right
c)not shift position at all as 2 mol gas <=> 2 mol gas
Answer:
A: Shift to the left
Explanation:
The reaction would try to use up the added heat by making more H 2 and Cl 2.
A molecule or ion that donates the hydrogen in a hydrogen bond is a hydrogen bond donor
a. True
b. False
Answer:
True
Explanation:
Hydrogen bonding is a type of intermolecular interaction that occurs when hydrogen is bonded to a highly electronegative atom.
We define the term ''hydrogen bond donor'' as the molecule that supplies the hydrogen atom in the hydrogen bond.
Hence, it is true that the molecule or ion that donates the hydrogen in a hydrogen bond is a hydrogen bond donor
An example of competition between members of two different species is
A) mold growing on a dead tree that has fallen in the forest
B) purple loosestrife plants growing in the same wet areas as cattail plants
C) a coyote feeding on the remains of a deer that died of starvation
D) two male turkeys displaying mating behaviors to attract a female turkey
Pls help
Answer:
B) purple loosestrife plants growing in the same wet areas as cattail plants
Explanation:
Competition between members of two species means that two different species are "fighting for" or desire the same resource. This resource could be shelter, food, territory, etc.
An organism cannot compete if it is already dead, so that eliminates answer choices A and C. Answer choice D is wrong because these are two organisms of the same species.
Answer choice B is correct because there are two different species competing for one area of land.
El agua del mar contiene aproximadamente un 3,0 % m/v de sal (NaCl, 58,44 g/mol), (asuma que es la única fuente de cloruros) si una fábrica de baterías para carro provoca un derrame de material con plomo(II). La concentración máxima (en g/L) de plomo(II) que puede contener el agua marina es:_______________
Kps=1,6x10^5
Answer:
s = 4.41 g/L.
Explanation:
¡Hola!
En este caso, considerando el escenario dado, se hace necesario para nosotros saber que la posible reacción de disociación la experimenta el cloruro de plomo (II) como se muestra a continuación:
[tex]PbCl_2(s)\rightleftharpoons 2Cl^-(aq)+Pb^{2+}(aq)[/tex]
Lo cual hace que la expresión de equilibrio se calcule como:
[tex]Ksp=[Pb^{2+}][Cl^-]^2[/tex]
Y que en términos de la solubilidad molar, s, se resuelve como:
[tex]1.6x10^{-5}=s(2s)^2\\\\1.6x10^{-5}=4s^3\\\\s=\sqrt[3]{\frac{1.6x10^{-5}}{4} } \\\\s=0.0159molPbCl_2/L[/tex]
Ahora, convertimos este valor a g/L al multiplicarlo por la masa molar del cloruro de plomo (II):
[tex]s=0.0159molPbCl_2/L*\frac{278.1gmolPbCl_2}{1molmolPbCl_2} \\\\s=4.41g/L[/tex]
¡Saludos!
HELP WITH MY 2 QUESTION
1-What type of packaging is used for milk?
2-How do the physical and chemical properties (material, reactivity, shape, hardness, color)
Answer:
Answer to number one
Explanation:
Polyethylene terephthalate (PET) is a plastic material used for milk packaging.