Answer
Explanation
Given:
2Al + 3Cl₂ 2AlCl₃
Molar mass of AlCl3 => 133 g/mol
Moles of Cl₂ = 25.0 mol
What to find:
The grams of aluminum chloride that can be produced.
Step-by-step solution:
Step 1: Determine the mole of AlCl₃ produced.
From the given chemical equation for the reaction,
3 mol Cl₂ produce 2 mol AlCl₃
So 25.0 mol Cl₂ will produce
[tex]\frac{25.0mol\text{ }Cl_2\times2mol\text{ }AlCl₃}{3mol\text{ }Cl_2}=16.67mol\text{ }AlCl₃[/tex]16.67 moles of AlCl₃ produced.
Step 2: Calculated the grams of AlCl₃ produced.
Using the mole formula
[tex]undefined[/tex]The synthesis of sodium chloride according to the reaction: ___ Na + ___ Cl2 → ___ NaCl.a. What volume of chlorine gas at STP is necessary for the complete reaction of 4.85 grams of sodium metal.
Answer:
[tex]2.36\text{ L }[/tex]Explanation:
Here, we want to know the volume of chlorine gas necessary to complete the reaction
We can start by writing a balanced equation of reaction as follows:
[tex]2Na\text{ + Cl}_2\text{ }\rightarrow2NaCl[/tex]From the balanced equation, we can see that 1 mole of chlorine gas requires 2 moles of sodium metal
Now, let us get the number of moles of sodium metal that actually reacted
We have that as the mass of sodium metal divided by the atomic mass
The atomic mass of sodium metal is 23 amu
Thus, we have the number of moles as:
[tex]\frac{4.85}{23}\text{ = 0.211 mole}[/tex]Since the mole ratio is 2 to 1, half of this number of moles is needed by the chlorine gas
Mathematically, 1 mole of a gas occupies a volume of 22.4 L at STP
the volume occupied by the calculated number of moles will be:
[tex]\frac{0.211}{2}\times\text{ 22.4 = 2.36 L }[/tex]Uranium-238 is an unstable nuclide that emits an alpha particle, three neutrons, and
gamma particles. A = 231 and Z = 90. What daughter nuclide is formed from this
reaction?
OA) Ra
232U2a + 2x + 3 n + Y
-
B) Pu
C) Th
D) Fr
Uranium-238 is an unstable nuclide that emits an alpha particle, three neutrons, and
gamma particles and produces a daughter nuclide Thorium.
The atomic mass of alpha particle is 4g and atomic number 2. Since an alpha particle emits, so there is reduction in the atomic mass 4 and atomic number 2.
The atomic number of neutron is 0 and the atomic mass of neutron is 1. When three neutrons emit the atomic mass reduces to 3 while atomic number remains same.
Gamma particles have no mass and charge. So, atomic number and atomic mass remains same.
Total reduction in atomic mass : 238 - 4 - 3 = 231.
Total reduction in atomic number = 92- 2 = 90
Therefore, the daughter nuclide is Thorium.
Thus, we concluded that the Uranium-238 is an unstable nuclide that emits an alpha particle, three neutrons, and gamma particles and produces a daughter nuclide Thorium.
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Explain how you would calculate the total change in bond energy for the reaction H2+Cl2->2HCI. How would you know if the reaction was endothermic or exothermic?
To calculate the total change in binding energy we must apply the following equation:
[tex]\Delta H=\Delta H_{f(\text{Reagents)}}-=\Delta H_{f(\text{Products)}}[/tex]This equation tells us that the energy change will be equal to the sum of the potential energy of the product bonds minus the sum of the potential energy of the reagent bonds.
Now, let's calculate each term separately.
[tex]\begin{gathered} \Delta H_{f(\text{Reagents)}}=n_{H2}\times\text{Energy bond H-H + }n_{Cl2}\text{Energy bond Cl-Cl} \\ \Delta H_{f(\text{Reagents)}}=1molH_2\times432\frac{kJ}{mol}+1molCl_2\times239\frac{kJ}{mol} \\ \Delta H_{f(\text{Reagents)}}=671kJ \end{gathered}[/tex][tex]\begin{gathered} \Delta H_{f(\text{Products)}}=n_{\text{HCl}}\times EnergyBond\text{ H-Cl} \\ \Delta H_{f(\text{Products)}}=2molHCl\times427\frac{kJ}{mol} \\ \Delta H_{f(\text{Products)}}=854kJ \end{gathered}[/tex]So, the change in bond energy will be:
[tex]\begin{gathered} \Delta H=\Delta H_{f(\text{Reagents)}}-=\Delta H_{f(\text{Products)}} \\ \Delta H=671kJ-854kJ=-183kJ \end{gathered}[/tex]We have a negative value in the result, when this happens it means that the reaction is exothermic, that is to say, that it releases heat and the energy of the products is greater than that of the reagents.
When we have a positive value the reaction will be endothermic, this means that it needs energy.
Phthalic acid ( H2C8H4O4) is a diprotic acid with a1=1.12×10−3 and a2=3.90×10−6. Determine the pH of a 0.294M phthalic acid ( H2C8H4O4 ) solution.
pH of the acid can be calculated from the first and second ionization constants of the acid. The pH of 0.294 M of Phthalic acid is 0.53.
What is pH?pH of a solution is the measure of its H+ ion concentration.It measures the acidity or basicity of the solution. Mathematically it is the negative logarithm of hydrogen ion concentration.
pH = -log [H+]
Give that the first ionization constant of the acid is 1.12 × 10 ⁻³ and second ionization constant is 3.90 × 10⁻⁶. The diprotic acid can be represented as H2A and after its first ionization it produce HA- and H+ On the second ionization the base A- and H+ ion is produced.
The second ionization constant a₂ is written as follows:
a₂ = [A-] [H+] / [HA - [A]] = 3.90 × 10⁻⁶
From this simply we can get [A] = 3.90 × 10⁻⁶
Thus [H+] = H₂A - [A]
The concentration of the diprotic acid H₂A is given 0.294 M.
Thus [H+] = 0.294 - 3.90 × 10⁻⁶
= 0.2933 M
Now, the pH is calculated as follows:
pH = -log (0.2933)
= 0.53.
Therefore, the pH of 0.294 M of Phthalic acid is 0.53.
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How many grams of Br are in 395 g CaBr2 ?
In order to answer this question we will need to use the molar mass of Calcium bromide (CaBr2), which is 199.89 g/mol, now we can find how much of this mass is only Bromide, so what is the percentage of Bromide that makes up this whole compound, to do so we will also use the molar mass of Bromide (Br2) which is 159.8 g/mol
199.89 g/mol = 100% of the compound
159.8 g/mol = x %
x = roughly 80% (is like 79.9%)
So, regardless of the mass of CaBr2, bromide needs to be 80% of its mass, now solving our question
395 g = 100%
x grams = 80%
x = 316 grams is the mass of Bromide in the compound
If carbon has undergone neutron capture what happened?
Answer:
It would generate a electro lithium nucleous
Two different samples, A and B, of zinc oxide were obtained from different sources. When heated in a stream of hydrogen, they were reduced to yield the following results: sample A mass of sample 10.0g mass of zinc residue 8.11g sample B mass of sample 13.2g mass of zinc residue 10.70g. Show that the figures illustrate the law of constant composition
-ZnO is the compound illustrated by the law of constant composition.
According to the rule of constant proportion, the elements are always present in certain amounts to one another by mass in chemical compounds. For instance, regardless of how or where water is obtained, it always contains H2O, hydrogen, and oxygen in a mass ratio of 1:8, according to an example.
for sample A
mass Zn = 8.11g
mass of O =10-8.11=1.89g
Zn= 8.11/65 = 0.1
O= 1.89/16 =0.1
ratio is 1:1
compound = ZnO
for sample b
Zn= 10.70g
O= 13.2-10.7=2.5g
Zn= 10.70/65 =1
O= 2.5/16 =0.16=1
ratio = 1:1
compound= -ZnO
Therefore, it is shown that the figures illustrate the law of constant composition.
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PLEASE HELP!! What is the WEIGHTED average atomic mass of this element?
An isotope with mass 185 amu is 19% abundant
An isotope with mass 180 amu is 81% abundant
Hello. This is Mr. Biederman, your science teacher at Insight School of Michigan. You are obviously one of my students. Posting content here is against school policy. Just ask me directly for help. I am always happy to help with anything!
Draw a flow sheet to show how you would separate the components of a mixture containing an acid substance, toluic acid, a basic substance, p-bromo-aniline, and anthracene, a neutral substance.
In order to separate the mixture of toluic acid, p-bromo-aniline, and anthracene, the difference in their physical properties such as solubility in organic or inorganic solvents is utilized.
The flow chart for the separation method is found in the attachment.
What are mixtures?Mixtures are substances that consist of two or more substances that are physically combined together.
The physical properties of the components of the mixture are employed in the separation of the constituents of the mixture
In separating the given mixture, it is first dissolved in dichloromethane, and then dilute NaOH is added to separate the toluic acid in the aqueous phase. The toluic acid is then precipitated by adding HCl.
Dilute HCl is then added to the organic phase to separate p-bromo-aniline into an aqueous phase and then precipitated by adding 10% NaOH.
Anthracene is left in the organic phase.
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A major component of gasoline is octane C8H18. When liquid octane is burned in air it reacts with oxygen O2 gas to produce carbon dioxide gas and water vapor. Calculate the moles of carbon dioxide produced by the reaction of 0.900mol of octane. Be sure your answer has a unit symbol, if necessary, and round it to 3 significant digits.
When 0.900mol of octane is burned in air it reacts with oxygen Oxygen gas to produce carbon dioxide gas and water vapor, then 7.20mole of carbon dioxide is produced.
What is combustion reaction?
Combustion reaction is a reaction in which a substance, basically hydrocarbons burns in presence of oxygen to produce carbon dioxide and water.
The balanced reaction is
[tex]2C_{8} H_{18}+25O_{2} \rightarrow16CO_{2} +18H_{2}O[/tex]
The mole ratio of octane to carbon dioxide is 2:16 that is 1:8
2 mole of octane produce =16 mole of carbon dioxide
1 mole of octane produce = 8 mole of carbon dioxide
0.900mol of octane produce= 8× 0.900
= 7.20mole
Thus the mole of carbon dioxide produced is 7.20mole.
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Give the names of following compounds (in aqueous solution):
HBr
7 HBrO
2 HCN
3 H₂CO3
4 HC₂H302
5 H₂SO4
6 H3ASO3
8 HNO2
9 HC104
10 H₂C2O4
11 H3PO4
12 H₂CrO4_
Here are all names of compounds in aqeous solution.
What is aqeous solution?
An aqeous solution is one in which the solvent is liquid water .
In this solution water act as a solvent.
Sol-
HBr- Hydrogen bromide
1-HBrO- hydrobromic acid
2-HCL-hydrochloric acid
3-H2O3- hydrogen peroxide
4- HC2H3O2- glacial acetic acid
5-H2SO4-sulfuric acid
6- H3ASO3- Arsenous acid
7- HNO2- Nitrous acid
8- HC104- Perchloric acid
9-H2C2O4- oxalic acid
10- H3pO4- phosporic acid
11-H2CrO4- chromic acid
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If 335g water at gains 102.3 J of heat, how much does the temperature of the water change? The specific heat of water is 4.184 J/g*C
Temperature is a measure of how hot a substance or radiation is expressed numerically.
There are three different types of temperature scales:
those that depend only on macroscopic properties and thermodynamic principles, like Kelvin's original definition;
those that depend on practical empirical properties of particles rather than theoretical principles;
and those that are defined by the average translational kinetic energy per freely moving microscopic particle, like an atom, molecule, or electron, in a body, like the SI scale.
How much heat is gained or lost by a sample can be calculated using the equation q = mcΔT, where m is the mass of the sample, c is the specific heat, and T is the temperature change (q).
Therefore,
q = m*c*ΔT
102.3 = 335 * 4.184 * ΔT
ΔT = 0.073 °c
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The Lewis dot model of a molecule is shown.
H
c=o
H
Based on the model, which of the following is true? (5 points)
O a
Ob
Oc
d
Oxygen is the least electronegative of the three atoms.
Carbon has a total of four bonded pairs of electrons around it.
Oxygen has four pairs of non-bonding Innermost shell electrons.
Carbon has an incomplete octet as it transfers an electron to each hydrogen.
Time left for this
Answer:
b.Carbon has a total of four bonded pairs of electrons around it.
Acetone has a density of 0.7857 g/cm^3. What is the volume in mL of 5.52 g of acetone?
Let's see that the formula to find the volume using density and mass is:
[tex]V=\frac{m}{d}\begin{cases}V=\text{volume} \\ m=\text{mass} \\ d=\text{density}\end{cases},[/tex]In this case, the problem is asking about the volume in mL but remember that mL is the same that cm^3, so using the formula we're going to obtain:
[tex]V=\frac{5.52\text{ g}}{0.7857\text{ }\frac{g}{mL}}=7.02\text{ mL.}[/tex]The volume of 5.52 g of acetone is 7.02 mL.
Following the Experiment 3 procedure, you combine 50.0 mL H2O, 50.0 mL of 2.2 M HCl(aq), and 4.039 g NaOH(s) (molar mass of NaOH = 40.00 g/mol) in a Styrofoam calorimeter, recording a maximum solution temperature change of 22.9 oC and a final solution mass of 104.153 g (csolution = 4.184 J/g.oC). Calculate the molar change in enthalpy of reaction
The molar change in enthalpy of reaction is 99.6 kJ/mol.
the formula for the specific heat capacity expressed as :
Q = mcΔT
where,
m , mass = 104.153 g
c, specific heat = 4.184 J/g °C
dt, change in temperature = 22.9 °C
Q = 104.153 × 4.184 × 22.9
Q = 9969.7 J = 9.96 kJ
now, the molar change in enthalpy is give by:
Q = ΔH / n
n is no. of moles
morality of HCl = 2.2 M
v = 50 mL = 0.05 L
n = 0.05 × 2.2
= 0.11 moles
no. of moles of NaOH = mass / molar mass
= 4.039 / 40
= 0.100
NaOH is limiting reactant.
using the formula we get:
Q = ΔH / n
ΔH = Q / n
ΔH = 9.96 kJ/ 0.10
ΔH, change in enthalpy = 99.6 kJ/mol
Thus, combine 50.0 mL H₂O, 50.0 mL of 2.2 M HCl(aq), and 4.039 g NaOH(s) (molar mass of NaOH = 40.00 g/mol) in a Styrofoam calorimeter, recording a maximum solution temperature change of 22.9 °C and a final solution mass of 104.153 g (c solution = 4.184 J/g°C). the molar change in enthalpy of reaction is 99.6 kJ/mol.
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Calculate the number of mol corresponding to 18.8 g Na2SO4.
Answer: 0.132 moles of Na2SO4
Explanation:
Multiply the grams of Na2SO4 by the 1/molar mass of Na2SO4, which is the sum of all the elements 2(Na)+S+4(O)= 142.04.
What are the strengths, weaknesses, and implications of risk/benefit analysis as a method for deciding whether a technology should be employed?
A risk-benefit analysis compares the risks and benefits of a situation and determines whether the advantages outweigh the disadvantages.
What is Risk-benefit analysis in technology?Strengths
Risk-benefit analysis calculates the amount of time will be worth it to the production of technology and whether the technology will have a healthy impact on the industry or not.
Weaknesses
Risk-benefit analysis cannon determine product implementation and the outcomes of real life experiences of individual customers. It has some drawbacks like benefits of customer is take under consideration but pollution in nature is not calculated.
Implication
Risk-benefit analysis is implied in almost all technical industry as it the decision maker of any developing team to work on a particular project or not. Some examples are automobile industry and smartphone industry.
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Lidocaine, a widely used local anesthetic, it’s available as a 0.5% (w/v) Solution for injection. Calculate the mass of lidocaine and 6.0 mL of the solution. Be sure your answer has a unit symbol and is rounded to the correct number of significant digits.
• We are given that Lidocaine is available as 0.5%(w/v) , 0.5%(w/v) represents the amount of Lidocaine in grams per 100mL. so, 0.5% is the same as 0.5g of Lidocaine in 100ml.
,• Since our solution has a volume of 6.0mL:
mass of Lidocaine = 6mL * (0.5g /100mL)
= 0.03g of Lidocaine in 6.0mLsolution.
This means that, in 6.0ml of this injection solution, we have 0.03g of Lidocaine .
How should the electrode of a pH meter be preserved? Explain your answer.
We can care for the pH meter electrode by;
1) Making sure that the electrode remains moist
2) The electrode should be stored in a 4M solution of KCl
3) Electrodes should not come in contact with deionized water.
What is pH?The term pH has to do with the negative logarithm of the hydrogen ion concentration. Now we know that the pH of a solution tells us the amount of the hydrogen ions or the hydroxyl ions that is present in the solution. The pH mete is the instrument that we could use to be able to measure the pH of the solution.
The pH scale runs between 0 - 14. The points on the scale that have been labeled from 0 - 6 tells us that the solution is an acidic solution and contains more hydrogen ions. If the solution has a pH of 7, then it is neutral and contains equal concentration of hydrogen and hydroxyl ions. A solution of pH 8 - 14 is basic and contains more hydroxyl than hydrogen ions.
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What is the mass of this piece of porcelain in oz?Show all work/use conversions/metric prefixes. I
The mass of the piece of porcelain is 27.3 oz.
1st) It is necessary to calculate the mass of the porcelain with the density formula:
[tex]\begin{gathered} \text{density}=\frac{mass}{\text{volume}} \\ 2.40g\mathrm{}mL^{-1}=\frac{mass}{325mL} \\ 2.40g.mL^{-1}\cdot325mL=\text{mass} \\ 780g=\text{mass} \end{gathered}[/tex]2nd) Now it is necessary to convert the gram unit to oz. Here we need to use the relation that 1 g is equal to 0.035 oz:
[tex]780g\text{.}\frac{0.035oz}{1g}=27.3oz[/tex]So, the mass of the piece of porcelain is 27.3 oz.
After an afternoon party, a small cooler full of ice is dumped onto the hot ground and melts. If the cooler contained 9.10 kg of ice and the temperature of the ground was 42.0 °C, calculate the energy that is required to melt all the ice at 0 °C. The heat of fusion for water is 80.0 cal/g
The heat energy that is required to melt all the ice at temperature of 0° C is 7.28x10^5 cal.
What is heat energy?
Heat energy is created by the movement of tiny atoms, molecules, or ions in solids, liquids, and gases. It is possible to transfer heat energy from one thing to another. Heat is a flow or transfer that happens when two objects have different temperatures from one another.
We know that the initial temperature of the ice (T) is 0° C
Latent heat of ice is given in the question as 80.0 cal/g
The mass of ice given in question is 9.10 kg
So, the heat required to change the phase of the ice is given as:
[tex]Q=mL[/tex]
[tex]Q =(9.10 \times 10^3)(80)[/tex]
[tex]Q = 728 \times 10^3[/tex]
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10) The chemistry lab has had a power cut and the electronic balance isn't working. You need to find the mass of a sample of zinc. You found
an old reference book and know that zine has a density of 7.14 g/mt. You found the volume of the sample using water displacement; 15.0
ml. of water was placed into a graduated cylinder, the zinc sample was added and the volume of the zinc with the water was 17.9 ml.
What is the mass of the zinc sample?
Atomic mass of zinc is 65. Simply dip one strip into the water for 3 seconds, remove it and shake once to remove excess water, wait 20 seconds for the color to develop, and compare it to the closest color match to determine your zinc concentration.
What is zinc in water?Water naturally contains zinc. In seawater, the average zinc concentration is 0.6-5 ppb. Zinc concentrations in rivers range between 5 and 10 ppb. Algae have 20-700 ppm, sea fish and shellfish have 3-25 ppm, oysters have 100-900 ppm, and lobsters have 7-50 ppm.Zinc has the chemical symbol Zn and the atomic number 30. At room temperature, zinc is a slightly brittle metal with a shiny-greyish appearance when oxidation is removed. It is the first element in periodic table group 12. Zinc is a trace mineral, which means that the body only requires trace amounts, but it is required by nearly 100 enzymes to carry out vital chemical reactions.To learn more about zinc, refer to:
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13. A gas of unknown molecular mass was allowed to effuse through a small opening under
constant pressure conditions. It required 72 s for the gas to effuse. Under identical
experimental conditions, it required 28 s for O₂ gas to effuse. Determine the molar mass
of the unknown gas.
The molar mass of unknown gas is 211.59 g/mol.
What is effusion of gases?Effusion occurs when a gas pass through an opening that is smaller than the mean free path of the particles, which is the average distance traveled between collisionsGraham's law is an empirical relationship which states that ratio of the rates of diffusion or effusion of two gases is the square root of the inverse ratio of their molar masses.Higher the molar mass of a gas, slower the effusionSince both the gases are present in identical experimental condition, using Grahams law of effusion:
[tex]\frac{time req for unknown gas}{time of oxygen}[/tex] = [tex]\sqrt{\frac{Molar mass of unknown gas}{molar mass of Oxygen} }[/tex]
Given:
Time of unknown gas = 72 sec
Time of Oxygen = 28 sec
we know the molar mass of oxygen = 32 g/mol
Now substituting:
[tex]\frac{72 sec}{28 sec} = \sqrt{\frac{M.M of unknown}{32g/mol} }[/tex]
M.M of unknown = (72 / 28)² × 32 g/mol
= 211.59 g/mol
So the molar mass of unknown gas is 211.59 g/mol
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Find the volume in liters of the 0.505 molar NaOH solution needed to react with 40 milliliters of the 0.505 molar H2SO4 solution.
A//: 80 mililiters
Please, it's urgent, I need it ASAP
The volume of NaOH required is 80ml.
Sulfuric acid is a dibasic acid and NaOH is monoacidic base.
M1V1 = M2V2
where, M1 = initial concentration,
V1 = initial volume,
M2 = concentration after mixing
V2 = total final volume.
This formula is used for calculating the final volume or molarity after mixing two solutions.
Given the question,
M1 = 0.505M
V1 =?
M2 = 0.505M
V2 = 40 ml
M1V1 = 2×M2V2
0.505 × V2 = 2 × 0.505 × 40
V2 = 2×0.505×40
0.505
V2 = 80ml
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In aqueous solution, 5.2 mg of iron(III) chloride reacts with excess ammonium hydroxide.A. Write and kalance the equation, including phasesB. Is this a limiting reactant problem? Why or why not?C. Calculate the moles of each product that are formedD. Calculate the grams of each product that are formedI already answered part a and b. I need help on part c and d.
With a balanced reaction, we can determine the moles of products. The balanced equation will be:
FeCl3 + 3NH4OH → Fe(OH)3 + 3NH4Cl
They says that NH4OH is in excess, so, the limiting reactant will be FeCl3 and we will do all the calculations with this reactant.
We have to calculate the moles of FeCl3, we will use the molar mass:
[tex]\begin{gathered} molFeCl_3=5.2mg\times\frac{1g}{1000mg}\times\frac{1molFeCl_3}{MolarMass,gFeCl_3} \\ molFeCl_3=5.2mg\times\frac{1g}{1,000mg}\times\frac{1molFeCl_3}{162.2gFeCl_3}=3.2\times10^{-5}molFeCl_3 \end{gathered}[/tex]Now, to calculate the moles of the products we must take into account the product/reactive ratios, for this we are guided by the coefficients that accompany the molecules.
Ratio Fe(OH)3 to FeCl3 = 1/1
Ratio NH4Cl to FeCl3 = 3/1
Moles of each product
Moles of Fe(OH)3
[tex]molFe(OH)_3=3.2\times10^{-5}molFeCl_3\times\frac{1molFe(OH)_3}{1molFeCl_3}=3.2\times10^{-5}molFe(OH)_3[/tex]Moles of NH4Cl
[tex]molNH_4Cl=3.2\times10^{-5}molFeCl_3\times\frac{3molNH_4Cl}{1molFeCl_3}=9.6\times10^{-5}molNH_4Cl[/tex]The grams of each product we will find by multiplying the moles by the molar mass. So we have.
g of Fe(OH)3
[tex]\begin{gathered} gFe(OH)_3=3.2\times10^{-5}molFe(OH)_3\times106.87g/molFe(OH)_3 \\ gFe(OH)_3=3.4\times10^{-3}g=3.4mg \end{gathered}[/tex]g of NH4Cl
[tex]\begin{gathered} gNH_4Cl=molNH_4Cl\times MolarMassNH_4Cl \\ gNH_4Cl=9.6\times10^{-5}molNH_4Cl\times53.491g/molNH_4Cl=5.1\times10^{-3}g=5.1mg \end{gathered}[/tex]Scientists are looking for ways to help increase the amount of calcium carbonate in oceans so that coral reefs can be healthy ecosystems again. Why is the amount of calcium carbonate in the ocean decreasing? (1 point)
Responses
Increased amounts of carbon dioxide cause reactions to happen in the water that decrease the pH and the amount of calcium carbonate.
Decreased amounts of carbon dioxide cause reactions to happen in the water that decrease the pH and the amount of calcium carbonate.
Increased amounts of carbon dioxide cause reactions to happen in the water that increase the pH and decrease the amount of calcium carbonate.
Decreased amounts of carbon dioxide cause reactions to happen in the water that increase the pH and decrease the amount of calcium carbonate.
Scientists are looking for ways to help increase the amount of calcium carbonate in oceans so that coral reefs can be healthy ecosystems again mount of calcium carbonate in the ocean decreasing because increased amounts of carbon dioxide cause reactions to happen in the water that decrease the pH and the amount of calcium carbonate
Calcium carbonate is a compound CaCO₃ found in nature as calcite and aragonite and in plant ashes, bones and shells and used especially in making lime and portland cement and as a gastric antacid and as ocean acidification increases available carbonate ion bond with excess hydrogen resulting in fewer carbonate ion available for calcifying organisms to build and maintain their shells, skeletons, and other calcium carbonate structures
Ocean acidification describe the lowering of sweater pH and carbonate saturation that result from increasing atmospheric CO₂ concentration and that's why increased amounts of carbon dioxide cause reactions to happen in the water that decrease the pH and the amount of calcium carbonate.
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Answer: Its C I did my research and that's what makes sense to me. because increased amount of carbon dioxide decreases the PH and the amount of calcium carbonate.
Gaseous ethane (CH3CH3) reacts with gaseous oxygen gas (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). if 2.46 g of carbon dioxide is produced from the reaction of 2.71 g of ethane and 16.7 g of oxygen gas, calculate the percent yield of carbon dioxide. Round to 3 sig figs
Taking into account definition of percent yield, the percent yield of carbon dioxide is 30.90%
Reaction stoichiometryIn first place, the balanced reaction is:
2 CH₃CH₃ + 7 O₂ → 4 CO₂ + 6 H₂O
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
CH₃CH₃: 2 molesO₂: 7 molesCO₂: 4 molesH₂O: 6 molesThe molar mass of the compounds is:
CH₃CH₃: 30 g/moleO₂: 32 g/moleCO₂: 44 g/moleH₂O: 18 g/moleThen, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
CH₃CH₃: 2 moles ×30 g/mole= 60 gramsO₂: 7 moles ×32 g/mole= 224 gramsCO₂: 4 moles ×44 g/mole= 176 gramsH₂O: 6 moles ×18 g/mole= 108 gramsLimiting reagentThe limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.
Limiting reagent in this caseTo determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 224 grams of O₂ reacts with 60 grams of CH₃CH₃, 16.7 grams of O₂ reacts with how much mass of CH₃CH₃?
mass of CH₃CH₃= (16.7 grams of O₂× 60 grams of CH₃CH₃)÷ 224 grams of O₂
mass of CH₃CH₃= 4.47 grams
But 4.47 grams of CH₃CH₃ are not available, 2.71 grams are available. Since you have less mass than you need to react with 16.7 grams of O₂, CH₃CH₃ will be the limiting reagent.
Percent yieldThe percent yield is the ratio of the actual return to the theoretical return expressed as a percentage.
The percent yield is calculated as the experimental yield divided by the theoretical yield multiplied by 100%:
percent yield= (actual yield÷ theoretical yield)×100%
where the theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.
Theoretical yield of CO₂Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 60 grams of CH₃CH₃ form 176 grams of CO₂, 2.71 grams of CH₃CH₃ form how much mass of CO₂?
mass of CO₂= (2.71 grams of CH₃CH₃× 176 grams of CO₂)÷ 60 grams of CH₃CH₃
mass of CO₂= 7.96 grams
Then, the theoretical yield of CO₂ is 7.96 grams.
Percent yield for the reaction in this caseIn this case, you know:
actual yield= 2.46 gramstheorical yield= 7.96 gramsReplacing in the definition of percent yields:
percent yield= (2.46 grams÷ 7.96 grams)×100%
Solving:
percent yield= 30.90%
Finally, the percent yield for the reaction is 30.90%.
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Which of the following changes are chemical changes?
Answer:
flammability, toxicity etc..
Explanation:
are chemical changes
The first-order rate constant for the decomposition of N2O5,
2N2O5(g)→4NO2(g)+O2(g)
at 70∘C is 6.82×10−3 s−1. Suppose we start with 2.50×10−2 mol of N2O5(g) in a volume of 1.6 L What is the half-life of N2O5 at 70 ∘C ?
The half-life of N₂O₅ at 70°C then it is 50s or 0.8min
Half life is the the interval of time required for one half of the atomic nuclei of a radioactive sample to decay
The first-order rate constant for the decomposition of N₂O₅ and reaction is
2N₂O₅(g) → 4NO₂(g)+O₂(g)
The half life is the time at which the concentration of N₂O₅ is the half initial concentration that is [ N₂O₅ ] = 1/2 [ N₂O₅ ]₀
Using this same equation then
ln ([N₂O₅]) = -2kt + ln ([N₂O₅]₀)
We replace [ N₂O₅ ] by 1/2[ N₂O₅ ]₀ and solve for t
ln (1/2 [N₂O₅]₀) = -2kt + ln ([N₂O₅]₀) applying logarithmic property
ln 1/2 + ln([N₂O₅]₀) = -2kt + ln ([N₂O₅]₀) solving for t and applying logarithmic property (ln 1/2 = ln 1 - ln 2 = 0 - ln 2 = - ln 2)
t = ln 2/2k
t = 50s or 0.8min
The half-life of N₂O₅ at 70°C then it is 50s or 0.8min
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PLEASE HELP!
In a monomolecular reaction A->B , at t =250C, the initial concentration decrease at 25% in t =52 min. Calculate:
a) the constant rate;
b) the time after the initial concentration decrease with 75%;
c) the initial reaction rate, if the initial concentration of the reactant is 2.5 mol/L·s
1) The calculate rate constant is 9.22 * 10^-5 s-1
2) In the period of is 15033 s 75% is used up
3) From the calculation, there is an initial 9.22 * 10^-5 M.
How can we find reaction rate?In chemistry, the rate of reaction would give the idea that the reaction is proceeding quickly or slowly. If a reaction has a large rate of reaction then it tends to move on to completion.
From the question, it is clear that there is 25% in t =52 min.
Initial concentration [A]o = [A]o
Final concentration = [A]o - 0.25 [A]o = 0.75 [A]o
Time taken = 52 min or 3120 s
From the formula that can be applied to a first order reaction;
ln[A] = -kt + ln[A]0
k = -(ln[A] - ln[A]0)/t
k = - (ln0.75 [A]o/A]0)/3120
k = 9.22 * 10^-5 s-1
b)
Then we have to find the time after the initial concentration decrease with 75%
[A] = [A]o - 0.75 [A]o = 0.25 [A]o
ln[A] = -kt + ln[A]0
t = -(ln[A] - ln[A]0)/k
t = - (ln0.25 [A]o/A]0)/9.22 * 10^-5
t = 15033 s
c) Given that in this case, the initial concentration of the reactant is 2.5 mol/L·s
k = - (ln0.25 (2.5)/ln(2.5))/15033
k = 9.22 * 10^-5 M
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