climate
Changes in the composition of the atmosphere have caused gradual changes in earth's climate throughout history, causing changes in plant and animal life that contributed to mass extinctions.
The following are some of the reasons:
UV lightclimatepollutantshydrofluorocarbonsheat
The surface of the Earth warms up as sunlight strikes it. Surface-emitted infrared light is absorbed in the atmosphere and transformed into heat. The temperature close to the surface rises as a result of this heat being trapped in the atmosphere.UV light:indirect impacts of climate change on UV radiation from the surface. By changing the concentrations of ozone, UV-absorbing tropospheric gases, aerosols, and clouds in the atmosphere, climate change may have indirectly affected UV radiation levels in the past. These influences are probably going to persist in the future.climate:People are at risk from food and water shortages, greater flooding, high heat, an increase in disease, and economic loss due to climate change. Conflict and human migration are potential outcomes. Climate change is the top hazard to world health in the twenty-first century, according to the World Health Organization (WHO).pollutants:these are also resulting in the increase of temperature of the Earth and is also damaging ozone layer.To learn more about the changes in earth visit:
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How much water would need to be added to 750 ml of a 2.8 m hcl solution to make a 1.0 m solution?
1.35 Litres of water need to be added to 750 ml of a 2.8 m hcl solution to make a 1.0 m solution.
Use the following relation:
M1V1=M2V2
Where M is molarity, V is volume and 1 is initial and 2 is the final conditions. Solving for V(2)
M1=2.8 M,V1=750 mL;M2=1.0 M
(2.8 M)×(750 mL)=(1.0 M)×V2
V2=(2.8 M)×(750 mL)(1.0 M) = 2100 mL = 2.1 L
Therefore, Volume of water to be added =2.1 L−0.75 L=1.35 L
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Al+HCl=AlCl6+H6
is this balanced or unbalanced
Answer:
unbalanced
Explanation:
balanced: Al+HCl=AlCl+H
What volume of hydrogen (in l) is produced from 3.143 mol of hydrochloric acid at stp
Volume of hydrogen (in l) is produced is 35.20 L
Given:
moles of hcl = 3.143 mol
To Find:
volume of hydrogen
Solution:
2HCl → H2 + Cl2
STP 2*22.4L 22.4L 22.4L
2 mol 1 mol 1 mol
73 gm 2gm 35.5gm
here we can see that 2 mol hcl is producing 22.4 litres of hydrogen
So, 3.143 mol hcl will produce hydrogen = 22.4/2*3.143 = 35.20 litres
Volume of hydrogen (in l) is produced is 35.20 L
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If butane had a volume of 500 mL. at a pressure of 1.5 atm and a temperature of
20 °C, what would the new volume of the gas be at a temperature of 30 °C and a pressure
of 500 Torr
Question:
If butane had a volume of 500 ml. at a pressure of 1.5 atm and a temperature of 20 °C, what would the new volume of the gas be at a temperature of 30 °C and a pressure of 500 Torr?
Solution Given:
Let P be the pressure V be volume and and T be temperature.
Volume of Butane [[tex]V_1[/tex]] = 500 ml
Pressure of Butane [[tex]P_1[/tex]] = 1.5 atm
Temperature [[tex]T_1[/tex]] =20°C=20+273=293K
New Volume of Butane [[tex]V_2[/tex]] = ?
New Pressure of Butane [[tex]V_2[/tex]] =500 Torr=500*0.00131579=0.657895 atm
Note: 1 Torr= 0.00131579 atm
New Temperature of Butane [[tex]V_2[/tex]] =30°C=30+273=303K
Now
By using combined gas law equation:
[tex]V_2=V_2*\frac{P_1*T_2}{P_2*T_1}[/tex]
[tex]V_2=500*\frac{1.5*303}{0.657895*293}=1178.9 ml[/tex]
The new volume of Butane is 1178.9 ml
The density of an unknown gas is 4. 20 grams per liter at 3. 00 atmospheres pressure and 127 °c. what is the molecular weight of this gas?
The molecular weight of this gas will be 45 g/mol .
The state equilibrium equation for a fictitious perfect gas is known as the ideal gas law, sometimes known as the generic gas equation. Although it has significant drawbacks, it represents a decent approximation of the activity of many gases under various conditions.
Ideal gas law can be expressed as:
PV =nRT
Calculation of molecular weight by using ideal as law.
Given data:
P = 3 atm
T = 127 °c
Density = 4. 20 grams per liter
PV =nRT
where p is pressure , T is temperature and R is gas constant.
PV = gram / molecular weight RT
Molecular weight = (g/v)( 1/P) RT
Putting the given data in above equation.
Molecular weight =4.20 × 1/ 3× 400 × 0.0831
Molecular weight = 45 g/mol.
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Chem Questions
1) Determine the mass of RbCN required to prepare a 370 mL solution which is 0.12 M RbCN (MW = 112 g/mol).
2) Determine the mass of RbCN required to prepare a 246 mL solution which is 0.12 M RbCN (MW = 112 g/mol).
3) What does a 1-molar solution of CaCl_2 contain?
4) Calculate the empirical formula for a chemical compound that gives the following analysis: 12.6 percent lithium, 29.2 percent sulfur, and 58.2 percent oxygen.
5) Calculate the empirical formula for a chemical compound that gives the following analysis: 65.2 g scandium and 34.8 g oxygen.
1. The mass of RbCN required to prepare the solution is 4.97 g
2. The mass of RbCN required to prepare the solution is 3.31 g
3. 1 molar solution of CaCl₂ contains 1 mole of CaCl₂ in 1 L of the solution
4. The empirical formula of the compound is Li₂SO₄
5. The empirical formula of the compound is Sc₂O₃
What is molarity?This is defined as the mole of solute per unit litre of solution. Mathematically, it can be expressed as:
Molarity = mole / Volume
1. How to determine the mass of RbCN requiredWe'll begin by calculating the mole of RbCN in the solution
Volume = 370 mL = 370 / 1000 = 0.37 L Molarity = 0.12 MMole = ?
Mole = molarity × volume
Mole = 0.12 × 0.37
Mole = 0.0444 mole
Finally, we shall determine the mass of RbCN required as illustrated below
Mole = 0.0444 moleMolar mass = 112 g/molMass of RbCN =?Mass = mole × molar mass
Mass of RbCN = 0.0444 × 112
Mass of RbCN = 4.97 g
2. How to determine the mass of RbCN requiredWe'll begin by calculating the mole of RbCN in the solution
Volume = 246 mL = 246 / 1000 = 0.246 L Molarity = 0.12 MMole = ?
Mole = molarity × volume
Mole = 0.12 × 0.246
Mole = 0.02952 mole
Finally, we shall determine the mass of RbCN required as illustrated below
Mole = 0.02952 moleMolar mass = 112 g/molMass of RbCN =?Mass = mole × molar mass
Mass of RbCN = 0.02952 × 112
Mass of RbCN = 3.31 g
3. What does a molar solution of CaCl₂ contains?This is a solution that contains 1 mole of CaCl₂ in 1 L of the solution
4. How to determine the empirical formulaLi = 12.6%S = 29.2%O = 58.2%Empirical formula =?Divide by their molar mass
Li = 12.6 / 7 = 1.8
S = 29.2 / 32 = 0.9125
O = 58.2 / 16 = 3.6375
Divide by the smallest
Li = 1.8 / 0.9125 = 2
S = 0.9125 / 0.9125 = 1
O = 3.6375 / 0.9125 = 4
Thus, the empirical formula of the compound is Li₂SO₄
5. How to determine the empirical formulaSC = 65.2 gO = 34.8 gEmpirical formula =?Divide by their molar mass
SC = 65.2 / 45 = 1.449
O = 34.8 / 16 = 2.175
Divide by the smallest
SC = 1.449 / 1.449 = 1
O = 2.175 / 1.449 = 3/2
Multiply by 2 to express in whole number
SC = 1 × 2 = 2
O = 3/2 × 2 = 3
Thus, the empirical formula of the compound is Sc₂O₃
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Please help me understand this!
Answer:
It's too short. Write at least 20 characters to explain it well.
Explanation:
It's too short. Write at least 20 characters to explain it well.
The chemoreceptors located on the tongue for the detection of taste are found in structures called blank______.
The chemoreceptors located on the tongue for the detection of taste are found in structures called taste buds.
What are chemoreceptors?
Receptors that respond to chemicals are known as chemoreceptors. Chemoreception depends primarily on the senses of taste and smell. Chemoreception relies on chemicals that act as signals to regulate cell function, without the chemical necessarily being taken into the cell for metabolic purposes.What are taste buds or taste receptors?
Taste receptors are confined to the oral cavity. They are most abundant on the tongue but also occur on the palate and epiglottis and in the upper part of the oesophagus.The taste receptor cells, with which incoming chemicals interact to produce electrical signals, occur in groups of 50–150. Each of these groups forms a taste bud. On the tongue, taste buds are grouped together into taste papillae. On average, the human tongue has 2,000–8,000 taste buds, implying that there are hundreds of thousands of receptor cells.To learn more about taste buds: https://brainly.com/question/8624931
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How wide in m is a single slit that produces its first minimum for 624-nm light at an angle of 18. 0°?
A single slit with a width of 2019 * 109 m creates its initial minimal for 624 nm light at an angle of 18°.
How does diffraction work?
Waves spreading outward around obstructions are known as diffraction. Diffraction happens with sound, electromagnetic radiation like light, X-rays, and gamma rays, as well as with incredibly minuscule moving particles like atoms, neutrons, and electrons that exhibit wavelike qualities. Diffraction prevents the creation of sharp shadows as one of its effects. In order to spread out and illuminate regions at which a shadow is anticipated, light must be bent around corners, which is known as diffraction.
Calculation:
Provided for a single slit, m=1
λ = 624 *10⁻⁹
sinθ = sin 18⁰
Therefore,
asinθ=mλ
a = [tex]\frac{1 * 624 *10^{-9} }{sin 18}[/tex]
⇒a = 2019 *10⁻⁹ m
Therefore the width of a single slit is 2019 *10⁻⁹ m.
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What descriptive term is applied to the type of diene represented by 1,5-octadiene?
The descriptive term that is applied to the type of diene represented by 1,5-octadiene is isolated diene. The correct option is C.
What is diene?Diene is a compound that contains two or more double bonds, usually carbon bonds, which are separated by a single bond. They are covalent compounds. Alkene units are surely present in these compounds, whose quantity is two.
1,5-octadiene is a polymer of diene, which are generally elastomers, and they made of vulcanized rubber. They are isolated diene.
Thus, the correct option is C) Isolated diene.
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The question is incomplete. The complete question is given below:
A) Conjugated diene.
B) Cumulated diene.
C) Isolated diene.
D) Alkynyl diene.
E) None of the above.
What can you say about an unknown compound when dissolved in distilled water and with ph paper, gave an estimated ph of about 13. 3?
The pH of the solution is used to estimate the acidic and the alkaline condition. The pH paper can be used to determine the conditions. The compound with pH 13.3 is basic.
What is pH?The concentration of the hydrogen or the hydroxide ion in the water gives the estimate of the pH. The potential and the amount of hydrogen decide the acidic and the basic compound.
The pH scale ranges from 0-14 where the scale of 0-6 is acidic, 7 is neutral and 8-14 is basic. If the substance shows a pH of 13.3 then it will lie in the basic range.
Therefore, the compound with a pH of 13.3 is basic.
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Project 1 Hypothesis Construction
Write a properly formatted hypothesis statement to answer this question: How does the amount of salt added to ice affect the rate at which the ice will melt?
Specify how you plan to change the independent variable by using terms such as increase or decrease. Also, specify how the dependent variable will change in response by using terms such as increase, decrease, or stays the same.
Criteria pts
Correct placement of IV 5
Correct placement of DV 5
If, then format 5
IV indicates either "increases" or "decreases" 5
DV indicates either "increases", "decreases", or "stays the same" 5
The hypothesis should be; "Increasing the amount of ice increases the rate at which the ice will melt"
What is a hypothesis?The term hypothesis refers to a tentative explanation that could be applied in order to explain an observation. When we give a hypothesis, it must be subjected to rigorous experimentation thus it can be confirmed or repelled by experiment.
Now, we can see that the question here is; "How does the amount of salt added to ice affect the rate at which the ice will melt?" This implies that the correct hypothesis here ought to have to do with the impact of the amount of ice in the affirmative.
Thus, the hypothesis should be; "Increasing the amount of ice increases the rate at which the ice will melt"
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Brainliest Available! Thank you in advance!
What is the molar solubility of aluminum hydroxide, Al(OH)3, if Ksp for Al(OH)3 is 3.0 × 10−34?
AL(OH)3 (s) ⇌ AL3+ (aq) + 3OH- (aq)
Answer:
3.0×10⁻¹³ M
Explanation:
The solubility product Ksp is the product of the concentrations of the ions involved. This relation can be used to find the solubility of interest.
EquationThe power of each concentration in the equation for Ksp is the coefficient of the species in the balanced equation.
Ksp = [Al₃⁺³]×[OH⁻]³
Solving for [Al₃⁺³]The initial concentration [OH⁻] is that in water, 10⁻⁷ M. The reaction equation tells us there are 3 OH ions for each Al₃ ion. If x is the concentration [Al₃⁺³], then the reaction increases the concentration [OH⁻] by 3x.
This means the solubility product equation is ...
Ksp = x(10⁻⁷ +3x)³
For the given Ksp = 3×10⁻³⁴, we can estimate the value of x will be less than 10⁻⁸. This means the sum will be dominated by the 10⁻⁷ term, and we can figure x from ...
3.0×10⁻³⁴ = x(10⁻⁷)³
Then x = [Al₃⁺³] will be ...
[tex][\text{Al}_3^{\,+3}]=\dfrac{3.0\times10^{-34}}{10^{-21}}\approx \boxed{3.0\times10^{-13}\qquad\text{moles per liter}}[/tex]
We note this value is significantly less than 10⁻⁷, so our assumption that it could be neglected in the original Ksp equation is substantiated.
__
Additional comment
The attachment shows the solution of the 4th-degree Ksp equation in x. The only positive real root (on the bottom line) rounds to 3.0×10^-13.
Equilibrium constant for the reaction is
[tex]\sf K_{sp}=[Al^{3+}][OH^-]^3[/tex]
Let [Al³+] be s[OH-] is 10^{-7}So
[tex]\\ \rm\Rrightarrow s=\dfrac{3.0\times 10^{-34}}{(10^{-7})^3}[/tex[
[tex]\\ \rm\Rrightarrow s=\dfrac{3.0\times 10^{-34}}{10^{-21}}[/tex]
[tex]\\ \rm\Rrightarrow s=3.0\times 10^{-13}M[/tex]
How many grams of calcium phosphate are theoretically produced if we start with 3. 40 moles of calcium nitrate and 2. 40 moles of lithium phosphate?
There are 372 grams of calcium phosphate are theoretically produced if we start with 3. 40 moles of calcium nitrate and 2. 40 moles of lithium phosphate .
Calculation,
The reactant or reagent that produces the least moles of the products is called limiting reagents. When limiting reagents used up , the reaction stops.
The balanced equation is given as,
[tex]3Ca(NO_{3} )_{2} + 2Li_{3} ( PO_{4} )[/tex] → [tex]3LiNO_{3} + Ca_{3}( PO_{4} )_{2}[/tex]
Multiply the moles of each reactant by the mole ratio between it and calcium phosphate in the balanced equation . so that the moles of the reactant cancel , leaving moles of calcium phosphate.
3.4 mol of calcium nitrate × 1 mol calcium phosphate / 3 mol calcium nitrate = 1.13 mol calcium phosphate
2.4 mol of lithium phosphate× 1 mol calcium phosphate / 2 mol lithium phosphate = 1.02 mol calcium phosphate
So, calcium nitrate is limiting reactant .
Calculation of mass of 1.02 mol calcium phosphate.
Multiply the moles of calcium phosphate by its molar mass.
molar mass of calcium phosphate = 3×40.078 g/mol calcium ion+2×30.9 g/mol phosphorus + 8×15.99 g/mol calcium phosphate = 310.178 g/mol calcium phosphate
1.20 mol calcium phosphate × 310.178 g/mol = 372 gram
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Hydrocarbons containing only single bonds between the carbon atoms are called?
Answer:
Alkanes
Explanation:
Alkanes are hydrocarbons containing only single bonds between the parent chain carbons.
Alkenes are hydrocarbons containing at least one double bond between the parent chain carbons.
Alkynes are hydrocarbons containing at least one triple bond between the parent chain carbons.
Starting with lead (ii) carbonate describe how a solid sample of lead (ii) sulphate can be prepared
The preparation of lead (ii) sulphate from lead (ii) carbonate occurs in two steps:
insoluble lead carbonate is converted to soluble lead (ii) nitratesoluble lead (ii) nitrate is reacted with sulphuric acid to produce lead (ii) sulphate.How can a solid sample of lead (ii) sulphate be prepared from lead (ii) carbonate?Lead (ii) carbonate and lead (ii) sulphate are both insoluble salts of lead.
In order to prepare lead (ii) sulphate, a two step process is performed.
In the first step, Lead (ii) carbonate is reacted with dilute trioxonitrate (v) acid to produce lead (ii) nitrate.
PbCO₃ + 2HNO₃ → Pb(NO₃)₂ + CO₂ + H₂OIn the second step, dilute sulfuric acid is reacted with the lead (ii) nitrate to produce insoluble lead (ii) sulphate which is filtered and dried.
Pb(NO₃)₂ + H₂SO₄ → PbSO₄ + 2HNO₃In conclusion, lead (ii) sulphate is prepared in two steps.
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A 20. 0 l cylinder of oxygen gas is at a temperature of 27. 0°c and a pressure of 5. 00 atm. what is the density of the oxygen gas in the cylinder?
A 20. 0 l cylinder of oxygen gas is at a temperature of 27. 0°c and a pressure of 5. 00 atm. The density of the oxygen gas in the cylinder is 1.24 atm
The physical force applied to an object is referred to as pressure. Per unit area, a perpendicular force is delivered to the surface of the objects. F/A is the fundamental formula for pressure (Force per unit area). Pascals are a unit of pressure (Pa). Absolute, atmospheric, differential, and gauge pressures are different types of pressure.
By pressing a knife on some fruit, one can get a straightforward illustration of pressure. The surface won't be cut if you press the flat section of the knife against the fruit. The force is dispersed over a wide area (low pressure).
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A student sets up the following equation to solve a problem in solution stoichiometry.
(The ? stands for a number the student is going to calculate.)
Enter the units of the student's answer.
Answer:
[tex]\frac{g}{mL}[/tex]
Explanation:
You can determine the units in the final answer by identifying which of the units cancel out. Units are eliminated (cancelled out) when they are located both in the numerator and denominator of proportions being multiplied.
In this case, these units are cancelled out.....
-----> milligrams (mg) = (1st and 2nd proportions)
-----> decaliters (dL) = (1st and 3rd proportions)
-----> liters (L) = (3rd and 4th proportions)
As these units are not cancelled out, you are left with grams (g) in the numerator and milliliters (mL) in the denominator.
The final units should be represented by:
[tex]? \frac{g}{mL}[/tex]
Landon thinks that taking care of one's family is very important. This is an example of a(n)?
Developing enduring relationships and sensing love - Taking care of others fosters empathy and the capacity to connect with others, especially through trying circumstances. Additionally, 77% of caregivers who are also employed claim that their relationship with their parents and/or in-laws has improved as a result of their caregiving.
Family supports us during both the best and worst of times, which makes these ties crucial. Family is crucial because they can provide unwavering love, support, and stability; they constantly strive to bring out the best in you even when you are unable to perceive it for yourself.
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A buffer is prepared by adding 300. 0 ml of 2. 0 mnaoh to 500. 0 ml of 2. 0 mch3cooh. what is the ph of this buffer? ka= 1. 8 10-5(ans. 4. 92)
The Henderson-Hasselbalch equation can be used to determine the pH of the buffer from the pKa value. The pH of the buffer will be 4.75.
What is the Henderson-Hasselbalch equation?Henderson-Hasselbalch equation is used to determine the value of pH of the buffer with the help of the acid disassociation constant.
Given,
Acid disassociation constant (ka) = 1. 8 10⁻⁵
Concentration of NaOH = 2.0 M
Concentration of CH₃COOH = 2.0 M
pKa value is calculated as,
pKa = -log Ka
pKa = - log (1. 8 x 10⁻⁵)
Substituting the value of pKa in the Henderson-Hasselbalch equation as
pH = - log (1. 8 x 10⁻⁵) + log [2.0] ÷ [2.0]
pH = - log (1. 8 x 10⁻⁵) + log [1]
= 4.745 + 0
= 4.75
Therefore, 4.75 is the pH of the buffer.
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What mass of CaCO3 is required to react completely with 25. 0 ml of 0. 750 m HCl?
There are 0.94 g mass of [tex]CaCO_{3}[/tex] is required to react completely with 25. 0 ml of 0. 750 m HCl .
Calculation ,
Mass of [tex]CaCO_{3}[/tex] = ?
The 1000ml of HCl = 27.375 g
then the 1 ml of solution contains HCl = 27.375 g/1000×1
25 ml of solution contains HCl = 27.375 g/1000 × 25 = 0.684 g
The chemical equation can be given as :
[tex]CaCO_{3} +HCl[/tex] → [tex]CaCl_{2} +CO_{2} +H_{2} O[/tex]
2 mol of HCl reacts with 1 mol of [tex]CaCO_{3}[/tex]
The amount of [tex]CaCO_{3}[/tex] reacted is given by ,
100/71 × 0.684 g = 0.9639 g = 0.94 g
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The burning of ethane produces both co2 and H2O. if 400 ml of CO2 is produced at 30°C and 740 torr, what volume of water vapor would be produced at 19°C and 780 torr?
The volume of water vapour would be produced at 19°C and 780 torr is 548.5mL.
If 400 ml of CO2 is produced at 30°C at 740 torr, then number of moles can be calculated as:
By using ideal gas equation:
P1V1 = N1R1T1
P1 = pressure = 740torr
V1 = 400 ml = volume of CO2
R = Gas constant = 8.314
T = 273+30 = 303 k
740×400 = N1×8.314×303
N1 = (740×400) /(8.314×303) =117.5.
Chemical equation
C2H6 ---- 2CO2 + 3H2O.
As we noticed from the equation that
2 moles of CO2 = 3 moles of H2O
1 moles of CO2 × 1 moles of H2O
Then N2 = 117.5 moles of CO2 = 3/2 × 117.5 moles of H2O
By using ideal gas equation:
P2V2 = N2RT2
V2 = 3/2 × 117.5 × 8.314 × 292/ 780
= 548.5ml.
Thus, we found that the volume of water vapour would be produced at 19°C and 780 torr is 548.5mL.
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some chemical reactions that are reversible (put some picture for better understanding)
Ammonium chloride is a white solid that breaks down when heated and produces ammonia and hydrogen chloride.
What is the reversible reaction?A reversible reaction is a reaction in which the conversion of reactants to products and products to reactants occur at the same time. In the above example, the chemical shows a reversible reaction because it moves both forward and backward direction. In reversible reaction, equal amount of reactant is converted into product and product into reactant.
So we can conclude that Ammonium chloride is a chemical that represents a reversible reaction.
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Which of the following can be mixed in solution with H₂CO3 to make a buffer?
OA. Na₂CO3
OB. HF
O C. NaHCO3
OD. NH3
SUBMIT
NaHCO3 is can be mixed in solution with H₂CO3 to make a buffer.
Sodium bicarbonate (IUPAC name: sodium bicarbonate, commonly known as baking soda or sodium bicarbonate) is a compound with the chemical formula NaHCO3. It is a salt composed of a sodium cation (Na+) and bicarbonate anion (HCO3-). Sodium bicarbonate is a crystalline white solid that often looks like a fine powder. It has a slightly salty, alkaline taste similar to soda carbonate (sodium carbonate). The natural mineral form is Narcolite. It is a component of sodium bicarbonate and is dissolved in many mineral springs.
Sodium bicarbonate is an antacid used to relieve heartburn and acid indigestion. Doctors may also prescribe sodium bicarbonate to reduce the acidity of blood and urine under certain conditions.
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The compound
is classified as a
Explanation:
secondary amine bro go leaen
Soil particles, soot, lead (pb), asbestos, sea salt, and sulfuric acid droplets are all types of?
Soil particles, soot, lead (pb), asbestos, sea salt, and sulfuric acid droplets are all types of particulate matter.
The atmosphere is dusty .The solid and liquid particles suspended in the atmosphere .
Examples : Soil particles, soot, lead (pb), asbestos, sea salt, and sulfuric acid droplets are all types of particulate matter
There are different categories of particulate matter. Such as mold spores , bacteria , dust , smoke and airborne viral paricles. The major sources of particulates matter are wildfires , dust storms , sea spray , volcanic eruption , and industrial process.
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Consider a 0.12 M solution of a weak polyprotic acid (H2A) with the possible values of Ka1 and Ka2 given here. Calculate the contributions to [H3O+] from each ionization step. At what point can the contribution of the second step be neglected?
A. Ka1=1.0×10−4 and Ka2=5.0×10−5
The [H3O+] in step 1 is 0.0034 M while the [H3O+] in step 2 is 0.00039 M
What is the contribution of each step?Let us set up the ICE table in each case, for K1;
H2A(aq) + H2O(l)--------> H3O^+(aq) + HA^-(aq)
I 0.12 0 0
C -x +x +x
E 0.12 - x x x
Ka1= [H3O^+] [HA^-]/[ H2A]
Ka1= x^2/ 0.12 - x
1.0×10^−4 = x^2/ 0.12 - x
1.0×10^−4(0.12 - x ) = x^2
1.2 * 10^-5 - 1.0×10^−4x = x^2
x^2 + 1.0×10^−4x - 1.2 * 10^-5 = 0
x =0.0034 M
[H3O+] = 0.0034 M
Again; [H3O+] = [HA^-] = 0.0034 M
HA^-(aq) + H20(l) -------> A^-(aq) + H3O^+
I 0.0034 0 0
C -x + x +x
E 0.0034 - x x x
Ka2= [A^-] [H3O^+]/[HA^-]
5.0×10^−5 = x^2/ 0.0034 - x
5.0×10^−5 (0.0034 - x ) = x^2
1.7 * 10^-7 - 5.0×10^−5x = x^2
x^2 + 5.0×10^−5x - 1.7 * 10^-7 = 0
x=0.00039 M
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________ is the enthalpy change when 1 mole of water is produced by the reaction of an acid and a base under standard conditions.
Standard enthalpy change of neutralization is the enthalpy change when 1 mole of water is produced by the reaction of an acid and a base under standard conditions.
Enthalpy change of neutralization:
Every time enthalpy change of neutralization is calculated, one mole of water is produced. Heat is released when an acid and an alkali react, hence enthalpy changes of neutralization are always negative. The values are always quite comparable for reactions involving strong acids and alkalis, falling between -57 and -58 kJ mol-1.If the reaction is the same in each case of a strong acid and a strong alkali, the enthalpy change is similar.Learn more about the Enthalpy of neutralization with the help of the given link:
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What is the reaction product of acetic acid and ethylamine at room temperature?
Ammonium salt of acetic acid is the reaction product of acetic acid and ethylamine at room temperature.
The acetic acid and ethylene amine form salts and this result is expected same as form in the reaction of oxalic acid and ethylene diamine at ambient temperature.
The reaction of acetic acid with ammonium forms ammonium salt of carboxylic acid which on heating undergoes elimination to form amide as acetamine. The salt is form due to deprotonation of acid by the base. The amide is form on heating of the ammonium carboxylate salt. Such type of reaction is called aminooacide reaction.
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How many particles are there in 5 grams of sodium carbonate?
3.01× 1024 particles are the number of particles are there in 5 grams of sodium carbonate.
How many particles are there in 5 grams of sodium carbonate?There are 6.022 × 1023 particles in one gram of a substance according to Avogadro's number. So when we find out for 5 grams, then we multiply 5 with 6.022 × 1023, we get 3.01 × 1024 particles. For one gram atomic weight of hydrogen, one mole of hydrogen contains 6.022 × 1023 hydrogen atoms.
So we can conclude that 3.01× 1024 particles are the number of particles are there in 5 grams of sodium carbonate.
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