Explanation:
The given balanced chemical equation is:
[tex]2 HI(g) <--> H_2 (g) + I_2 (g)[/tex]
The value of Kc at 445oC is 0.020.
[HI]=1.5M
[H2]=2.50M
[I2]=0.05M
The value of Qc(reaction quotient ) is calculated as shown below:
Qc has the same expression as the equilibrium constant.
[tex]Qc=\frac{[H_2][I_2]}{[HI]^2} \\Qc=(2.50Mx0.05M)/(1.5M)^2\\Qc=0.055[/tex]
Qc>Kc,
Hence, the backward reaction is favored and the formation of Hi is favored.
Among the given options, the correct answer is option d. Qc is greater than Kc; more HI will be produced.
Enough of a monoprotic weak acid is dissolved in water to produce a 0.0118 M solution. The pH of the resulting solution is 2.32 . Calculate the Ka for the acid.
Answer:
1.94 × 10⁻³
Explanation:
Step 1: Calculate the concentration of H⁺ ions
We will use the definition of pH.
pH = -log [H⁺]
[H⁺] = antilog -pH = antilog -2.32 = 4.79 × 10⁻³ M
Step 2: Calculate the acid dissociation constant (Ka) of the acid
For a monoprotic weak acid, whose concentration (Ca) is 0.0118 M, we can use the following expression.
Ka = [H⁺]²/Ca
Ka = (4.79 × 10⁻³)²/0.0118 = 1.94 × 10⁻³
Predict the products (if any) that will be formed by the reaction below. If no reaction occurs, write NR after the reaction arrow.
2HClO4(aq) + Co(s) -->
Answer:
The product is aqueous [tex]CO(HCl)_2[/tex] and [tex]O_2(g)[/tex].
Explanation:
Given:
⇒ [tex]2HClO_4(aq) +CO(s)[/tex]
then,
The reaction will be:
⇒ [tex]2HClO_4(aq)+CO(s) \rightarrow CO(HCl)_2 +O_2 (g)[/tex]
In the above reaction, we can see that
The products is:
aqueous [tex]CO(HCl)_2[/tex] and [tex]O_2(g)[/tex]
Thus the above is the correct answer.
A complex ion that forms in solution has a structure that:____.
a. can be determined simply by stoichiometry.
b. can be predicted on the basis of electrical charge.
c. can only be determined experimentally.
d. cannot be determined.
Answer:
can only be determined experimentally.
Explanation:
In the early days of inorganic chemistry, the structure of complex ions remained a mystery hence the name ''complex''.
These ions appear to have structures that defied accurate elucidation. However, by diligent laboratory investigation, Alfred Werner was able to accurately determine the structure of cobalt complexes. As a result of this, he is regarded as a pathfinder in coordination chemistry.
Hence, the structure of complex ions can only be determined experimentally.
Answer:
c. can only be determined experimentally
Explanation:
It is not possible to know for certain the structure of a complex ion on the basis of stoichiometry or by the electrical charges on the components. The structure of the resulting complex ion can only be known by experiment.
Adding more than one equivalent of HCl to pent-1-yne will lead to which product:______.
a. 1,2-dichloro-1-butene.
b. 1,1-dichloropentane.
c. 2,2-dichloropentane.
d. 2,2-dichlorobutane.
Answer:
c. 2,2-dichloropentane.
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to firstly draw the structure of the reactant, pent-1-yne:
[tex]CH\equiv C-CH_2-CH_2-CH_2[/tex]
Now, we infer the halogen is added to the carbon atom with the most carbon atoms next to it, in this case, carbon #2, in order to write the following product:
[tex]CH\equiv C-CH_2-CH_2-CH_2+2HCl\rightarrow CH_3- CCl_2-CH_2-CH_2-CH_2[/tex]
Whose name is 2,2-dichloropentane.
Regards!
A sample of gas contains 0.1200 mol of H2(g) and 0.1200 mol of O2(g) and occupies a volume of 11.5 L. The following reaction
takes place:
H2(g) + O2(g)>H2O2(g)
Calculate the volume of the sample after the reaction takes place, assuming that the temperature and the pressure remain constant.
L
Answer:
5.75L is the volume of the sample after the reaction
Explanation:
Based on the reaction, 1 mole of H2 reacts with 1 mole of O2 to produce 1 mole of H2O2.
As in the reaction, 0.1200 moles of H2 and 0.1200 moles of O2 are added, 0.1200 moles of H2O2 are produced.
Before the reaction, the moles of gas are 0.2400 moles and after the reaction the moles are 0.1200 moles of gas.
Based on Avogadro's law, the moles of a gas are directly proportional to the volume under temperatura and pressure constant. The equation is:
V1/n1 = V2/n2
Where V is volume and n are moles of 1, initial state and 2, final state.
Replacing:
V1 = 11.5L
n1 = 0.2400 moles
V2 = ?
n2 = 0.1200 moles
11.5L*0.1200 moles / 0.2400 moles = V2
V2 = 5.75L is the volume of the sample after the reaction
Write chemical equations for the reactions that occur when solutions of the following substances are mixed:
a. HNO₂ (nitrous acid) and C₂H₇NO (aq) ethanolamine, a base.
b. H₃O+ and F-
a) HNO₂ + C₂H₇NO → N₂ + C₂H₆O + H₂O
b) H₃O⁺ + F⁻ → HF + H₂O
[tex]\large\color{lime}\boxed{\colorbox{black}{Answer : - }}[/tex]
a) HNO₂ + C₂H₇NO → N₂ + C₂H₆O + H₂O
b) H₃O⁺ + F⁻ → HF + H₂O
Cesium- is radioactive and has a half life of years. Calculate the activity of a sample of cesium-. Give your answer in becquerels and in curies. Round your answer to significant digits.
The question is incomplete, the complete question is;
Cesium-137 is radioactive and has a half life of 30. years. Calculate the activity of a 6.8 mg sample of cesium-137. Give your answer in becquerels and in curies. Round your answer to 2 significant digits Bq Ci
Answer:
See explanation
Explanation:
The formula for activity is;
R= 0.693N/t1/2
N= 6.02 ×10^23 mol × 6.8 ×10^-3g/137 g/mol = 3 × 10^19
Substituting into the formula;
R= 0.693 × 3 × 10^19/30 years
R= 6.93 ×10^17 y^-1
In Bq;
6.93 ×10^17 y^-1 × 1.00y/3.16 ×10^7 seconds
= 2.19 ×10^10 Bq
In Ci;
2.19 ×10^10 Bq/3.7 ×10^10 Bq/Ci
= 0.59 Ci
For the following reaction, 4.77 grams of carbon (graphite) are allowed to react with 16.4 grams of oxygen gas.
carbon (graphite) (s) + oxygen (g) → carbon dioxide (g)
1. What is the maximum amount of carbon dioxide that can be formed?
2. What is the FORMULA for the limiting reagent?
3. What mass of the excess reagent remains after the reaction is complete?
Answer:
1. 17.5 g of CO₂
2. The limiting reactant is carbon (graphite), and its formula is C(graphite)
3. 3.7 g of O₂
Explanation:
First, we have to write the chemical equation for the reaction. For this, we have to know the chemical formula of each reactant and product:
Reactants: carbon(graphite) ⇒ C(graphite) ; oxygen gas ⇒ O₂(g)Products: carbon dioxide ⇒ CO₂(g)Thus, we write the chemical equation:
C(graphite) + O₂(g) → CO₂(g)
The equation is already balanced because it has the same number of C and O atoms on both sides. Thus, we can see that 1 mol of C(graphite) reacts with 1 mol of O₂ and produce 1 mol of CO₂ (mole-to-mole reaction).
Now we convert the grams of reactants to moles by using the molecular weight (Mw) of each compound:
Mw(C) = 12 g/mol
moles of C(graphite) = 4.77g/(12 g/mol) = 0.3975 mol
Mw(O₂) = 16 g/mol x 2 = 32 g/mol
moles of O₂ = 16.4 g/(32 g/mol) = 0.5125 mol
Now, we can compare the stoichiometric ratio (given by the moles of reactants in the equation) with the actual ratio (given by the mass of reactants we have):
stoichiometric ratio ⇒ 1 mol C(graphite)/mol O₂
actual ratio ⇒ 0.3975 mol C(graphite)/0.5125 mol O₂
We can see that we need 0.3975 moles of O₂ to react with C(graphite) and we have more moles (0.5125 mol) so the excess reactant is O₂. Thus, the limiting reactant is C(graphite).
The amount of product (CO₂) that is formed is calculated from the amount of limiting reactant. We can see in the chemical equation that 1 mol of CO₂ is produced from 1 mol of C(graphite) ⇒ stoichiometric ratio = 1 mol CO₂/mol C(graphite).
Thus, we multiply the moles of C(graphite) we have by the stoichiometric ratio to calculate the moles of CO₂ produced:
moles of CO₂ = 0.3975 mol C(graphite) x 1 mol CO₂/mol C(graphite) = 0.3975 mol CO₂
Now, we convert the moles of CO₂ to mass by using the Mw:
Mw(CO₂) = 12 g/mol + (16 g/mol x 2) = 44 g/mol
mass of CO₂ = 0.3975 mol CO₂ x 44 g/mol CO₂ = 17.5 g
Therefore, the maximum amount of carbon dioxide (CO₂) formed is 17.5 g.
Since this is a mole-to-mole reaction, the moles of excess reactant that remains after the reaction is complete is calculated as the difference between the moles of excess reactant and limiting reactant:
remaining moles of O₂ = 0.5125 mol - 0.3975 mol = 0.115 mol O₂
Finally, we convert the moles of O₂ to mass with the Mw (32 g/mol) :
mass of O₂ = 0.115 mol O₂ x 32 g/mol = 3.68 g
Therefore, the mass of the excess reagent that remains after the reaction is complete is 3.7 g.
How many Noble gases we have in Periodic Table???
Answer:
Six
Explanation:
2. XC12 is the chloride of metal X. The formulae of its sulphate is
Answer:
XSO₄
Explanation:
XCl₂ is the chloride of metal X. The sum of the charges of the cation and the anion must be zero because the salt is electrically neutral. The charge of the cation of X is:
1 × X + 2 × Cl = 0
1 × X + 2 × (-1) = 0
X = +2
X has a charge +2 and sulphate (SO₄²⁻) a charge -2. The neutral salt they form is XSO₄.
Hydrogengasand oxygengas react to form water vapor. Suppose you have of and of in a reactor. Calculate the largest amount of that could be produced. Round your answer to the nearest .
The question is incomplete. The complete question is :
Hydrogen [tex](H_2)[/tex] gas and oxygen [tex](O_2)[/tex] gas react to form water vapor [tex](H_2O)[/tex]. Suppose you have 11.0 mol of [tex]H_2[/tex] and 13.0 mol of [tex]O_2[/tex] in a reactor. Calculate the largest amount of [tex]H_2O[/tex] that could be produced. Round your answer to the nearest 0.1 mol .
Solution :
The balanced reaction for reaction is :
[tex]$2H_2(g) \ \ \ \ + \ \ \ \ \ O_2(g)\ \ \ \rightarrow \ \ \ \ 2H_2O(g)$[/tex]
11.0 13.0
11/2 13/1 (dividing by the co-efficient)
6.5 mol 13 mol (minimum is limiting reagent as it is completely consumed during the reaction)
Therefore, [tex]H_2[/tex] is limiting reagent. It's stoichiometry decides the product formation amount from equation above it is clear that number of moles for [tex]H_2O[/tex] will be produced = number of moles of [tex]H_2[/tex]
= 11.0 mol
explain why it is important not to correct any gas from the first few seconds of the experiment
Answer:
gu kha fuschhehdjdvdbeodbr
When 3-methyl-1-pentene is treated with in dichloromethane, the major product is 1-bromo-3-methyl-2-pentene.
a. True
b. False
Answer:
True
Explanation:
When Methyl Pentene is introduced in a chemical reaction with dichloromethane then the major product will be bromomethylpentene. There can be small amount of bromo methyl pentene than the amount of methyl pentene introduced for reaction.
8) Determine whether mixing each pair of the following results in a buffera. 100.0 mL of 0.10 M NH3 with 100.0 mL of 0.15 MNH4Cl b. 50.0 mL of 0.10 M HCL with 35.0 mL of 0.150 M NaOHc. 50.0 mL of 0.15 M HF with 20.0 mL of 0.15 M NaOHd. 175.0 mL of 0.10 M NH3 with 150.0 mL of 0.12 M NaOH
Answer:
a. 100.0 mL of 0.10 M NH₃ with 100.0 mL of 0.15 M NH₄Cl.
c. 50.0 mL of 0.15 M HF with 20.0 mL of 0.15 M NaOH.
Explanation:
A buffer system is formed in 1 of 2 ways:
A weak acid and its conjugate base.A weak base and its conjugate acid.Determine whether mixing each pair of the following results in a buffer.
a. 100.0 mL of 0.10 M NH₃ with 100.0 mL of 0.15 M NH₄Cl.
YES. NH₃ is a weak base and NH₄⁺ (from NH₄Cl ) is its conjugate base.
b. 50.0 mL of 0.10 M HCl with 35.0 mL of 0.150 M NaOH.
NO. HCl is a strong acid and NaOH is a strong base.
c. 50.0 mL of 0.15 M HF with 20.0 mL of 0.15 M NaOH.
YES. HF is a weak acid and it reacts with NaOH to form NaF, which contains F⁻ (its conjugate base).
d. 175.0 mL of 0.10 M NH₃ with 150.0 mL of 0.12 M NaOH.
NO. Both are bases.
Which of the following statements is true about what happens in all chemical reactions? A. The ways in which atoms are joined together is not changed. B. Bonds between atoms are broken and new bonds are formed. C. The final substances are called reactants.
Answer:
B.bonds are broken and new bonds are formed
A quantity of ideal gas requires 800 kJ to raise the temperature of the gas by 10.0 K when the gas is maintained at constant volume. The same quantity of gas requires 900 kJ to raise the temperature of the gas by 10.0 K when the gas is maintained at constant pressure. What is the adiabatic gas constant of this gas
Answer:
[tex]\gamma=1.125[/tex]
Explanation:
From the question we are told that:
Initial Heat [tex]Q_1=800kJ[/tex]
initial Temperature [tex]T_1=10.0K[/tex]
Final Heat [tex]Q_2=800kJ[/tex]
Final Temperature [tex]T_2=10.0K[/tex]
Generally the equation for Adiabatic constant is mathematically given by
[tex]\gamma=\frac{Cp}{Cv}[/tex]
Since
Equation for Heat [tex]dQ=nCdT[/tex]
Where
[tex]n_1=n_2\\\\T_1=T_2[/tex]
Therefore
[tex]Q_1=Cv\\\\Cv=800[/tex]
And
[tex]Cp=900[/tex]
Therefore
[tex]\gamma=\frac{900}{800}\\\\\gamma=\frac{9}{8}[/tex]
[tex]\gamma=1.125[/tex]
Carbon dioxide gas is collected at 27.0 oC in an evacuated flask with a measured volume of 30.0L. When all the gas has been collected, the pressure in the flask is measured to be 0.480atm. Calculate the mass and number of moles of carbon dioxide gas that were collected.
Answer:
[tex]M_{CO_2}= 25.7g[/tex]
Explanation:
From the question we are told that:
Temperature [tex]T=27.0[/tex]
Volume [tex]V=30L[/tex]
Pressure [tex]P=0.480atm[/tex]
Generally the equation for Ideal gas is mathematically given by
PV=nRT
Therefore
[tex]n=\frac{0.480 x 30}{0.08205 x 300}[/tex]
[tex]n=0.59moles[/tex]
Generally Mass of CO2 is given as
[tex]M_{CO_2}= 0.59 * 44 g/mol[/tex]
[tex]M_{CO_2}= 25.7g[/tex]
90
1
39
is the
In the following decay equation,
90
Sy →
38
et
90
39
-1
A. alpha particle
B. parent element
C. daughter element
D. beta particle
Answer:
D. beta particle
Explanation:
Number of protons increases from 38 to 39 indicating beta decay (only one proton up from parent isotope to daughter isotope) Also atomic mass (on top of an isotope), 90 stays the same as beta particle is very small.
Balance the redox reaction Al(s) + MnO4^- (aq) --> MnO2 (s) + Al(OH)4^- (aq) in aqueous basic solution
Answer:
Al + MnO4- + 2H2O → Al(OH)4- + MnO2
Explanation:
First of all, we out down the skeleton equation;
Al + MnO4- → MnO2 + Al(OH)4-
Secondly, we write the oxidation and reduction equation in basic medium;
Oxidation half equation:Al + 4H2O + 4OH- → Al(OH)4- + 4H2O + 3e-
Reduction half equation:MnO4- + 4H2O + 3e- → MnO2 + 2H2O + 4OH-
Thirdly, we add the two half reactions together to obtain:
Al + MnO4- + 8H2O + 4OH- + 3e- → Al(OH)4- + MnO2 + 6H2O + 3e- + 4OH-
Lastly, cancel out species that occur on both sides of the reaction equation;
Al + MnO4- + 8H2O→ Al(OH)4- + MnO2 + 6H2O
The simplified equation now becomes;
Al + MnO4- + 2H2O → Al(OH)4- + MnO2
By process of incineration, a mystery substance is empirically determined to contain 40.00% carbon by weight, 6.67% hydrogen, and 53.33% oxygen. Its molecular weight ranges between 55 and 62 g/mole. a. (6 points) Determine the chemical formula of this substance
Answer:
C₂H₄O₂
Explanation:
Step 1: Divide each percentage by the atomic mass of the element
C: 40.00/12.01 = 3.331
H: 6.67/1.01 = 6.60
O: 53.33/16.00 = 3.333
Step 2: Divide all the numbers by the smallest one
C: 3.331/3.331 = 1
H: 6.60/3.331 ≈ 2
O: 3.333/3.331 ≈ 1
The empirical formula is CH₂O, with a molecular weight of 12 g/mol + 2 × 1 g/mol + 16 g/mol = 30 g/mol. The molecular weight of the compound must be a product of 30, such as 60 (between 55 and 62 g/mol). Since we have to multiply by 2 (30 to 60) to get to the molecular weight of the compound, we also have to multiply the empirical formula by 2 to get the chemical formula of the compound.
CH₂O × 2 = C₂H₄O₂
Suppose you are studying the Ksp of CaCl2, which has a molar mass of 110.98 g/mol, at multiple temperatures. You dissolve 4.99 g of CaCl2 in 10.0 mL of water at 100 oC and cool the solution. At 90 oC, a solid begins to appear. What is the Ksp of CaCl2 at 90 oC
Answer:
Hence the Solubility product,
Ksp = [Ca2+] [Cl-]2
or, Ksp = (4.5) (9)2
or, Ksp = 364.5
Explanation:
Mass of CaCl2 = 4.99 g
Molar mass of CaCl2 = 110.98 g/mol
Moles of CaCl2
= given mass/ molar mass
= 4.99/ 110.98
= 0.045
Volume = 10.0 mL = 0.01 L
CaCl2 dissociates into its ion as:
CaCl2 (s) \rightleftharpoons Ca2+ (aq) + 2 Cl- (aq)
At 90°C, the solution is saturated with Ca2+ and Cl- ions.
Moles of Ca2+ = Moles of CaCl2 dissolved = 0.045
Moles of Cl- = 2 x ( Moles of CaCl2 dissolved) = 2 x 0.045 = 0.09
[Ca2+] = Moles/ Volume = 0.045/ 0.01 = 4.5 M
[Cl-] = 0.09/ 0.01 = 9 M
Solubility product,
Ksp = [Ca2+] [Cl-]2
or, Ksp = (4.5) (9)2
or, Ksp = 364.5
A sample of neon gas occupies 105 L at 27°C under a pressure of
985 torr. What volume would it occupy at standard condition
Answer: Volume occupied by given neon sample at standard condition is 123.84 L.
Explanation:
Given: [tex]V_{1}[/tex] = 105 L, [tex]T_{1} = 27^{o}C = (27 + 273) K = 300 K[/tex], [tex]P_{1}[/tex] = 985 torr
At standard conditions,
[tex]T_{2}[/tex] = 273 K, [tex]P_{2}[/tex] = 760 K, [tex]V_{2}[/tex] = ?
Formula used to calculate the volume is as follows.
[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}[/tex]
Substitute the values into above formula as follows.
[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{985 torr \times 105 L}{300 K} = \frac{760 torr \times V_{2}}{273 K}\\V_{2} = \frac{94116.75}{760} L\\= 123.84 L[/tex]
Thus, we can conclude that volume occupied by given neon sample at standard condition is 123.84 L.
Arrange the following in order of increasing melting point: NaCl, H2O, CH4, C6H4(OH)2.
a. NaCl < H2O < CH4 < C6H4(OH)2
b. CH4 < H2O < NaCl < C6H4(OH)2
c. CH4 < H2O < C6H4(OH)2 < NaCl
d. CH4 < C6H4(OH)2 < H2O < NaCl
e. CH4 < NaCl < C6H4(OH)2 < H2O
Explanation:
one thing to know is that higher surface area = higher boiling point.
NaCl has the smallest surface area, so it's the first one.
H2O has less surface area than methane, so it's second.
Methane has more surface area than H20, so it's third.
The big molecule has the most surface area, so it's last
Temperature measures the average kinetic energy of particles of the substances. Melting point is directly proportional to surface area. Therefore, the correct option is option C.
What is temperature?Temperature is used to measure degree or intensity of heat of a particular substance. Temperature is measured by an instrument called thermometer.
Temperature can be measured in degree Celsius °c, Kelvin k or in Fahrenheit. Temperature is a physical quantity. Heat always flow from higher temperature source to lower temperature source.
We can convert these units of temperature into one another. The relationship between degree Celsius and Fahrenheit can be expressed as:
°C={5(°F-32)}÷9
Melting point is directly proportional to surface area. NaCl has the smallest surface area. Water has less surface area than methane. Methane has more surface area than H[tex]_2[/tex]O.
Therefore, the correct option is option C.
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A 0.204 g sample of a CO3 2- antacid is dissolved with 25.0ml of 0.0981 M HCL. The hydrochloric acid that is not neutralized by the antacid is titrated to a bromophenol blue endpoint with 5.83 ml of 0.104 M NaOH. Assuming the active ingredient in the antsacid sample is CaCO3, calculate the mass of CaCO3 in the sample.
Answer:
0.0922 g
Explanation:
Number of moles of acid present = 25/1000 × 0.0981
= 0.00245 moles
Number of moles of base = 5.83/1000 × 0.104
= 0.000606 moles
Since the reaction of HCl and NaOH is 1:1
Number of moles of HCl that reacted with antacid = 0.00245 moles - 0.000606 moles
= 0.001844 moles
From the reaction;
CaCO3 + 2HCl ----> CaCl2 + H2O + CO2
1 mole of CaCO3 reacts with 2 moles of HCl
x moles of CaCO3 reacts with 0.001844 moles ofHCl
x = 1 × 0.001844/2
= 0.000922 moles
Mass of CaCO3 = 0.000922 moles × 100 g/mol
= 0.0922 g
Why does nitrogen not show allotropy?
Answer:
Nitrogen does not show allotropy because of its small size and high electronegativity. The single N-N bond is weaker than P-P bond because of high inter electronic repulsions among non-bonding electrons due to the small bond distance. Hence it does not show allotropy.
Answer:
The nitrogen atom has short inter-bond distance, hence highly electronegative in terms of magnitude. This creates no relation in energy varieties hence no allotropes formed.
Nitrogen atom is also very small.
Calculate the percent error in the atomic weight if the mass of a Cu electrode increased by 0.4391 g and 6.238x10-3 moles of Cu was produced. Select the response with the correct Significant figures. You may assume the molar mass of elemental copper is 63.546 g/mol. Refer to Appendix D as a guide for this calculation.
Answer:
10.77%
Explanation:
Molar mass of Cu = mass deposited/number of moles of Cu
Molar mass of Cu = 0.4391 g/6.238x10^-3 moles
Molar mass of Cu = 70.391 g/mol
%error = 70.391 g/mol - 63.546 g/mol/63.546 g/mol × 100
%error = 10.77%
As discussed in class, the Fischer esterification reactants and products are at equilibrium. How was the equilibrium of the reaction that you performed shifted towards the products
Answer:
See explanation
Explanation:
The particular reactants in the Fischer esterification reaction were not stated.
Generally, a Fischer esterification is a reaction that proceeds as follows;
RCOOH + R'OH ⇄RCOOR' + H2O
This reaction occurs in the presence of an acid catalyst.
We can shift the equilibrium of this reaction towards the products side in two ways;
I) use of a large excess of either of the reactants
ii) removal of one of the products as it is formed.
Any of these methods shifts the equilibrium of the Fischer esterification reaction towards the products side.
Determine whether each of the examples represents a colligative property or a non-colligative property. boiling point elevation Choose... color Choose... freezing point depression Choose... vapor pressure lowering Choose... density Choose...
Answer:
boiling point elevation - colligative property
color - non-colligative property
freezing point depression - colligative property
vapor pressure lowering - colligative property
density - non-colligative property
Explanation:
A colligative property is a property that depends on the number of particles present in the system.
Freezing point depression, boiling point elevation and vapour pressure lowering are all colligative properties of solutions.
Colour and density do not depend on the number of particles present hence they are not colligative properties.
The boiling point elevation, freezing point depression, and vapor pressure lowering are colligative properties. And color and density are non-colligative properties.
Explanation:
The colligative properties are the properties depending upon the number of particles of solute not on the nature of the solute.Example of colligative properties:Vapor pressure loweringElevation boiling pointDepression in freezing pointOsmotic pressureThe non-colligative properties are the properties depending upon the nature of solute and solvent.Example of non-colligative properties :ViscositySurface tensionDensitySolubilitySo, from this, we can conclude that boiling point elevation, freezing point depression, and vapor pressure lowering are colligative properties. And color and density are non-colligative properties.
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World leaders met in Copenhagen, Denmark in December 2009, to try to come up with an agreement that
would lead to reducing greenhouse gas emissions. They agreed that we need to reduce carbon dioxide
emissions 80% by the year 2050. How does Lester Brown feel about that?
Select one:
O a. He thinks it is a great achievement.
b. He thinks it is not fast enough
O c. He thinks that we do not need international cooperation
d. He thinks that greenhouse gas emissions are not the most important factor.
Clear my choice
Lester Brown thinks reducing carbon dioxide emissions 80% by 2050 is not fast enough.
Lester Brown is an American environmentalist who has focused on studying the environment and its protection. In recent years, he has made alerts for world leaders and large industries to strive to stop CO2 emissions because this greenhouse gas has a massive influence on global warming.
Therefore, Lester Brown considers that the projections of reduction of greenhouse gases (especially CO2) made by the world powers for the year 2050, ignore the reality because he considers that CO2 emissions must decrease by at least one 80% in 2020 to avoid drastic consequences in current living conditions. Therefore, the answer is B.
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Identify the element for each of these electron configurations. Then determine whether this configuration is the ground state or an excited state.
a. 1s2 2s2 2p6 3s2 3p6 4s2 3d 9
b. 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f 14 5d7
Answer:
a. 1s2 2s2 2p6 3s2 3p6 4s2 3d 9 - the element is zinc and this is an excited state configuration
b. 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f 14 5d7 -the element is iridium and this is a ground state configuration
Explanation:
The ground state is the lowest energy state of an atom while excite states of Zn atom are higher energy states.
The configuration, 1s2 2s2 2p6 3s2 3p6 4s2 3d 9 applies to zinc atom in excited state while the configuration 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f 14 5d7 applies to iridium atom in ground state.
