Answer:
yes it is corrwect iyt is absolitle correct
Explanation:
Melanie has completed the analysis of her data for the reaction of KMnO4 with malonic acid and data for a reaction of KMnO4 with tartaric acid. She compared the activation energies, Ea, she calculated for the two reactions and found the Ea for the malonic acid reaction to be greater than the Ea for the tartaric acid reaction.
Required:
What does this mean about the magnitude of the rate constant, k, and the rate of the reaction?
Answer:
See explanation
Explanation:
The relationship between the activation energy and rate of reaction is best captured by the Arrhenius equation;
k= Ae^-Ea/RT
Where;
k= rate constant
A= pre-exponential factor
Ea=activation energy
R= gas constant
T= temperature
We can see from the foregoing that, as the activation energy increases, the rate of reaction decreases and vice versa. reactions that have a very high activation energy are markedly slow.
Since the activation energy for the malonic acid reaction is found to be greater than the activation energy for the tartaric acid reaction, then the rate of the malonic acid reaction(k) will be slower than that of the tartaric acid reaction.
The study of chemistry and bonds is called chemistry. There are two types of elements metal and nonmetals.
The correct answer is mentioned below.
What is the Arrhenius equation?The relationship between the activation energy and rate of reaction is best captured by the Arrhenius equationThe equation is as follows:-
[tex]k= Ae^{-Ea/RT[/tex] Where;
k= rate constantA= pre-exponential factorEa=activation energyR= gas constantT= temperatureWe can see from the foregoing that, as the activation energy increases, the rate of reaction decreases and vice versa. reactions that have very high activation energy are markedly slow. Since the activation energy for the malonic acid reaction is found to be greater than the activation energy for the tartaric acid reaction, then the rate of the malonic acid reaction(k) will be slower than that of the tartaric acid reaction.
Hence, the correct answer is mentioned above.
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How many neutrons does Carbon- 14 and Carbon -15 have? *
Answer: 8 for both
Explanation:
A 1 liter solution contains 0.370 M hypochlorous acid and 0.493 M sodium hypochlorite. Addition of 0.092 moles of barium hydroxide will: (Assume that the volume does not change upon the addition of barium hydroxide.)
In the original solution you have the mixture of a weak acid (Hypochlorous acid) and its conjugate base (Sodium hypochlorite). That is a buffer.
The barium hydroxide will react with hypochlorous acid. If this reaction cause the complete reaction of hypochlorous acid, the buffer break its capacity and the pH change in several units. In this case:
The addition of barium hydroxide will raise the pH slightly because the buffer still working.
The initial moles of those species are:
Hypochlorous acid:
[tex]1L * \frac{0.370mol}{1L} = 0.370 moles[/tex]
Sodium hypochlorite:
[tex]1L * \frac{0.493mol}{1L} = 0.493 moles[/tex]
Now, a strong acid as barium hydroxide (Ba(OH)₂) reacts with a weak acid as hypochlorous acid (HClO) as follows:
Ba(OH)₂ + 2HClO → Ba(ClO)₂ + 2H₂O
For a complete reaction of 0.092 moles of barium hydroxide are required:
[tex]0.092 moles Ba(OH)_2*\frac{2mol HClO}{1molBa(OH)_2} = 0.184 moles HClO[/tex]
As there are 0.370 moles, the moles of HClO after the reaction are:
0.370 moles - 0.184 moles = 0.186 moles of HClO will remain
As you still have hypochlorite and hypochlorous acid you still have a buffer.
Thus, the pH will raise slightly because the amount of acid is decreasing and slightly because the buffer can keep the pH.
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Which of the following are examples of single replacement reactions? Select all that apply.
Answer:
Na2S(aq)+Cd(No3)2(aq)=CdS(s)+2NaNo3(aq)
Answer: it’s checkbox 2&3
A central atom has two lone pairs on opposite sides and four single bonds. What is the molecule geometry of the result?
A. octahedral
B. tetrahedral
C. square planar
D. linear
The correct answer is C. square planar
According to the Valence Shell Electron Pair Repulsion Theory(VSEPR), The shape of a molecule depends on the number of electron pairs in the molecule.
VSEPR theory was first coined by Gillespie and Nyhlom in 1957 as an improvement over the Sidgwick - Powell theory.
According to this theory, the shape of a molecule is determined by the number of electron pairs that surround the valence shell of the central atom in the molecule. The electron pairs are positioned as far apart in space as possible to minimize repulsion of electron pairs.
However, the presence of lone pairs distorts the shape anticipated for the molecule on the basis of VSEPR.
For a molecule having six electron pairs, an octahedral geometry is expected(electron domain geometry). However, the presence of two lone pairs which are positioned at opposite side of the four single bonds leads to an observed square planar molecular geometry.
https://brainly.com/question/13591921
Answer:
square planar
Explanation:
In the graphic, 195 represents the _______.
195 Pt
78
A. Atomic Mass
B. Atomic Number
C. Neutron Number
Answer:
ITS ANSWER IS
OPTION B. ATOMIC NUMBER
HI HAVE A NICE DAY
Heating water makes most solids in it
soluble, and it makes gases
soluble.
Increasing the pressure on a gas above water makes the gas
soluble. Compounds with comparatively stronger ionic bonds are
soluble.
Answer:
1. more
2. less
3. more
4. less
Explanation:
Consider the reaction: NaNO3(s) + H2SO4(l) NaHSO4(s) + HNO3(g) ΔH° = 21.2 kJ
How much heat must absorbed by the reaction system to convert 100g of NaNO3 into NaHSO4(s)?
Answer:
endet nach selam nw
4gh7
PLEASE HELP ASAP!!!
Answer:
I don't know What can I do.
What volume of each solution contains 0.14 mol of KCl? Express your answer using two significant figures.
1.8 M KCl
Answer:
Solution given:
1 mole of KCl[tex]\rightarrow [/tex]22.4l
1 mole of KCl[tex]\rightarrow [/tex]74.55g
we have
0.14 mole of KCl[tex]\rightarrow [/tex]74.55*0.14=10.347g
74.55g of KCl[tex]\rightarrow [/tex]22.4l
10.347 g of KCl[tex]\rightarrow [/tex]22.4/74.55*10.347=3.11litre
volume of each solution contains 0.14 mol of KCl contain 3.11litre.
[tex]\:[/tex]
1 mole of KCl → 22.4l
1 mole of KCl → 74.55g
we have
0.14 mole of KCl → 74.55*0.14=10.347g
74.55g of KCl → 22.4l
10.347 g of KCl → 22.4/74.55*10.347=3.11litre
volume of each solution contains 0.14 mol of KCl contain 3.11litre.
The density of toluene (C7H8) is 0.867 and that of thiophene (C4H4S) is 1.065 g/ml. A solution is made by dissolving 10.00g thiophene in 250.00ml of toluene. a)Calculate the molarity of the solution
b)Assuming the volume are addictive ,calculate the molarity of the solution
Answer:
Calcular la molaridad de una solución que se preparó disolviendo 14 g de KOH en suficiente
agua para obtener 250 mL de solución. (masa molar del KOH = 56 g/mol).
Resolución: de acuerdo a la definición de “molaridad” debemos calcular primero, el número de mol de soluto (KOH) que
se han disuelto en el volumen dado, es decir, “se transforma g de soluto a mol de soluto” por medio de la masa molar,
así:
56 g de KOH 14 g de KOH
----------------- = ------------------- X = 0,25 mol de KOH
1 mol X
Ahora, de acuerdo con la definición de molaridad, el número de mol debe estar contenido en 1000 mL (o 1 L) de
solución, que es el volumen estándar para esta unidad de concentración, lo que se determina con el siguiente planteamiento:
0,25 mol X
----------------------- = ------------------------- X = 1 mol de KOH
250 mL de solución 1000 mL de solución
Explanation:
Hãy cho biết giá trị và ý nghĩa của số lượng tử n, l, m, ms khi mô tả trạng thái của electron trong nguyên tử?
which of the following is is a chemical property of pure water
Answer:
Pure water has an acidity of about 7 on the pH scale. -is a chemical property of pure water. Pure water has an acidity of about 7 on the pH scale
Answer: không màu , không mùi không vị
Explanation:
Draw a formula for Thr-Gly-Ala (T-G-A) in its predominant ionic form at pH 7.3. You may assume for the purposes of this question that the pKa values of the acidic groups of amino acid residues in the peptide are the same as in the amino acid itself.
Answer:
gggggggggg
Explanation:
gggggggg
The tripeptide formed from threonine, glycine and alanine is neutral at the pH of 7.3. The carboxylic end is negative charged by donating its proton to form the NH₃⁺ group.
What is peptide?Peptides are protein units formed from two or more amino acids bonded through peptide bonds. There are essential and non-essential amino acids. Essential amino acids have to be uptake from food and non-essential amino acids are synthesized inside the body.
Threonine is an essential amino acid with a CH₃CHOH side group. Glycine has the simplest side group hydrogen and alanine has CH₃ side chain. Both glycine and alanine are non-essential amino acids.
Each amino acids are represented with a three letter code or one letter symbol. Thus threonine is T, G for glycine and A for alanine. At a pH of 7.3 the peptide formed from these amino-acids contains a negatively charged carboxylic end.
A positively charged amino end made by protonation from the acid group make the overall charge zero. The structure of the peptide is given in the uploaded image.
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Forcus on the yellow highlighted texts, your help is appreciated.
[tex]{ \sf{ \red{no \: pranks}}}[/tex]
Answer:
Transition temperature is the temperature at which a substance changes from one state to another.
Allotropy is the existence of an element in many forms.
What volume of 1.50 mol/L stock solution is needed to make 125 mL of 0.60 mol/L solution?
Chemistry 11 Solutions
978Ͳ0Ͳ07Ͳ105107Ͳ1Chapter 8 Solutions and Their Properties • MHR | 85
Amount in moles, n, of the NaCl(s):
NaCl
2.5 g
m
n
M
58.44 g
2
4.2778 10 m l
ol
o
/m
u
Molar concentration, c, of the NaCl(aq):
–2 4.2778 × 10 mol
0.100
0.42778 mol/L
0.43 mol
L
/L
n
c
V
The molar concentration of the saline solution is 0.43 mol/L.
Check Your Solution
The units are correct and the answer correctly shows two significant digits. The
dilution of the original concentrated solution is correct and the change to mol/L
seems reasonable.
Section 8.4 Preparing Solutions in the Laboratory
Solutions for Practice Problems
Student Edition page 386
51. Practice Problem (page 386)
Suppose that you are given a stock solution of 1.50 mol/L ammonium sulfate,
(NH4)2SO4(aq).
What volume of the stock solution do you need to use to prepare each of the
following solutions?
a. 50.0 mL of 1.00 mol/L (NH4)2SO4(aq)
b. 2 × 102 mL of 0.800 mol/L (NH4)2SO4(aq)
c. 250 mL of 0.300 mol/L NH4
+
(aq)
What Is Required?
You need to calculate the initial volume, V1, of (NH4)2SO4(aq) stock solution
needed to prepare each given dilute solution.
The dilution gives the relationship between the molarity and the volume of the solution. The volume of stock solution with a molarity of 1.50 mol/L is 50 mL.
What is dilution?Dilution is said to be the addition of more volume to the concentrated solution to make it less in molar concentration. This tells about the inverse and indirect relationship between the volume and the molar concentration of the solution.
Given,
Initial volume = V₁
Initial molar concentration (M₁) = 1.50 mol/L
Final volume (V₂) = 125 mL = 0.125 L
Final molar concentration (M₂)= 0.60 mol/L
The dilution is calculated as:
M₁V₁ = M₂V₂
V₁ = M₂V₂ ÷ M₁
Substituting the values in the above formula as
V₁ = M₂V₂ ÷ M₁
V₁ = (0.60 mol/L × 0.125 L) ÷ 1.50 mol/ L
V₁ = 0.05 L
= 50 mL
Therefore, 50 mL of stock solution is needed to make a 0.60 mol/L solution.
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Determine the effect each given mutation would have on the rate of glycolysis in muscle cells.
a. loss of binding site for fructose 1 ,6-bisphophate in pyruvate kinase.
b. loss of allosteric binding site for ATP in pyruvate kinase.
c. loss of allosteric binding site for AMP in phosphofructokinase.
d. loss of regulatory binding site for ATP in phosphofructokinase.
1. Increase
2. decrease
3. No effect
Answer:
a. Decrease
b. Increase
c. Increase
d. No effect
Explanation:
Glycolysis is present in muscle cells which converts glucose to pyruvate, water and NADH. It produces two molecules of ATP. Cellular respiration produces more molecules of ATP from pyruvate in mitochondria. Glycolysis increases in pyruvate kinase.
a. Loss of binding site for fructose 1,6-bisphosphate in pyruvate kinase: Decrease
b. Loss of allosteric binding site for ATP in pyruvate kinase: No effect
c. Loss of allosteric binding site for AMP in phosphofructokinase: Increase
d. Loss of regulatory binding site for ATP in phosphofructokinase: Increase
A. An important substrate in the glycolysis pathway is fructose 1,6-bisphosphate. It stimulates pyruvate kinase, an essential enzyme in glycolysis. The amount of pyruvate kinase that is activated will decrease if the fructose 1,6-bisphosphate binding site in pyruvate kinase is eliminated. As a result the rate of glycolysis in the muscle cells will probably decrease.
B. The allosteric ATP binding site of pyruvate kinase controls how active the enzyme is. However, pyruvate kinase is not significantly regulated by ATP in muscle cells. Therefore, it is unlikely that deletion of the ATP-binding allosteric site in pyruvate kinase would have no effect on the rate of glycolysis in muscle cells.
C. The rate-limiting enzyme in glycolysis, phosphofructokinase, is activated from all forms by AMP. It increases the rate of glycolysis by stimulating the activity of phosphofructokinase. If the allosteric binding site for AMP is eliminated, phosphofructokinase activation will be reduced. As a result, the rate of glycolysis in muscle cells will decrease.
D. Phosphofructokinase is inhibited allosterically by ATP. It regulates the rate of glycolysis by a feedback mechanism. High ATP concentrations cause phosphofructokinase to bind to its regulatory site, limiting its activity and delaying glycolysis. If the regulatory binding site for ATP is eliminated, the inhibitory action of ATP on phosphofructokinase would be lost. As a result, muscle cells will glycolysis at a faster rate.
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A sample of Kr gas is observed to effuse through a pourous barrier in 8.15 minutes. Under the same conditions, the same number of moles of an unknown gas requires 4.53 minutes to effuse through the same barrier. The molar mass of the unknown gas is ____________ g/mol.
Answer:
25.88 g/mol
Explanation:
Graham's law is a famous law which states that the diffusion rate or the effusion rate of any gas varies inversely to the square root of the molecular weight the gas.
So from Graham's law, we have,
[tex]$\frac{\text{time}}{M^{1/2}}=\text{constant}$[/tex]
Using the sample of Kr gas having M = 83.8
[tex]$\frac{8.15}{(83.8)^{0.5}}= \frac{4.53}{M^{0.5}}$[/tex]
[tex]$M^{0.5}= 5.088$[/tex]
M = 25.88 g/mol
does anyone know how to solve this and what the answer would be?
Dynamic equilibrium is showed at the point at which solid liquid and gas intersect.
At the point at which solid liquid and gas intersect represents a system that shows dynamic equilibrium. There is equal amount of reactants and products at the point of dynamic equilibrium because the transition of substances occur between the reactants and products at equal rates, means that there is no net change. Reactants and products are formed at the rate that no change occur in their concentration.
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There are three isotopes of carbon. They have mass number of 12, 13 and 14. The average atomic mass of carbon is 12.0107 amu. What does this say about the relative abundances of the three isotopes?
Answer:
lots more of the carbon 12 than the others
havent calculated it percentage-wise but you can see its very close to 12 meaning it is of far greater abundance that carbon 13 and 14
Explanation:
which of the following measurements is equivalent to 5.461x10^-7m?
Answer:
B. 0.0000005461m
I used the method of moving the decimal.
Calculate the no. of moles in 15g of CaCl2
Answer:
[tex]\boxed {\boxed {\sf 0.14 \ mol \ CaCl_2}}[/tex]
Explanation:
We are asked to calculate the number of moles of 15 grams of calcium chloride (CaCl₂).
To convert from grams to moles, we use the molar mass, or the mass of 1 mole of a substance. Molar masses are found on the Periodic Table because they are equivalent to the atomic masses, but the units are grams per mole instead of atomic mass units.
Look up the individual elements in the compound: calcium and chloride.
Ca: 40.08 g/mol Cl: 35.45 g/molNotice the chemical formula has a subscript of 2 after Cl or chlorine. There are 2 moles of chlorine in every 1 mole of calcium chloride. We must multiply chlorine's molar mass by 2 before adding calcium's molar mass.
Cl₂: 35.45 * 2 = 70.9 g/mol CaCl₂= 40.08 + 70.9 = 110.98 g/molWe will convert using dimensional analysis, so we must create a ratio using the molar mass.
[tex]\frac {110.98 \ g \ CaCl_2}{ 1 \ mol \ CaCl_2}[/tex]
We are converting 15 grams of calcium chloride to moles, so we must multiply the ratio by this value.
[tex]15 \ g \ CaCl_2 *\frac {110.98 \ g \ CaCl_2}{ 1 \ mol \ CaCl_2}[/tex]
Flip the ratio so the units of grams of calcium chloride cancel.
[tex]15 \ g \ CaCl_2 *\frac { 1 \ mol \ CaCl_2}{110.98 \ g \ CaCl_2}[/tex]
[tex]15 *\frac { 1 \ mol \ CaCl_2}{110.98}[/tex]
[tex]\frac { 15}{110.98} \ mol \ CaCl_2[/tex]
[tex]0.1351594882\ mol \ CaCl_2[/tex]
The original measurement of grams (15) has 2 significant figures, so our answer must have the same. For the number we calculated, that is the hundredth place. The 5 in the thousandth place tells us to round the 3 up to a 4.
[tex]0.14 \ mol \ CaCl_2[/tex]
15 grams of calcium chloride is approximately 0.14 moles of calcium chloride.
Write balanced equations for the reaction of each of the following carboxylic acids with NaOH. Part A formic acid Express your answer as a chemical equation. A chemical reaction does not occur for this question. Request Answer Part B 3-chloropropanoic acid Express your answer as a chemical equation. nothing A chemical reaction does not occur for this question.
Answer:
Part A
HCOOH(aq) + NaOH(aq) → HCOONa(aq) + H2O(l)
Part B
ClCH2CH2CO2H(aq) + NaOH(aq) ------> ClCH2CH2CO2Na(aq) + H2O(l)
Explanation:
The reaction between an alkanoic acid and a base is a neutralization reaction. The reaction occurs as follows;
RCOOH + NaOH ----> RCOONa + H2O
We have to note the fact that the net ionic reaction still remains;
H^+(aq) + OH^-(aq) ---> H2O(l)
In both cases, the reaction can occur and they actually do occur as written.
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
How did Kepler's discoveries contribute to astronomy?
O They supported the heliocentric model.
O They established the laws of planetary motion.
O They explained how the Sun rises and sets.
O They made astronomy accessible to people who spoke Italian.
They made astronomy accessible to people who spoke italian
Answer:
"They established the laws of planetary motion"
Explanation:
Mr. Kepler was the astronomer who came up with the "Laws of Planetary Motion."
name a factor tht affects the value of electron affinity
Answer:
Atomic sizeNuclear chargesymmetry of the electronic configurationConsider the Fischer ester synthesis of methyl benzoate from benzoic acid and methanol in the presence of sulfuric acid as a catalyst. A reaction was performed in which 3.3 g of benzoic acid was reacted with excess methanol to make 1.7 g of methyl benzoate. Calculate the theoretical yield and percent yield for this reaction.
Answer:
46.2%
Explanation:
Number of moles benzoic acid reacts = 3.3g/122.12 g/mol = 0.027 moles
Since the reaction is 1:1, 0.027 moles of methyl benzoate is formed.
Hence;
Theoretical yield of methyl benzoate = 0.027 moles × 136.15 g/mol = 3.68 g
% yield = actual yield/theoretical yield × 100
% yield = 1.7 g/3.68 g × 100
% yield = 46.2%
A skydiver slows down from 65 m/s to 5 m/s by opening the parachute. If this
takes 0.75 seconds, what is the skydiver's acceleration?
A. 45 m/s2 up
B. 80 m/s2 up
C. 45 m/s2 down
D. 80 m/s2 down
Answer:
D. -80m/s^2
Explanation:
V = u + at
5 = 65 + a (0.75)
0.75a = -60
a = -60/0.75
a = -80m/s^2
Therefore, is decelerating at 80m/s^2
Answer:
[tex]\boxed {\boxed {\sf D. \ 80 \ m/s^2 \ down}}[/tex]
Explanation:
We are asked to find the acceleration of a skydiver. Acceleration is the change in velocity over the change in time, so the formula for calculating acceleration is:
[tex]a= \frac{v_f-v_i}{t}[/tex]
The skydiver was initially traveling 65 meters per second, then he slowed down to a final velocity of 5 meters per second. He slowed down in 0.75 seconds.
[tex]\bullet \ v_f = 5 \ m/s \\\bullet \ v_i= 65 \ m/s \\\bullet \ t= 0.75 \ s[/tex]
Substitute the values into the formula.
[tex]a= \frac{ 5 \ m/s - 65 \ m/s}{0.75 \ s}[/tex]
Solve the numerator.
[tex]a= \frac{-60 \ m/s}{0.75 \ s}[/tex]
Divide.
[tex]a= -80 \ m/s^2[/tex]
The acceleration of the skydiver is -80 meters per second squared or 80 meters per second squared down. The skydiver is slowing down or decelerating, so the acceleration is negative or down.
How many grams of H₂SO₄ are contained in 2.00 L of 6.0 M H₂SO₄?
Please explain and show work.
Answer:
1176 grams
Explanation:
nH2SO4 =2*6=12 mol
mH2SO4=12*98=1176 grams
Answer:
solution given:
molarity of H₂SO₄=6 M
volume=2L
no of mole =6M*2=12mole
we have
mass =mole* actual mass=12*98=1176g
the mass is 1176g.
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
