Answer:
D
Explanation:
there is no other answer choice that does not move. This is a fool-proof question because all the other answer choices contain movement except the butterfly resting
g The boiling of water is a Question 4 options: chemical and physical damage chemical change because a gas (steam) is given off chemical change because heat is needed for the process to occur physical change because the water merely disappears physical change because the gaseous water is chemically the same as the liquid
Answer:
physical change because the gaseous water is chemically the same as the liquid
Explanation:
Matter can be defined as anything that has mass and occupies space. Any physical object that is found on earth is typically composed of matter. Matter are known to be made up of atoms and as a result has the property of existing in states.
Generally, matter exists in three (3) distinct or classical phases and these are; solid, liquid and gas.
A physical change can be defined as a type of change that only affects the physical form of a chemical substance (matter) without having any effect on its chemical properties. Thus, a physical change would only affect the physical appearance and properties of a chemical substance (matter) but not its chemical properties.
This ultimately implies that, a physical change result in a change of matter from one form or phase (liquid, solid or gas) to another without a corresponding change in chemical composition.
Hence, the boiling of water is considered to be a physical change because the gaseous water is chemically the same as the liquid i.e there isn't any changes in chemical composition of water when boiling.
What is Markanikov rule?
Answer:
Regla de Markovnikov En fisicoquímica orgánica, la regla de Markovnikov es una observación respecto a la reacción de adición electrófila.Regla de Markovnikov En fisicoquímica orgánica, la regla de Markovnikov es una observación respecto a la reacción de adición electrófila.
Explanation:
Toothpastes containing sodium hydrogen carbonate (sodium bicarbonate) and hydrogen peroxide are widely used. Write Lewis structures for the hydrogen carbonate ion and hydrogen peroxide molecule, with resonance forms where appropriate.
Answer:
See explanation and images attached
Explanation:
The first image shows the structure of hydrogen peroxide. It does not exist as resonance structures. The structure and properties of the molecule can wholly be explained on the basis of a single Lewis structure.
However, the structure of the bicarbonate ion in sodium bicarbonate can not be completely described by a single Lewis structure. Hence, two resonance structures are shown for the bicarbonate ion. In each case, Na^+ is the counter ion.
Andy needs to determine the freezing point depression of 30.0 mL of 0.50 m AlCl3 aqueous solution. When dissolved, AlCl3 dissociates into Al3 and Cl-. Andy realizes that he may need to reduce the effect of ion-pairing in his 0.50 m AlCl3 aqueous solution. Which of the following would reduce the ion-pairing effect in his AlCl3 solution?A. Add some solid AICIz to his solution.B. Add some water to his solution.C. Use only half of his initial volume.D. Use 45 mL of the 0.50 m AlCl3 solution.
Answer:
Add some water to his solution
Explanation:
Ion pair effect refers to strong electrostatic interaction between oppositely charged ions in solution. Such strong interaction affects solute- solvent interaction when an ionic substance is dissolved in water.
High solute concentration may lead to ion-pair effect. Hence, the ion pair effect may be minimized by adding more water (decreasing the concentration of the solution).
To reduce the ion- pair effect, Andy needs to add some water to his solution (dilution).
The activity that would reduce the ion-pairing effect in his AICI₃ solution is : ( B ) add some water to his solution
Ion pair effectIon pair effect is a strong electrostatic interaction seen between ions with opposite charges, when an ionic substance is been dissolved in a solvent such as water this effect will affect the dissolution of the substance.
High concentration of the solute in a solution also leads to the ion pair effect therefore for Andy to reduce the ion-pairing effect in his solution he has to add more water to reduce the concentration of the solute.
Hence we can conclude that The activity that would reduce the ion-pairing effect in his AICI₃ solution is to add some water to his solution
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Discuss the unique properties of carbon as a group 4a element
Answer:
Carbon is dull, have a low density, and are poor conductors of heat.
Explanation:
Carbon is a non-metal, therefore bearing properties of a non-metal element.
An experiment was performed under identical conditions as yours. The absorbance of the penny solution was recorded as 0.219 absorbance units. A calibration plot of absorbance vs. concentration of Cu(II) (M) yielded the following trendline equation:
y= 11.589x - 0.0002
Required:
What is the concentration (mol/L) of the penny solution?
Answer:
Concentration C = 0.0189 mol/L
Explanation:
From the given information:
Let consider the formula used in calculating the concentration according to Beer's law:
[tex]\mathtt{A =\varepsilon \times L \times C}[/tex] --- (1)
here;
A = absorbance
ε = coefficient of molar absorptivity
L = path length
C = concentration (mol/L)
Also, from Beer law plot:
y = mx+b
where,
y represent absorbance A
b represents intercept
m represents the coefficient of molar absorptivity ε
and x represents the concentration(C).
replacing the substituted entities
A = ε × C + b ---- (2)
Making the concentration the subject of the above formula:
[tex]C = \dfrac{A-b}{\varepsilon}[/tex]----(3)
From y = 11.589x - 0.0002
A = 11.589 *C - 0.0002
Given that:
A = 0.219
∴
0.219 = 11.589 *C - 0.0002
0.219 + 0.0002 = 11.589 *C
C = 0.2192/11.589
C = 0.0189 mol/L
If chromium(II) iodide is dissolved in water you can say that the equilibrium concentrations of chromium(II) and iodide ions are: __________
Answer:
See explanation.
Explanation:
Hello there!
In this case, according to the given information, it turns out firstly necessary for us to set up the equilibrium reaction by which chromium (II) iodide is ionized as follows:
[tex]CrI_2(s)\rightarrow Cr^{2+}(aq)+2I^-(aq)[/tex]
Thus, it is possible to notice that the the moles, and therefore concentrations, of chromium(II) and iodide ions are going to be different, by a factor of 2 due to the mole ratio in the aforementioned chemical equation.
Regards!
what is the characteristics of tropical air mass
Answer:
Explanation:
Continental tropical air masses are extremely hot and dry. Arctic, Antarctic, and polar air masses are cold. The qualities of arctic air are developed over ice and snow-covered ground. Arctic air is deeply cold, colder than polar air masses.
on the right it says molecules H2o in questions 1 and 2
Answer:
For 1: 12.0 moles of water will contain [tex]7.2264\times 10^{24}[/tex] number of molecules
For 2: The given amount of water molecules will be present in 5.41 moles of water.
Explanation:
According to the mole concept:
1 mole of a compound contains [tex]6.022\times 10^{23}[/tex] number of molecules
For 1:We are given:
Moles of water = 12.0 moles
Using above concept:
If 1 mole of a compound contains [tex]6.022\times 10^{23}[/tex] number of molecules
So, 12.0 moles of water will contain = [tex](12.0\times 6.022\times 10^{23})=7.2264\times 10^{24}[/tex] number of molecules
Hence, 12.0 moles of water will contain [tex]7.2264\times 10^{24}[/tex] number of molecules
For 2:We are given:
Molecules of water = [tex]3.26\times 10^{24}[/tex] molecules
Using above concept:
If [tex]6.022\times 10^{23}[/tex] number of molecules are present in 1 mole of a compound
So, [tex]3.26\times 10^{24}[/tex] number of molecules will be present in = [tex]\frac{1mol}{6.022\times 10^{23}}\times 3.26\times 10^{24}=5.41[/tex] moles of water
Hence, the given amount of water molecules will be present in 5.41 moles of water.
What is the conjugate base of the following acids:
1. HCIO,
2. PH4^+
Answer:
ECUACIÓN:HClO 2 + H 2O → ClO− 2 + H 3O
ACIDO: HClO2
BASECONJUGADA:ClO-2
Explanation:
In a titration to find the concentration of 30ml of a H2SO4 solution, a student found that 40ml of 0.2M KOH solution was needed to reach the endpoint. What's the concentration of the H2SO4?
Question 21 options:
A) 0.27M
B) 0.53M
C) 0.4M
D) 1.1M
Answer:
it's B
Explanation:
Write the balanced equation: H2SO4 + 2KOH → K2SO4 +2H2O. So 2(moles KOH) = (moles H2SO4); 2(volume KOH)(concentration KOH) = (volume H2SO4)(concentration H2SO4); 2(40ml)(0.2M) = (30ml)(x); x = 0.53M
The concentration of H₂SO₄ solution is equal to 0.133 M.
What is a neutralization reaction?A neutralization reaction can be described as a chemical reaction in which an acid and base react together to form respective salt and water. When a strong acid such as HCl will react with a strong base such as NaOH the salt can be neither acidic nor basic.
When H₂SO₄ (a strong acid) reacts with KOH, the resulting salt will be K₂SO₄ and water.
H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O
Given, the concentration of KOH solution = 0.2 M
The volume of the KOH solution = 40 ml = 0.040 ml
The number of moles of KOH, n = M × V = 0.2 × 0.04 = 0.008 mol
The volume of the H₂SO₄ = 30 ml = 0.03 L
The number of moles of H₂SO₄, n = 0.008/2 = 0.004 mol
The concentration of H₂SO₄ solution = 0.004/0.03 = 0.133 M
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Half-life
Nickel-63 has a half life of 92 hours. If a 1000 gram
sample decayed for 368 years
, how much
Nickel -63 remains?
A chemist adds 370.0mL of a 1.41/molL potassium iodide KI solution to a reaction flask. Calculate the millimoles of potassium iodide the chemist has added to the flask. Be sure your answer has the correct number of significant digits.
Answer: The millimoles of potassium iodide the chemist has added to the flask is 522 millimoles.
Explanation:
Given: Volume of KI = 370.0 mL (1 mL = 0.001 L) = 0.37 L
Molarity of KI solution = 1.41 mol/L
Now, moles of KI (potassium iodide) is calculated as follows.
[tex]Moles = Volume \times Molarity \\= 0.37 L \times 1.41 M\\= 0.5217 mol[/tex]
Convert moles into millimoles as follows.
1 mol = 1000 millimoles
0.5217 mol = [tex]0.5217 mol \times \frac{1000 millimoles}{1 mol} = 521.7 millimoles[/tex]
This can be rounded off to the value 522 millimoles.
Thus, we can conclude that the millimoles of potassium iodide the chemist has added to the flask is 522 millimoles.
Based on amino acid metabolism in mammals, classify the descriptions and amino acids is ketogenic or glucogenic. a. serine b. aspartate c. leucine d. converted to citric acid cycle intermediates e. converted to pyruvate f. converted to acetyl COA g. converted to fatty acids1) ketogenic2) glucogenic
Answer:
a. Serine is converted to pyruvate. It is glucogenic.
b. Aspartate is converted citric acid cycle intermediates. It is glucogenic.
c. Leucine is converted to AcetylCoA. It is ketogenic.
Explanation:
The degradation of amino acids usually account for a significant amount of energy production by the human body. The carbon skeletons of amino acids after deamination are either channelled to gluconeogenesis, ketogenesis or are completely oxidized to carbon dioxide and water.
The amino acids that are channelled to gluconeogenesis are said to be glucogenic. The glucogenic amino acids are those that are either degraded to ∝-ketoglutarate, succinylCoA, fumarate or oxaloacetate which are citric acid cycle intermediates that can be converted to glucose and glycogen. Also, amino acids that are degraded to pyruvate are glucogenic as pyruvate can be converted to glucose via oxaloacetate. Examples of these amino acids are serine which is converted to pyruvate and aspartate which is converted to the citric acid cycle intermediate, oxaloacetate.
On the other hand, amino acids that are channelled to ketogenesis are said to be ketogenic. The ketogenic amino acids are degraded to either acetylCoA or acetoacetylCoA. AcetylCoA and acetoacetylCoA are used in the formation of key one bodies. An example of a ketogenic amino acid is leucine which is converted both to AcetylCoA and acetoacetylCoA.
heating, the particle _______________ increases as more __________ __________ is added
Answer: what are the choices?!.
Explanation:
Compound A and compound B are constitutional isomers with molecular formula C3H7Cl. When compound A is treated with sodium methoxide, a substitution reaction predominates. When compound B is treated with sodium methoxide, an elimination reaction predominates.
Required:
Propose structures A and B.
Answer:
Compound A and compound B are constitutional isomers with molecular formula C3H7Cl.
When compound A is treated with sodium methoxide, a substitution reaction predominates. When compound B is treated with sodium methoxide, an elimination reaction predominates.
Explanation:
Constitutional isomers are the one which differs in the structural formula.
When compound A is treated with sodium methoxide, a substitution reaction predominates.
That means sodium methoxide is a strong base and a strong nucleophile.
But when it reacts with primary alkyl halides it forms a substitution product and when it reacts with secondary alkyl halide it forms mostly elimination product.
The reaction and the structures of A and B are shown below:
Equilibrium constants for gases can be expressed in terms of concentrations, Kc, or in terms of partial pressures, Kp. Which one of the following statements regarding Kc and Kp is correct?
a. Kc and Kp are equal when all stoichiometric coefficients in the balanced reaction equation equal one.
b. Kc and Kp are equal when the conditions are standard (P= 1 atm, T=298 K)
c. Kc and Kp are equal when the sum of the stoichiometric coefficients for the products equals the sum of the stoichiometric coefficients for the reactants.
d. Kc and Kp can never be equal.
e. Kc and Kp have the same values but different units.
Answer:
c. Kc and Kp are equal when the sum of the stoichiometric coefficients for the products equals the sum of the stoichiometric coefficients for the reactants.
Explanation:
Hello there!
In this case, according to the given information, it turns out firstly necessary for us to write the relationship between Kc and Kp in terms of the temperature and the change in the stoichiometric coefficients:
[tex]Kp=Kc(RT)^{\Delta \nu}[/tex]
Thus, we can see that the Kp=Kc just when Δυ=0, which is the sum of the coefficients of products minus that of reactants; and therefore, the correct answer will be c. Kc and Kp are equal when the sum of the stoichiometric coefficients for the products equals the sum of the stoichiometric coefficients for the reactants.
Regards!
When H2S(g) reacts with O2(g) to form H2O(g) and SO2(g), 124 kcal of energy are evolved for each mole of H2S(g) that reacts. Write a balanced equation for the reaction with an energy term in kcal as part of the equation.
Answer:
2H2S(g) + 3O2(g) → 2H2O(g) + 2SO2(g) + 248kcal
Explanation:
The reaction of the problem occurs as follows:
H2S(g) + O2(g) → H2O(g) + SO2(g)
To balance the reaction we must balance oxygens:
H2S(g) + 3O2(g) → 2H2O(g) + 2SO2(g)
To balance the complete reaction:
2H2S(g) + 3O2(g) → 2H2O(g) + 2SO2(g)
As the energy is evolved, 124kcal are as product in the reactio per mole of H2S. As the balanced reaction contains 2 moles of H2S, the heat evolved is:
124kcal*2 = 248kcal:
2H2S(g) + 3O2(g) → 2H2O(g) + 2SO2(g) + 248kcal
And this is the balanced equation
Concentration data is commonly monitored during a reaction to determine the order with respect to a reactant. Consider the types of observations listed, and determine which order is likely for that reactant. Assume all other factors are held constant. The reaction rate increases in direct proportion to the concentration of the reactant in solution.
a. The reaction rate is constant regardless of the amount of reactant in solution.
b. An increase in the concentration of the reactant in solution causes the reaction rate to increase exponentially.
Answer:
The reaction rate increases in direct proportion to the concentration of the reactant in solution - second order reaction
The reaction rate is constant regardless of the amount of reactant in solution - zero order reaction
An increase in the concentration of the reactant in solution causes the reaction rate to increase exponentially - first order reaction
Explanation:
In a second order reaction, the rate of reaction is directly proportional to the concentration of reactants. This implies that, reaction rate varies as the concentration of the reactant in solution varies.
For a zero order reaction, the rate of reaction is independent of the concentration of the reactants in solution. This means that reaction rate is constant regardless of the amount of reactant in solution.
For a first order reaction, the rate of reaction varies exponentially as the concentration of reactants. Hence, an increase or decrease in the concentration of the reactant in solution causes the reaction rate to increase or decrease exponentially.
A chemist requires a large amount of 1-bromo-4-phenyl-2-butene as starting material for a synthesis and decides to carry out the following NBS allylic bromination reaction in the presence of UV light. Draw the structures of all of the observed products.
NBS
(C6H5)CH2CH = CHCH3 → ?
CCI4
Draw one additional resonance structure for the species below:
CH3CH=CHCH=CHCH=CHCH2
Answer:
CH 2 CH 3 CHCH = C6H5 + 5 Hydrogen Atoms
I hope I helped you.
Deepest apologies if I was wrong!
Bye!
~ Myaka O.
how many atoms of one formula unit in Fe2O3
Answer:
5 atoms form one formula unit of Fe2O3
Explanation:
2 atoms of Fe (Iron)
and 3 atoms of O ( Oxygen)
so total = 3 + 2
= 5
Answer:
fe203 the right answer is thus
How to find the number of protons
Answer:
No of protons is equal to the atomic number.
Answer:
atomic number
Explanation:
examples of isotones
Answer:
Examples of isotones include carbon-12, nitrogen-13 and oxygen-14. These atoms all have six neutrons and six, seven and eight protons respectively. A mnemonic that can be used to differentiate isotones from isotopes and isobars is as follows: same Z (number of protons) = isotopes.
Which of the following can be mixed in solution with NH3 to make a buffer?
Answer:
c. NaOH
Explanation:
just took the quiz
Sodium fluoroacetate (NaO₂C₂H₂F) is a common poison used in New Zealand to control invasive species, such as rats. It can be prepared by the substitution of a C-Cl bond in sodium chloroacetate (NaO₂C₂H₂Cl) for a C-F bond. What is the approximate enthalpy change for this substitution reaction on a 1.500 mole sample, based on the following bond energies?
C-Cl = 339.0 kJ/mol
C-F = 485.0 kJ/mol
Answer:
-219
Explanation:
1.5(339) - 1.5(485) = -219
The approximate enthalpy change for this substitution reaction is -219 kJ.
In the case of this substitution reaction, we need to find the enthalpy change when NaO₂C₂H₂Cl is converted to NaO₂C₂H₂F.
This reaction involves the breaking of the C-Cl bond and the formation of the C-F bond.
We have to subtract the bond energy of the C-F bond from that of the C-Cl bond and multiply by the number of moles involved.
So we will have;
ΔH= 1.500 mole [339.0 kJ/mol - 485.0 kJ/mol]
ΔH= -219 kJ
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Calculate the mass of 2.5 mols of Sodium Oxide
Answer:
8
Explanation:
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I am holding a balloon containing 439 mL of gas over my fireplace. The temperature and pressure of the gas inside the balloon is 317.15 K and 0.959 atm, respectively. Suppose I don't want the pressure to change, but I want to the volume to go down to 0.378 L. What is the temperature that I need to reach when I cool down the balloon?
To what temperature (in Celsius) must the balloon be cooled to reduce its volume to 0.378 L if the pressure doesn't change (remained constant)?
Answer:
-0.08 °C
Explanation:
We can solve this problem by using Charles' law, which states that at constant pressure:
V₁T₂ = V₂T₁Where in this case:
V₁ = 439 mLT₂ = ?V₂ = 0.378 L ⇒ 0.378 * 1000 = 378 mLT₁ = 317.15 KWe input the data:
439 mL * T₂ = 378 mL * 317.15 KAnd solve for T₂:
T₂ = 273.08 KFinally we convert 273.08 K to Celsius:
273.08 K - 273.16 = -0.08 °CWhich of the following is considered a standard unit of length in the United States?
O square inch
O acre
O cubic yard
O yard
Answer:
Yard . I hope this helped:))
You decide to share some sugar sweetened fruit juice with your friend. You divide the fruit equally into two glasses. Then an additional equal volume of water is added to glass 1. Which glass would have the lower molarity?
Answer:
Glass 1
Explanation:
Molarity is measured in moles of substance per liter.
For the sake of calculations, let's say that each glass contains 1 mole of juice and 1 liter after it is divided between the glasses. If you add an equal amount of water to glass 1 ( another liter), you now have:
Glass 1 = 1 mole / 2 liters = 0.5 M
Glass 1 = 1 mole/ 1 liter = 1 M
So glass 1 will have a lower molarity
A student is adding DI water to a volumetric flask to make a 50% solution. Unfortunately, he was not paying attention and filled the flask past the calibration mark. Instead of starting over, the student decides to continue with the experiment. How will the student's results be affected by his mistakes
Answer:
His results will be skewed because there was more water than stock solution. Which would cause the percentage solution to be less than 50% therefore the density would be less than the actual value.
Explanation:
The solution will have percentage less than that of 50%. Therefore the density would be less than the actual value.
Suppose there should be 50 mL of the solution, and he added 60 mL. So 10 mL of the solution is added more.
Suppose the mass of the solute is m.
Originally, the density is = [tex]$\frac{m}{50}$[/tex] [tex]\left(\frac{\text{mass}}{\text{volume}}\right)[/tex]
Now after adding extra 10 mL , the density becomes [tex]$\frac{m}{60}$[/tex].
Therefore, [tex]$\frac{m}{50}>\frac{m}{60}$[/tex]
So the density decreases when we add more solution.