The correct answer is; "The frozen metal was discovered through biased research practices."
Scientific research is based on standard practices and not on personal bias hence they should not have taken a decision based on the frozen metals. Bias prevents a researcher from taking an objective look at experimental observations before arriving at a conclusion.
The greatest pitfall that must be avoided when conducting a scientific study is bias.
A bias is introduced when there is a disparity between observation and interpretation of data in a scientific study.
In the case of this question, the researchers already wanted to establish the possibility of obtaining a carbonless fuel. Their minds were so clouded with this bias that they forgot to thoroughly investigate the pieces of frozen metal thereby arriving at a lobe-sided, unscientific conclusion. This makes their work, "bad" science.
For more about bias in research, see:
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Answer:
it would be A The conditions of the experiment are not replicate able or repeatable.
Explanation:
A team of researchers is working on a project to make a new kind of carbonless fuel. During their experiment, there is an explosion that destroys the lab. While they are cleaning the debris, they discover three pieces of frozen metal. The researchers were excited to report that they had discovered a fuel that burns so clean it becomes cold. Other scientists attempt to make the carbonless fuel but are unsuccessful. Which statement best explains why this scenario represents "bad" science?
The conditions of the experiment are not replicable or repeatable.
The process of making clean fuel is known due to the explosion.
The frozen metal was discovered through biased research practices.
The result of clean fuel production, the frozen metal, is not observable.
Give an example of a microscope problem that could be prevented by cleaning
For which of the following reactions is the enthalpy change equal to the second ionization energy of nitrogen?
Answer:
"[tex]N^+(g) \rightarrow N^{2+}(g) + e^-[/tex]" is the appropriate answer.
Explanation:
Whenever one electron or particle must be removed from some kind of gas atom or molecule, it requires that the very first amount of energy necessary.Two electrons must be removed from such a mono-positive exhaust gases structure or position of ion before they may become a dipositive gaseous ion.Thus the above is the correct answer.
A solution has a OH- concentration of 7.7x10-3. What is the pH of this solution?
Answer:
11.9 pH
Explanation:
First, we need to find pOH
To find that, we use the formula -log[OH]
-log[7.7x10^-3] = 2.11351
To find the pH, we'll use this formula: 14 = pH + pOH
14 = pH + 2.11351
Subtract boths sides by 2.11351
14 = pH + 2.11351
-2.11351 -2.11351
pH = 11.88649
instruments that have been soaking in cold sterilization for blank minutes are considered aseptic and can be used during non sterile procedures
a. 60
b. 30
c. 15
d. 10
Answer:
Its C you can check
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?
What is normality in chemistry?
Answer:
a measure of concentration equal to the gram equivalent weight per liter of solution.
Explanation:
Gram equivalent weight is the measure of the reactive capacity of a molecule. The solute's role in the reaction determines the solution's normality. Normality is also known as the equivalent concentration of a solution.
hope it helped
Based on periodic properties, choose the more metallic element from each of the following pairs.
Match the words in the left column to the appropriate blanks in the sentences on the right.
Between Sr and Sb, the more metallic element is ______
Between \rm Sr and \rm Sb, the more metallic element is _______
Between As and Bi, the more metallic element is ______
Between \rm As and \rm Bi, the more metallic element is _______
Between Cl and O, the more metallic element is ______
Between \rm Cl and \rm O, the more metallic element is ______
Between S and As, the more metallic element is ______
Between \rm S and \rm As, the more metallic element is _______
Answer:
Sr is the more metallic element
Bi is the more metallic element
O is the more metallic element
As is the more metallic element
Explanation:
One thing should be clear; metallic character increases down the group but decreases across the period.
Hence, as we move across the period, elements become less metallic. As we move down the group elements become more metallic.
This is the basis upon which decisions were made about the metallic character of each of the elements listed above.
Consider the reaction between CaCO3 and HCl. Which of the following could speed up the reaction?
I. Increasing concentration of the HCl
II. Increasing size of the CaCO3 pieces
III. Increasing temperature
a) I and III only
b) I, II, and III
c) I only
d) II and III only
what is the difference between red phosphorus and white phosphorus?
Answer:
White phosphorusRed PhosphorusIt is insoluble in water but soluble in carbon disulphide.It is insoluble in both water and carbon disulphide.It undergoes spontaneous combustion in air.It is relatively
Explanation:
I hope it will help you
If we slowly add a solution of mercury(II) ions to a solution of aqueous halide ions with roughly equal concentrations, a precipitate will form. Explain what the precipitate will consist of initially. g
Water, mercury chloride and nitrogen oxide.
Water, mercury chloride and nitrogen oxide will present in the precipitate when we slowly add a solution of mercury(II) nitrate to a solution of aqueous hydrochloric acid having halide ions both in equal concentrations. The equation of this reaction is Hg2(NO3)2 + 4 HCl ----> 2 HgCl2 + 2 H2O + 2 NO so it is concluded that from this reaction we get precipitate of water, mercury chloride and nitrogen oxide.
https://brainly.com/question/24261598.
FILL IN THE BLANK:
The rate of a reaction is measured by how fast a (Product Or Reactant)
is used up or how fast a
(Reactant Or Product) is formed?
Answer:
the rate of a reaction is measured by how fast a REACTANT is used up or how fast a PRODUCT is formed
a cup of coffee changes from a temperature of75ﾟC to 12ﾟC. What is the change in temperature in ﾟC
Explanation:
why is because I never did this class
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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.
Different steps of the oxidative decarboxylation of pyruvate by the pyruvate dehydrogenase PDH complex are given. Place these five steps in the correct order. Note that thiamine pyrophosphate, TPP, is sometimes called thiamine diphosphate, TDP.
1. FADH2 is reoxidized to FAD reducing NAD* to NADH.
2. The lipoamide arm of E2 moves to the active site of E1, enabling the transfer of the acetyl group to the lipoamide.
3. Pyruvate reacts with TPP and is decarboxylated forming hydroxyethyl-TPP.
4. The lipoamide arm moves to the active site of E3 where the reduced lipoamide is oxidized by FAD forming the active lipoamide and FADH2.
5. The acetyl lipoamide arm of E2 moves to the active site of E2, where the acetyl group is transferred to CoA, forming acetyl-CoA and the reduced form of lipoamide.
Answer:
1. Pyruvate reacts with TPP and is decarboxylated forming hydroxyethyl-TPP.
2. The lipoamide arm of E2 moves to the active site of E1, enabling the transfer of the acetyl group to the lipoamide.
3. The acetyl lipoamide arm of E2 moves to the active site of E2, where the acetyl group is transferred to CoA, forming acetyl-CoA and the reduced form of lipoamide.
4. The lipoamide arm moves to the active site of E3 where the reduced lipoamide is oxidized by FAD forming the active lipoamide and FADH2.
5. FADH2 is reoxidized to FAD reducing NAD+ to NADH.
Explanation:
The oxidation of pyruvate to AcetylCoA is catalyzed by the pyruvate dehydrogenase complex. The reaction is an irreversible oxidative decarboxylation process in which the carboxyl group of pyruvate is removed as a molecule of carbon dioxide, CO₂, while the remaining two carbons are attached to a CoASH molecule to form acetylCoA.
The pyruvate dehydrogenase complex contains three enzymes - Pyruvate dehydrogenase known as E₁, dihydrolipoyl transacetylase known as E₂, and dihydrolipoyl dehydrogenase known as E₃. It also requires five coenzymes namely: thiamine pyrophosphate (TPP), flavine adenine dinucleotide (NAD), coenzyme A (CoA-SH), nicotinamide adenine dinucleotide (NAD) and lipoate.
Oxidative decorbyxylation of pyruvate takes place in the pyruvate dehydrogenase complex in five steps:
1. Pyruvate reacts with TPP and is decarboxylated forming hydroxyethyl-TPP.
2. The lipoamide arm of E2 moves to the active site of E1, enabling the transfer of the acetyl group to the lipoamide.
3. The acetyl lipoamide arm of E2 moves to the active site of E2, where the acetyl group is transferred to CoA, forming acetyl-CoA and the reduced form of lipoamide.
4. The lipoamide arm moves to the active site of E3 where the reduced lipoamide is oxidized by FAD forming the active lipoamide and FADH2.
5. FADH2 is reoxidized to FAD reducing NAD+ to NADH.
Let's assume you were given 2.0 g benzil, 2.2 g dibenzyl ketone, 50 mL 95% ethanol and 0.3 g potassium hydroxide to synthesize tetraphenylcyclopentadienone. You isolated 3.0 g of tetraphenylcyclopentadienone. What is the % yield
Answer:
the % yield is 82%
Explanation:
Given the data in the question,
we know that;
Molar mass of benzil is 210.23 g·mol−1
Molar mass of dibenzyl ketone is 210.27 g·mol−1
Molar mass of tetraphenylcyclopentadienone is 384.5 g·mol−1
Now,
2.0 g benzil = 2 g / 210.23 g·mol−1 = 0.0095 mole
2.2 g dibenzyl ketone = 2.2 / 210.27 = 0.0105 mole
3.0 g of tetraphenylcyclopentadienone = 3 / 384.5 = 0.0078 mole
Now, the limiting reagent is benzil. 0.0095 mole can reacts wiyh 0.0095 mole of dibenzyl ketone
percentage yield = ( 0.0078 mole / 0.0095 mole ) × 100%
= 0.82 × 100%
= 82%
Therefore, the % yield is 82%
Good afternoon everyone, Define Radicals?Thank You
Explanation:
A radical is is an atom, molecule, or ion that isnt paired up with at least one valence electron.
Answer:
a radical also called a free radical is a molecule that contains at least one unpaired electron.
I hope this helps
Name the following cycloalkane:
H3C-
CH2CH3
A. 1-ethyl-4-methylcyclohexane
B. 4-ethyl-1-methylcyclohexane
C. 1-methyl-4-ethylcyclohexane
Answer:
A
Explanation:
1-ethyl-4-methylcyclohexane
i believe it is A
hope that helped :)
In a quantitative analysis, a methanol (CH3OH) contaminated water sample was titrated with 0.0021 mol L- potassium permanganate (KMnO4). 50.00 mL samples of the water to be tested were acidified by sulfuric acid, then titrated with the permanganate solution. The results are shown below. Burette reading, ml 1st titration 2nd titration 3rd titration 4th titration Final volume 12.40 19.60 26.60 17.25 Initial volume 4.45 12.50 19.60 10.15 Titre 7.95 7.10 7.00 7.10 The complete equation for the redox titration reaction is: 4MnO4- + 12H+ + 5CH3OH → 4Mn2+ + 11H2O + 5HCOOH a. [5] Calculate the concentration of the methanol in mol L-1.
In a REDOX titration, one specie is oxidized while the other is reduced. The concentration of methanol is 0.012 mol L-1. Methanol is the oxidizing agent while permanganate is the reducing agent.
The average titre value is; [tex]\frac{7.95 + 7.10 + 7.00 + 7.10}{4}[/tex] = 7.29 mL
Equation of the reaction is:
[tex]4MnO4- + 12H+ + 5CH3OH ----> 4Mn2+ + 11H2O + 5HCOOH[/tex]
Concentration of oxidizing agent = CA = ?
Concentration of reducing agent = CB = 0.0021 mol L-1
Volume of oxidizing agent = VA= 7.29 mL
Volume of reducing agent = VB = 50.00 mL
Number of moles of oxidizing agent NA = 4
Number of moles of reducing agent NB = 5
Note that NA and NB are obtained from the balanced reaction equation
CAVA/CBVB = NA/NB
CAVANB = CBVBNA
CA = CBVBNA/VANB
CA = 0.0021 mol L-1 * 50.00 mL * 4/7.29 mL * 5
CA= 0.012 mol L-1
For a comprehensive definition of redox titration see
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this is on USAtestprep
a)
b)
c)
d)
Urea, CH4N2O (s), is manufactured from NH3 (g) and CO2 (g). H2O (l) is another product of this reaction. An experiment is started with 2.6 grams of NH3 (g) added into a reaction vessel with CO2 (g).
Write the balanced equation for this reaction, being sure to include physical states. Based on the balanced equation above, calculate the following:
a. the theoretical yield of urea in grams that can be made from the NH3
b. the actual amount of urea made if the percent yield for this reaction is 34%.
Answer:
a. 4.41 g of Urea
b. 1.5 g of Urea
Explanation:
To start the problem, we define the reaction:
2NH₃ (g) + CO₂ (g) → CH₄N₂O (s) + H₂O(l)
We only have mass of ammonia, so we assume the carbon dioxide is in excess and ammonia is the limiting reactant:
2.6 g . 1mol / 17g = 0.153 moles of ammonia
Ratio is 2:1. 2 moles of ammonia can produce 1 mol of urea
0.153 moles ammonia may produce, the half of moles
0153 /2 = 0.076 moles of urea
To state the theoretical yield we convert moles to mass:
0.076 mol . 58 g/mol = 4.41 g
That's the 100 % yield reaction
If the percent yield, was 34%:
4.41 g . 0.34 = 1.50 g of urea were produced.
Formula is (Yield produced / Theoretical yield) . 100 → Percent yield
1) Recall the two written definitions of an oxidation-reduction reaction provided in our lessons. Which of these definitions is
most inclusive of redox reactions? Explain your answer:
A redox reaction is where the oxidation and reduction reaction takes place at the same time, the oxidation half
reaction involves losing electrons and in the reduction half reaction involves gaining electrons. So in a redox
reaction an electron is lost by the reducing agent.
Explain how the reaction below meets these definitions. Which substance is being oxidized and which is
being reduced?
4Ag(s) + 2H2S(g) + O2(g)
2Ag2S(s) + 2H20(9)
Answer:
A redox reaction in which a change of oxidation number of the reacting species occurs either by oxidation or by reduction which occur simultaneously and to the same extent during the reaction.
Silver atoms are oxidized while oxygen atoms are reduced by a loss of electrons and a gain of electrons respectively.
Explanation:
A redox reaction in which a change of oxidation number of the reacting species occurs either by oxidation or by reduction which occur simultaneously and to the same extent during the reaction.
In a redox reaction,the two reactions occurring simultaneously can be divided into two half reactions; an oxidation half-reaction and a reduction half-reaction.
The oxidation half-reaction involves losing electrons and thus an increase in oxidation number of the species being oxidized. Whereas, the reduction half reaction involves gaining electrons and thus, a reduction innthe oxidation number of the species being reduced.
The species which oxidizes another species is known as an oxidizing agent and isnitself reduced due to its accepting electrons from the species being oxidized. Th reducing agent reduces another species and is itself oxidized as it loses electrons to the oxidized agent.
In the given reaction as shown below:
4 Ag (s) + 2 H₂S (g) + O₂ (g) ---> 2 Ag₂S (s) + 2 H₂0 (g)
The reaction is a redox reaction as a change innthe oxidation number of the reacting species; both oxidation and reduction occurs simultaneously and to the same extent.
The metallic silver atoms, have an oxidation number of zero initially. However, each of the four moles of atoms give up one mole of a electrons each to become oxidized to silver (i) ions, Ag+.
On the other hand, molecular oxygen gas also having oxidation number of zero becomes reduced to oxygen ion, O²-. Each of the two moles of atom in the oxygen gas molecule accept two electrons each donated by the metallic silver atoms to become reduced to oxygen ion, O²-.
The oxidation numbers of hydrogen ion and sulfide ion do not change.
H2SO4 ????????????????
Explanation:
Sulfuric Acid (H2SO4) is a strong mineral acid that has is colorless when pure. This chemical is used as a chemical intermediate to manufacture other chemicals and cleaning metal surfaces. The formula for sulfuric acid is H2SO4. The molar mass of sulfuric acid is 98.07848 g mol.
what characterizes a homogeneous mixture?
Answer:
a mixture that doesn't really show the ingredients or things put into the material or food.
The fact that a beam of particles was deflected in the presence of an electric
or magnetic force led J.J. Thomson to conclude that the particles had a(n)
O A. large mass
B. electric charge
O C. negligible mass
O D. neutral charge
Answer:
electric charge
Explanation:
Charged particles are deflected in an electric or a magnetic field. The particles discovered by J.J. Thomson were charged particles.
When these charged particles are passed through electric and magnetic fields, deflection occurs depending on the nature of the charge.
A positive charge is deflected towards the negative part of an electric field or the south pole of a magnetic field.
A negative charge is selected towards the positive end of an electric field or the north pole of a magnetic field.
a 67.5 L sample of gas at 27.6 °c and 383.1mm hg expands to 244.2 L at 4.7 °c. what is the new gas pressure.
a 97.8
b 18.0
c 115
d 1.65
Answer:
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)Calculate the molar mass of glucose (C6H12O6)
Answer:
Molar mass = 180 g/mol
Explanation:
Relative Atomic Mass of C = 12
of H = 1
of O =16
Let Molar mass be mm
mm of C6H12O6 = 6(12) + 12(1) + 6(16)
= 72 + 12 + 96 = 180 g/mol
A sample of gas is held at constant volume. If the number of moles of this sample of gas is doubled and the pressure of this sample of gas is halved, what happens to the absolute temperature of the gas?
Select one
a. The absolute temperature is doubled.
b. The absolute temperature is halved.
c. The absolute temperature is quadrupled.
d. The absolute temperature is quartered.
e. The absolute temperature stays the same.
Answer:
number of moles of gas increases the volume also increases.
Isotopes are: A. are only theoretical. B. only formed in laboratories. C. found in nature. D. found in the nuclear reactions in stars but not on Earth.
Answer:
B. only formed in laboratories
Explanation:
i know
Isotopes are only formed in Laboratories. hence, Option (B) is correct.
What are Isotopes ?
Each of two or more forms of the same element that contain equal numbers of protons but different numbers of neutrons in their nuclei, and hence differ in relative atomic mass but not in chemical properties; in particular, a radioactive form of an element is known as Isotope.
Isotopes are two or more types of atoms that have the same atomic number and position in the periodic table, and that differ in nucleon numbers due to different numbers of neutrons in their nuclei.
Therefore, Isotopes are only formed in Laboratories. hence, Option (B) is correct.
Learn more about Isotopes here ;
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The specific heat capacity of lead is 0.13 J/g-K. How much heat (in J) is required to raise the temperature of 15 g of lead from 22 °C to 37 °C? a. 5.8 × 10-4 J b. 0.13 J c. 29 J d. 2.0 J e. -0.13 J
Answer:
c. 29 J
Explanation:
Step 1: Given data
Specific heat capacity of Pb (c): 0.13 J/g.K (= 0.13 J/g.°C)Mass of Pb (m): 15 gInitial temperature: 22 °CFinal temperature: 37 °CStep 2: Calculate the temperature change
ΔT = 37 °C - 22 °C = 15 °C
Step 3: Calculate the heat (Q) required to raise the temperature of the lead piece
We will use the following expression.
Q = c × m × ΔT
Q = 0.13 J/g.°C × 15 g × 15 °C = 29 J
chemistry
Which of the following is a characteristic of solid silver?
O A. High electrical conductivity
O B. Brittleness
C. Low melting point
O D. Good insulator
SUBM
Answer:
A. High electrical conductivity
Explanation:
solid silver isn't brittle, it has a high melting point, and its not a good insulator.
