The oxidation numbers for the elements Co, Al, C, N, Cl, and Cr in the given compounds are +3, -3, -2, -3, +1, and +6 respectively.
Explanation:The oxidation numbers for the elements in the compounds are determined as follows:
(a) In LiCoO2, Co has an oxidation number of +3, as lithium contributes +1 and each of the two oxygen atoms contribute -2.
(b) In NaAlH4, Al has an oxidation number of -3, as sodium contributes +1, while hydrogen as a metal hydride contributes -1 each adding up to -4.
(c) In CH3OH, C has an oxidation number of -2, as hydrogen contributes +1 each from the three hydrogen atoms and the fourth hydrogen attached to oxygen contributes -1 and oxygen contributes -2.
(d) In GaN, N has an oxidation number of -3, as gallium contributes +3.
(e) In HClO2, Cl has an oxidation number of +1, as hydrogen contributes +1, and each of the two oxygen atoms contribute -2.
(f) In BaCrO4, Cr has an oxidation number of +6, as barium contributes +2 and each of the four oxygen atoms contribute -2.
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The oxidation numbers for the given elements in their compounds are as follows: Co in LiCoO₂ is +3, Al in NaAlH₄ is +3, C in CH₃OH is -2, N in GaN is -3, Cl in HClO₂ is +3, and Cr in BaCrO₄ is +6.
(a) Co in LiCoO₂: The oxidation number of Co is +3. This is because Lithium (Li) has an oxidation number of +1, and Oxygen (O) has an oxidation number of -2. Since the molecule LiCoO₂ is neutral, the sum of oxidation numbers is zero. Letting 'x' be the oxidation number of Co, we get: 1(+1) + x + 2(-2) = 0, which simplifies to x = +3.
(b) Al in NaAlH₄: The oxidation number of Al is +3. This is because Sodium (Na) has an oxidation number of +1, and Hydrogen (H) has an oxidation number of -1. Since the molecule NaAlH₄ is neutral, the sum of oxidation numbers is zero. Letting 'x' be the oxidation number of Al, we get: 1(+1) + x + 4(-1) = 0, which simplifies to x = +3.
(c) C in CH₃OH (methanol): The oxidation number of C is -2. This is because Hydrogen (H) has an oxidation number of +1, and Oxygen (O) has an oxidation number of -2. Since the molecule CH₃OH is neutral, the sum of oxidation numbers is zero. Letting 'x' be the oxidation number of C, we get: 4(+1) + x + 1(-2) = 0, which simplifies to x = -2.
(d) N in GaN: The oxidation number of N is -3. This is because Gallium (Ga) has an oxidation number of +3. Since the molecule GaN is neutral, the sum of oxidation numbers is zero. Letting 'x' be the oxidation number of N, we get: 1(+3) + x = 0, which simplifies to x = -3.
(e) Cl in HClO₂: The oxidation number of Cl is +3. This is because Hydrogen (H) has an oxidation number of +1, and Oxygen (O) has an oxidation number of -2. Since the molecule HClO₂ is neutral, the sum of oxidation numbers is zero. Letting 'x' be the oxidation number of Cl, we get: 1(+1) + x + 2(-2) = 0, which simplifies to x = +3.
(f) Cr in BaCrO₄: The oxidation number of Cr is +6. This is because Barium (Ba) has an oxidation number of +2, and Oxygen (O) has an oxidation number of -2. Since the molecule BaCrO₄ is neutral, the sum of oxidation numbers is zero. Letting 'x' be the oxidation number of Cr, we get: 1(+2) + x + 4(-2) = 0, which simplifies to x = +6.
Calculate the standard entropy of vaporization of ethanol at its boiling point
The question is incomplete, here is a complete question.
Calculate the standard entropy of vaporization of ethanol at its boiling point 352 K. The standard molar enthalpy of vaporization of ethanol at its boiling point is 40.5 kJ/mol.
Answer : The standard entropy of vaporization of ethanol is, 115 J/mol.K
Explanation :
Formula used :
[tex]\Delta S=\frac{\Delta H_{vap}}{T_b}[/tex]
where,
[tex]\Delta S[/tex] = change in entropy
[tex]\Delta H_{vap}[/tex] = change in enthalpy of vaporization = 40.5 kJ/mol
[tex]T_b[/tex] = boiling point temperature = 352 K
Now put all the given values in the above formula, we get:
[tex]\Delta S=\frac{\Delta H_{vap}}{T_b}[/tex]
[tex]\Delta S=\frac{40.5kJ/mol}{352K}[/tex]
[tex]\Delta S=\frac{40.5\times 10^3J/mol}{352K}[/tex]
[tex]\Delta S=115J/mol.K[/tex]
Therefore, the standard entropy of vaporization of ethanol is, 115 J/mol.K
Apply the Law of Conservation of Mass to the following problem: During a combustion reaction, 12.2 grams of methane reacts with 14 g of oxygen. The reaction produces carbon dioxide and water. If 20 grams of water are produced, how many grams of carbon dioxide are produced?
Answer:
5.8 g of carbon dioxide are produced
Explanation:
The Law of Conservation of Mass states that the mass of the reactants must equal the mass of the products in all chemical reactions.
This is the chemical reaction (combustion)
CH₄ + 2O₂ → CO₂ + 2H₂O
12.2 g 14 g x 20g
Mass in reactants = 12.2 g + 14 g = 26.2 g
Mass in products = x + 20 g
26.2 g = x + 20g
26.2 g - 20g = x
5.8 g = x
which of the following compounds is (are) hydrolyzed tobutanoic acid upon heating in H20, H2SO4?
a. ethyl butanoate
b. butyl acetate
c. N-methylbutanamide
d. both A and B
e. both A and C
Answer:
The correct option is option e, both A and C
Explanation:
Ethyl butanoate is an ester. Ester undergoes hydrolysis when heated with water containing dilute acids. Ester hydrolysis gives carboxylic acid and alcohol.
Therefore, ethyl butanoate on hydrolysis gives butanoic acid and ethanol.
Amide also undergoes acidic hydrolysis and yields carboxylic acid and amine.
N-methylbutanamide is an amide. It gives butanoic acid and methylamine when undergoes hydrolysis with H2O in the presence of H2SO4.
Hence, the correct option is option e.
Final answer:
Hydrolysis of ethyl butanoate and N-methylbutanamide in acidic conditions with H₂O and H₂SO₄ will yield butanoic acid; hence, option e) both A and C is correct.
Explanation:
The student's question pertains to the identification of compounds which, upon hydrolysis in an acidic condition with H₂O and H₂SO₄, will yield butanoic acid. To solve this, we need to understand the process of ester hydrolysis under acidic conditions.
Ethyl butanoate (a) is an ester formed from butanoic acid and ethanol. When hydrolyzed, it reverts back to butanoic acid and ethanol. Hence, option (a) is correct. Butyl acetate (b), similarly, is an ester of butanoic acid with butanol and would also yield butanoic acid upon hydrolysis.
N-methylbutanamide (c) is an amide, and on hydrolysis, it gives butanoic acid and a methylamine derivative. Therefore, upon heating with water and sulfuric acid, all listed compounds will yield butanoic acid as a hydrolysis product.
The correct answer to the student's question is e) both A and C.
To make an effective buffer, 25.00 mL of 0.025 M HF should be mixed with ______mL of 0.050 M NaOH
Answer: The volume of NaOH will be, 12.5 mL
Explanation:
To calculate the volume of base, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]HF[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is NaOH.
We are given:
[tex]n_1=1\\M_1=0.025M\\V_1=25.00mL\\n_2=1\\M_2=0.050M\\V_2=?[/tex]
Putting values in above equation, we get:
[tex]1\times 0.025M\times 25.00mL=1\times 0.050M\times V_2\\\\V_2=12.5mL[/tex]
Hence, the volume of NaOH will be, 12.5 mL
Consider atoms of the following elements. Assume that the atoms are in the ground state.
(A) S(B) Ca(C) Ga(D) Sb(E) Br2. The atom that contains only one electron in the highest occupied energy sublevel
Answer:
C
Explanation:
Gallium is in group thirteen with outermost electron configuration ns2 np1. The highest occupied sub-level is np1 having only one electron which is the situation required in the question.
Answer:
C. Ga
Explanation:
Organic Chemistry, 7e by L. G. Wade, Jr. Reactions of Alkenes Christine Hermann Radford University Radford VA Copyright © 2010 Pearson Education
Answer: Christine Herman & L.G Wade Jr., "2010". Organic Chemistry: Reaction of Alkane, 7e, Pearson Education, Radford University, Radford, VA.
Explanation:
This is an edited book. The Harvard reference style was used in the following order:
Authors name
Year of publication
Title
Edition
Publisher
Place of publication.
Note that the title of book should be italicized with capitalization of first word.
Which of the following is a macronutrient? Select one:
a. nitrogen
b. manganese
c. zinc
d. boron
Answer:
a. nitrogen is the correct answer.
Explanation:
Nitrogen is a macronutrient because nitrogen present in the plant in large quantities.Macronutrients are the nutrients which are needed in large amount for the proper growth and development,as they are required in large amounts they are called macronutrients.Macronutrients are the most essential elements for plants and required so that body functions properly.Macronutrients comprise carbon, nitrogen, sulfur, oxygen, potassium, hydrogen, calcium, phosphorus and magnesium.Final answer:
Among the given options, nitrogen is a macronutrient required in large amounts for plant growth and development, and is part of vital biomolecules such as carbohydrates, proteins, and nucleic acids.
Explanation:
The macronutrient in question can be identified by knowing that macronutrients are elements that organisms need in relatively large amounts compared to micronutrients which are needed in smaller amounts. The list of macronutrients includes nitrogen (N), phosphorus (P), potassium (K), calcium (Ca), magnesium (Mg), and sulfur (S), while micronutrients or trace elements, such as manganese (Mn), iron (Fe), zinc (Zn), and boron (B), are needed in smaller quantities.
Given the options provided (a. nitrogen b. manganese c. zinc d. boron), the correct answer is a. nitrogen because it is one of the primary macronutrients essential for plant growth and is a part of carbohydrates, proteins, and nucleic acids.
Element Z has 2 natural isotopes. One isotope has a mass of 15.0amu and has a relative abundance of 30%. The other isotope has a mass of 16.0amu and has a relative abundance of 70%. Estimate the average atomic mass for this element to one decimal place.
Answer:
The answer to your question is 15.7 amu
Explanation:
Abundance Mass
Isotope 1 30% 15
Isotope 2 70% 16
Average atomic mass = (Abundance isotope 1 x abundance) +
(Abundance isotope 2 x abundance)
Substitution
Average atomic mass = (0.30 x 15) + (0.70 x 16)
Simplify
Average atomic mass = 4.5 + 11.2
Result
Average atomic mass = 15.7 amu
Calcium has a cubic closest packed structure as a solid. Assuming that calcium has an atomic radius of 197 pm, calculate the density of solid calcium.
Answer:
[tex]\rho=1.54\ g/cm^3[/tex]
Explanation:
The expression for density is:
[tex]\rho=\frac {Z\times M}{N_a\times {{(Edge\ length)}^3}}[/tex]
[tex]N_a=6.023\times 10^{23}\ {mol}^{-1}[/tex]
M is molar mass of Calcium = 40.078 g/mol
For cubic closest packed structure , Z= 4
[tex]\rho[/tex] is the density
Radius = 197 pm = [tex]1.97\times 10^{-8}\ cm[/tex]
Also, for fcc, [tex]Edge\ length=2\sqrt{2}\times radius=2\sqrt{2}\times 1.97\times 10^{-8}\ cm=5.572\times 10^{-8}\ cm[/tex]
Thus,
[tex]\rho=\frac{4\times \:40.078}{6.023\times \:10^{23}\times \left(5.572\times 10^{-8}\right)^3}\ g/cm^3[/tex]
[tex]\rho=\frac{160.312}{10^{23}\times \:6.023\left(10^{-8}\times \:5.572\right)^3}\ g/cm^3[/tex]
[tex]\rho=\frac{160.312}{10^{23}\times \:1.04195E-21}\ g/cm^3[/tex]
[tex]\rho=\frac{160.312}{104.19483}\ g/cm^3[/tex]
[tex]\rho=1.54\ g/cm^3[/tex]
The density of solid calcium can be calculated by determining the density of its unit cell using the face-centered cubic (FCC) structure. The mass and volume of the unit cell can be calculated using the atomic radius and atomic mass of calcium. Dividing the mass by the volume gives the density of solid calcium.
Explanation:The density of solid calcium can be calculated by determining the density of its unit cell, which is a face-centered cubic (FCC) structure. In an FCC structure, each unit cell contains 4 atoms. The mass of 4 calcium atoms can be calculated using the atomic mass of calcium, and the volume of the unit cell can be calculated using the atomic radius of calcium. Dividing the mass by the volume gives the density of solid calcium.
The atomic radius of calcium is given as 197 pm, which can be converted to cm by multiplying by 10^-10. The volume of the unit cell can be calculated using the formula V = (edge length)^3. The edge length can be calculated using the diagonal of the face, which is 4 times the atomic radius. The mass of 4 calcium atoms can be calculated using the atomic mass of calcium, which is 40.08 g/mol. Dividing the mass by the volume gives the density of solid calcium.
Density of solid calcium = mass of 4 Ca atoms / volume of unit cell
Keywords: density, solid calcium, unit cell, face-centered cubic (FCC) structure, atomic radius, atomic mass
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. If 84 grams of sodium chloride reacts with an excess amount of magnesium oxide, how many grams of sodium oxide will be produced? Question 2 options: 23.2 g Na20 45g MgCl2 107g MgO 44.5g Na20
Answer:
44.5 g of Na₂O
Explanation:
The reaction is this one:
2NaCl + MgO → Na₂O + MgCl₂
Moles of NaCl = Mass / Molar mass
84 g / 58.45 g/m = 1.43 moles
Ratio is 2:1, so if we produce 1 mol of Na₂O, from 2 moles of NaCl; If we have 1.43 moles, we 'll produce the half of moles
1.43 / 2 = 0.72 moles
Molar mass Na₂O = 62 g/m
Mol . molar mass = 0.72 m . 62 g/m = 44.5 g
Answer:
There will be 44.5 grams of sodium oxide (Na2O) produced
Explanation:
Step 1: Data given
Mass of Sodium chloride (NaCl) = 84.00 grams
Magnesium oxide = in excess
Molar mass of NaCl = 58.44 g/mol
Molar mass of sod)ium oxide (Na2O = 61.98 g/mol
Step 2: The balanced equation
2NaCl + MgO → Na2O + MgCl2
Step 3: Calculate moles of NaCl
Moles NaCl = Mass / Molar mass
Moles NaCl = 84.00 grams / 58.44 g/mol
Moles NaCl = 1.437 moles
Step 4: Calculate moles of Na2O
The limiting reactant is NaCl.
For 2 moles NaCl consumed, we need 1 mol MgO to produce 1 mol Na2O and 1 mol of MgCl2
For 1.437 moles of NaCl we'll have 1.437/2 = 0.7185 moles of Na2O
Step 5: Calculate mass of Na2O
Mass Na2O = Moles Na2O * Molar mass Na2O
Mass Na2O = 0.7185 moles * 61.98 g/mol
Mass Na2O = 44.53 grams of Na2O
There will be 44.5 grams of sodium oxide (Na2O) produced
After gathering sufficient evidence to generate an _____for a nutrient, that value is used to establish an RDA for the same nutrient.
Answer:
Estimated Average Requirement (EAR).
Explanation:
Hello,
In this case, it is important to consider that Dietary Reference Intakes (DRIs) are reference values to quantitatively estimate the nutrient necessities to be taken for planning and assessing diets for healthy people. On the other hand, the Recommended Dietary Allowance (RDA) is the average daily dietary intake level that is enough to know the nutrient necessity of nearly all (about 98%) healthy individuals in a particular population. The answer is Estimated Average Requirement (EAR) which is a nutrient intake value that is considered to meet the necessity of half (50%) the healthy individuals in a particular population.
Best regards.
if all of the carbon atoms are linked by single covalent bonds and there are no branches, the compounds are called_____________.
Answer: If all of the carbon atoms are linked by single covalent bonds and there are no branches, the compounds are called homologous series.
Explanation:
A series of carbon atoms which include different number of carbon atoms but have same functional group are known as homologous series.
Generally, these type of series have a chemical formula as [tex]C_{n}H_{2n+2}[/tex].
No branches are present in this type of series.
For example, [tex]CH_{4}[/tex], [tex]C_{2}H_{6}[/tex], [tex]C_{3}H_{8}[/tex] etc are all homologous series.
Thus, we can conclude that if all of the carbon atoms are linked by single covalent bonds and there are no branches, the compounds are called homologous series.
The half-life of iodine-131 is about 8 days. How much of a 50mg sample will be left in 25 days? Write your answer rounded to the nearest tenth.
Final answer:
Approximately 20.7mg of the 50mg sample will be left after 25 days.
Explanation:
The half-life of iodine-131 is 8 days. To determine how much of the 50mg sample will be left after 25 days, we can use the formula:
Amount remaining = Initial amount x (0.5)^(time elapsed / half-life)
Plugging in the values:
Amount remaining = 50mg x (0.5)^(25 / 8) = 50mg x 0.4142 = 20.7mg
Therefore, approximately 20.7mg of the 50mg sample will be left after 25 days.
Cindy predicts that plastic foam insulates cold drinks better than metal or ceramic materials do. To test the hypothesis, she fills cups made from these materials with equal amounts of cold water. She records the temperature of the water in each cup, using scientific thermometers, every 10 minutes until the water reaches room temperature. Which of these conditions must be the same for this experiment to be valid?
A. the thermometer that is in each cup
B. the starting temperature of the water in each cup
C. the ending temperature of the water in each cup
D. the material that makes up each cup
Answer:
B. The starting temperature of the water in each cup
Explanation:
Cindy is trying to see if foam or ceramic is a better insulator. Those are her independent variables.
The other variables , like the starting temperature of the water in each cup, must be controlled variables. If she uses different starting temperatures in each cup, she won't know if it was the temperature or the materials that caused her results.
A. is wrong. The thermometers should be identical but, if they aren't, it will make little difference in the results.
C. is wrong. The ending temperature is room temperature, so it is automatically the same for each cup.
D. is wrong. She is trying to measure the effect of different materials.
Answer:the answer is B(the starting temperature of the water in each cup)
Explanation:
10.0 mL of a Cu2+ solution of unknown concentration was placed in a 250 mL Erlenmeyer flask. An excess of KI solution was added. Indicator was added and the solution was diluted with H2O to a total volume of 75 mL. For rxn 2, the solution from rxn 1 was titrated with 0.15 M Na2S2O3. The equivalence point of the titration was reached when 13.05 mL of Na2S2O3 had been added. What is the molar concentration of Cu2+ in the original 10.0 mL solution?
Answer:
Molar concentration: 0,0489M
Explanation:
In this titration of Cu²⁺ you add an excess of I⁻ that reacts with Cu²⁺ producing I₂, this I₂ reacts with Na₂S₂O₃. If you know the I₂ that reacts with Na₂S₂O₃ you can know the I⁻ that reacts with Cu²⁺ and, thus, the quantity of Cu²⁺. The reactions are:
2Cu²⁺ + 4I⁻ → 2CuI + I₂
I₂ + 2S₂O₃⁻ → S₄O₆ + 2I⁻
Moles of S₂O₃⁻ are:
0,01305L×0,15M = 1,96x10⁻³ moles of S₂O₃⁻.
Moles of I₂ are:
1,96x10⁻³ moles of S₂O₃⁻× ( 1 mole of I₂ / 2 moles of S₂O₃⁻) = 9,79x10⁻⁴ moles of I₂
Moles of Cu²⁺ are:
9,79x10⁻⁴ moles of I₂×( 2 moles of Cu²⁺ / 4 moles of I₂) = 4,89x10⁻⁴ moles of Cu²⁺
As volume of the solution was 10,0mL = 0,0100L, the molar concentration of the original solution is:
4,89x10⁻⁴ moles of Cu²⁺ / 0,0100L = 0,0489M
The unknown concentration of the Cu2+ solution can be found by determining the moles of Na2S2O3 at the equivalence point, using this to calculate the moles of Cu2+ from stoichiometry, and then dividing by the volume of the Cu2+ solution in liters.
Explanation:In this question, a Cu2+ solution, of unknown concentration, is titrated with 0.15 M Na2S2O3. The Cu2+ solution's concentration can be calculated using the data provided. The titration of this Cu2+ solution is complete, or the equivalence point is reached, when 13.05 mL of Na2S2O3 is added. The reaction that occurs is 2Na2S2O3 + Cu2+ -> CuS2O3 + 2Na+, and from the stoichiometry of the reaction, we know that two moles of Na2S2O3 react with one mole of Cu2+ ion.
Using the moles of Na2S2O3 that reacted (moles = Molarity x Volume (in liters), so moles = 0.15 M x 13.05 mL/1000), we can find out the moles of Cu2+ that were present in the 10 mL sample. We can then calculate the molarity of the Cu2+ solution by dividing the moles of Cu2+ by the volume of the solution in liters (0.01 L).
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Why did J.J. Thomson reason that electrons must be a part of the atoms of all element
Answer:
Same particles (electrons) were emitted even after changing the cathode material.
Explanation:
In his famous experiment, Thompson tested the properties of atomic particles. He used a cathode ray tube to apply voltage on the cathode. This generated beam of electrons, also called cathode rays. He bombarded the rays on phosphorus on the other end of the tube, to observe the pathway it took.
When he noticed the deflection of cathode rays when it passes through the electric and magnetic field, he repeated the experiment by changing the cathode material. To his surprise, rays emitted from all the materials exhibited the same behavior.
He concluded that these rays comprising of electrons, are a fundamental part of atoms of every element.
What mass of gold is produced when 17.6^A of current are passed through a gold solution for 37.0 min ?
Answer:
There is 26.58 grams of gold formed
Explanation:
Step 1: Data given
17.6 A of current are passed through a gold solution for 37.0 min
Molar mass of Au = 196.967 g/mol
Step 2: The equation
Au^3+ + 3e- → Au
Step 3: Calculate coulombs
17.6 Coulomb/s * 37.0 min * 60 sec/min = 39072 Coulombs
1 Faraday = 96500 Coulombs
Step 4: Calculate faraday
39072 Coulombs / 96500 Coulombs / Faraday = 0.40489 Faraday
Step 5: Calculate mass of gold formed
For every 3 Faraday of electricity used up , 1 mole Au is formed
0.40489 Faraday * 1 mole Au/ 3 Faraday = 0.13496 mole Au
196.967 g/mol * 0.13496 mol = 26.58 g Au
There is 26.58 grams of gold formed
The mass of gold that is produced is 26.59 g
Using the formula
[tex]m = \frac{Atomic\ mass}{nF}\times It[/tex]
Where m is the mass
n is the number of equivalents
F is the Faraday constant ( F = 96485 C)
I is the current
and t is the time
From the given information
I = 17.6 A
t = 37.0 min = 37.0 × 60
t = 2220 secs
For gold
Atomic mass = 196.97 g/mol
and n = 3
Putting these parameters into the formula, we get
[tex]m = \frac{196.97}{3 \times 96485} \times 17.6 \times 2220[/tex]
[tex]m = \frac{7696011.84}{289455}[/tex]
m = 26.59 g
Hence, the mass of gold that is produced is 26.59 g
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capable of bonding to surfaces with the application of light pressure is a chemical or physical reaction?
Answer:
Capable of bonding to surfaces with the application of light pressure is a chemical reaction.
Explanation:
Here, bonding occurs . Bonding results in destruction of old bond and formation of new bonds.Hence new substance with completely different properties is formed. These changes occur only in a chemical reaction .
In physical processes no new substance is formed (no bonding).So,no change in properties of a substance . ]
This process (capability of bonding to surfaces with the application of light pressure) results in chemical reaction.
This type of substances are called Pressure-sensitive Adhesives.(PSA)
Analysis
1. Alpha Particles
a. What is the mass number of an alpha particle?
b. What kind of charge does an alpha particle have?
c. What is the identity of an alpha particle? (Hint: What atomic symbol is used?)
d. Write the balanced equation for the alpha decay that is below the “Show Equation.” Label the parent, daughter, and alpha particle.
Please help
Answer:
Below.
Explanation:
1.
a. An alpha particle has 2 protons and 2 neutrons ( a helium nucleus).
b. The charge on an alpha particle is 2+.
c. Atomic symbol used is α or He2+.
d. (no equation shown).
Americanum 241 decays to Neptunium 237 with the loss of 1 alpha particle
Am 241 = Np 237 + Ne2+
- note the atomic mass is 4 less under the decay.
Describe light with respect to its speed and its dual nature as both a wave and a particle.
Answer:
Scientists have been debating over light being a wave or particle since its recognition.
Sir Issac Newton discovered that light had frequency and other properties. Newton described light to be a particle because it created shadows which were sharp and very clear.
Francesco Maria Grimaldi, claimed that light was a wave. This was because this scientist observed the diffraction of light and hence, claimed light to be a type of wave.
The speed of light is 299 792 458 m / s. Nothing can travel faster than light.
Some insects can glide across the surface of water due to water's: hydrogen bonds. viscosity. capillarity. polarity. surface tension
Answer: surface tension
Explanation: the tiny weight of insect is not strong enough to break the surface tension of water. So when insects stands or move on water, their feets creates something like dimples on the surface of water which then spring back to propel the insect forward thereby preventing them from sinking.
When scientists use one of their five senses to gather information, they are A. making an observation. B. making an inference. C. predicting a relationship D. drawing a conclusion.
The scientists use one of their five senses to gather information, they are making an observation.
What is an observation ?The active gathering of data from a primary source is observation. Observation of living things makes use of the senses. Using scientific tools to perceive and record data is another way that observation may be used in science.
A technique for gathering information through observation of individuals, situations, or physical qualities as they are occurring naturally.
Using one or more senses, our observational abilities provide us with knowledge about things, occasions, attitudes, and occurrences. Being able to see and learn about the outside world is crucial because it forms the basis of effective communication.
Thus, option A is correct.
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In scientific endeavors, when one of the five senses is used to gather information, it constitutes making an observation. Observations can involve both qualitative and quantitative data, and they provide evidence that scientists use to form conclusions.
Explanation:When scientists use one of their five senses to gather information, they are A. making an observation. Scientists usually perform observations by using one or more of their five senses to gain an understanding of the properties or behaviors of a substance or a system. These data can be qualitative (descriptive) or quantitative (consisting of numbers). For instance, seeing a plant growing (observation) and measuring its growth over weeks (data collection). From these observations, the scientist can infer conclusions based on the evidence collected. In brain studies, the activity in the brain during specific tasks is observed, which could be the basis for conclusions about brain functionality.
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Over 200 items are scored on the driving test. You may only make ______ or fewer errors to pass the test.
Answer:
The correct answer is 15.
Explanation:
The driving test must be passed with more than 200 points. This means that you can only make a maximum of 15 mistakes to pass the test. This is possible as long as no critical mistakes are made, which are enough to fail the test. These critical mistakes are for example driving faster than allowed, too slow, driving distracted, etc.
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An element is in Group 15. The last electron added to an atom of this element will be in a(n) __ sublevel
a. p
b. d
c. f
d. s
Answer:
a
Explanation:
Group 15 form trihydrides with the non metal atoms like phosphine, ammonia
Astor Manufacturing stores hazardous and volatile chemicals in its warehouse. The warehouse has state of the art equipment to make sure the chemicals do not explode. An unexpected earthquake shakes the warehouse, causing the chemicals to explode and injure William, a passer-by on a nearby sidewalk. Astor Manufacturing is
Answers. The correct option is A
Explanation:
The Astronomy manufacturing company is liable to Williams injury only if the company was grossly negligent.
Analyze feasibility of given reactions based on electrode potentials at standard conditions and nonstandard conditions.
Answer and Explanation:
In order to predict the feasibility of redox processes, standard electrode potentials are majorly employed. Generally, if the electrode potential for the reaction is positive, it is considered to be feasible. However, some conditions affect this statement
The value of E° talks about the feasibility of the reaction under standard conditions only and says nothing about the reaction rate.
A positive value of E° means, the equilibrium constant K is greater than 1; while a negative value of E° means, that it is less than 1.
The attachment below shows the simple analysis of the feasibility of two different reactions A and B, at standard and non standard conditions respectively.
NOTE: Standard conditions for Redox reaction: 298.15K(Temperature), 1 atm(Pressure), 1.0M(Concentration) for both anode and cathode.
Non standard conditions for Redox reaction: Any of the 3 conditions above are changed, especially the concentration.
According to the VSEPR model, the progressive decrease in the bond angles in the series of molecules CH4, NH3, and H2O is best accounted for by the:_________
Answer:
Presence of lone pairs of electrons
Explanation:
According to VSEPR theory, the presence of lone pairs caused increased repulsion of electron pairs on the valence shell of the centeral atom of the molecule. This decreases or distorts the bond angle. The decrease in bond ange depends on the number of lone pairs present on the valence shell of the central atom of the molecule. Ammonia has only one lone pair hence the bond angle is 107°, water has two lone pairs and the bond angle is 104°. Compare this this with the bond angle of 109° in methane which has only bond pairs and no lone pairs.
The decrease in bond angles from CH4 to NH3 to H2O is due to increasing number of lone pairs on the central atom which cause greater repulsion and smaller bond angles, as per the VSEPR theory.
Explanation:According to the Valence shell electron-pair repulsion (VSEPR) theory, the progressive decrease in the bond angles in the series of molecules CH4, NH3, and H2O is best accounted for by the presence and arrangement of lone electron pairs on the central atom. CH4 has no lone pairs on the central carbon atom and has a tetrahedral shape with bond angles of 109.5°. In NH3, the nitrogen atom has one lone pair, which decreases the bond angle to less than 109.5°. In H2O, the oxygen atom has two lone pairs, leading to an even smaller bond angle of 104.5°. The VSEPR theory predicts that the electron pairs will arrange themselves to minimize repulsion, and the lone pair-lone pair repulsion is greatest, followed by lone pair-bonding pair, and finally, bonding pair-bonding pair repulsion is the least.
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What alteration to an HCN channel will reduce the transport of K+ in favor of Na+ transport? HCN channels are permeable to both K+ and Na+ ions.
Answer:
Decreasing the diameter of the channel by removing aa's. Since Na+ is chemically similar to K+, one can assume the difference must be due to the size of the atom. The K+ ion is larger than the Na+ ion, so reducing the diameter of the channel can allow Na+ to enter while preventing K+ entry. This explains clearly and perfectly how reducing the diameter reduces the
transport of K+ in favor of Na+ transport.
If 200. mL of 0.60 M MgCl2(aq) is added to 400 mL of distilled water, what is the concentration of Mg and Cl in the resulting solution?
A. 0.20 M Mg ion & 0.20 M Cl ionB. 0.40 M Mg ion & 0.40 M Cl ionC. 0.20 M Mg ion & 0.40 M Cl ionD. 2.0 M Mg ion & 2.5 M Cl ion
Answer:
C. 0.20 M Mg ion & 0.40 M Cl ion
Explanation:
MgCl₂ is a ionic salt which is dissociated as this
MgCl₂ → Mg²⁺ + 2Cl⁻
First of all, we have a solution of 200 mL, with [MgCl₂] = 0.6M
Molarity . volume = moles.
0.6 mol/l . 0.2l = 0.12 mol
MgCl₂ → Mg²⁺ + 2Cl⁻
0.12mol 0.12 0.24
This moles are also in 400mL of water, so the new concentration is
[Mg²⁺] = 0.12 m/0.6L = 0.2M
[Cl⁻] = 0.24 m/0.6L = 0.4M
Remember we initially have 200mL and then, we add 400 mL, so we supose aditive volume. (600mL)
The concentration of magnesium ion, Mg²⁺ in the resulting solution is 0.2 M while that of the chloride ion, Cl¯ is 0.4 M
The correct answer to the question is Option C. 0.20 M Mg ion & 0.40 M Cl ion
we'll begin by calculating the molarity of the diluted solution. This can be obtained as follow:
Volume of stock solution (V₁) = 200 mL
Molarity of stock solution (M₁) = 0.60 M
Volume of diluted solution (V₂) = 200 + 400 = 600 mL
Molarity of diluted solution (M₂) = ?The molarity of the diluted solution can be obtained as follow:
M₁V₁ = M₂V₂0.6 × 200 = M₂ × 600
120 = M₂ × 600
Divide both side by 600
M₂ = 120 / 600
M₂ = 0.2 MThus, the molarity of the diluted (i.e resulting) solution of MgCl₂ is 0.2 M
Next, we shall determine the concentration of magnesium ion, Mg²⁺ in the diluted solution. This is illustrated below:
MgCl₂(aq) —> Mg²⁺(aq) + 2Cl¯(aq)
From the balanced equation above,
1 mole MgCl₂ dissolves to produce 1 mole Mg²⁺.
Therefore,
0.2 M MgCl₂ will also produce 0.2 M Mg²⁺.
Thus, the concentration of magnesium ion, Mg²⁺ in the resulting solution is 0.2 M.
Finally, we shall determine the concentration of the chloride ion, Cl¯ in the resulting solution.
MgCl₂(aq) —> Mg²⁺(aq) + 2Cl¯(aq)
From the balanced equation above,
1 mole MgCl₂ dissolves to produce 2 moles of Cl¯
Therefore,
0.2 M MgCl₂ will also produce = 2 × 0.2 = 0.4 M Cl¯
Thus, the concentration of chloride ion, Cl¯ in the resulting solution is 0.4 M.
From the calculations made above:
The concentration of magnesium ion, Mg²⁺ in the resulting solution is 0.2 M while that of the chloride ion, Cl¯ is 0.4 M
Option C. 0.20 M Mg ion & 0.40 M Cl ion gives the correct answer to the question.
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How many grams of water will absorb a total of 2400 joules of energy when the temperature changes from 10.0°C to 30.0°C?
Answer:
28.7 grams of water
Explanation:
Calorimetry problem:
Q = C . m . ΔT
2400 J = 4.18 J/g°C . m . (30°C - 10°C)
2400 J = 4.18 J/g°C . m . 20°C
2400J = 83.6 J/g . m
2400J / 83.6 g/J = m
28.7 g = m
Final answer:
To calculate the amount of energy absorbed by water when the temperature changes, use the formula: q = mass * specific heat capacity * temperature change.
Explanation:
To calculate the amount of energy absorbed by water, we can use the formula: q = mass * specific heat capacity * temperature change. The specific heat capacity of water is approximately 4.184 J/g°C. In this case, the temperature change is from 10.0°C to 30.0°C, which gives us a ΔT of 20.0°C. We need to convert the mass from grams to kilograms by dividing it by 1000.
So, q = (mass / 1000) * 4.184 J/g°C * 20.0°C.
Substituting the given mass of water (in grams) into the equation, we have: q = (2400 / 1000) * 4.184 J/g°C * 20.0°C. Simplifying the expression gives us the amount of energy absorbed by the water in joules.