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Cesium-137 is part of the nuclear waste produced by uranium-235 fission. The half-life of cesium-137 is 30.2 years. How much time is required for the activity of a sample of cesium-137 to fall to 20.0 percent of its original value?
Answer:
There are required 70.1 years for the activity of a sample of cesium-137 to fall to 20.0 percent of its original value
Explanation:
The radioactive decay follows always first-order kinetics where its general law is:
Ln[A] = -Kt + ln[A]₀
Where [A] is actual concentration of the atom, k is rate constant, t is time and [A]₀ is initial concentration.
We can find rate constant from half-life as follows:
Rate constant:
t(1/2) = ln 2 / K
As half-life of Cesium-137 is 30.2 years:
30.2 years = ln 2 / K
K = 0.02295 years⁻¹
Replacing this result and with the given data of the problem:
Ln[A] = -Kt + ln[A]₀
Ln[A] = -0.02295 years⁻¹* t + ln[A]₀
Ln ([A] / [A₀]) = -0.02295 years⁻¹* t
As you want time when [A] is 20% of [A]₀, [A] / [A]₀ = 0.2:
Ln (0.2) = -0.02295 years⁻¹* t
70.1 years = t
There are required 70.1 years for the activity of a sample of cesium-137 to fall to 20.0 percent of its original valueCompound A is an alkene that was treated with ozone to yield only (CH3CH2CH2)2C=O. Draw the major product that is expected when compound A is treated with a peroxy acid (RCO3H) followed by aqueous acid (H3O+).
Answer:
2,2,3,3-tetrapropyloxirane
Explanation:
In this case, we have to know first the alkene that will react with the peroxyacid. So:
What do we know about the unknown alkene?
We know the product of the ozonolysis reaction (see figure 1). This reaction is an oxidative rupture reaction. Therefore, the double bond will be broken and we have to replace the carbons on each side of the double bond by oxygens. If [tex](CH_3CH_2CH_2)_2C=O[/tex] is the only product we will have a symmetric molecule in this case 4,5-dipropyloct-4-ene.
What is the product with the peroxyacid?
This compound in the presence of alkenes will produce peroxides. Therefore we have to put a peroxide group in the carbons where the double bond was placed. So, we will have as product 2,2,3,3-tetrapropyloxirane. (see figure 2)
There are 2.4g of calcium hydroxide reacted with nitric acid. Calculate the number of moles of calcium hydroxide used. Write your answer using proper significant digits and units. Show all your work.
Answer:
0.032 moles
Explanation:
no of moles =
[tex] \frac{mass \: in \: grams}{relative \: molecular \: mass} [/tex]
=
[tex] \frac{2.4}{40 + 32 + 2} [/tex]
= 0.032
Calcium hydroxide reacted with nitric acid the total number of moles will be 0.032 moles.
What is a mole?
A mole is Avogadro's number of particles, which is exactly 6.02214076×1023.
The mole is widely used in chemistry as a convenient way to express amounts of reactants and products of chemical reactions. For example, the chemical equation 2H2 + O2 → 2H2O can be interpreted to mean that for each 2 mol dihydrogen (H2) and 1 mol dioxygen (O2) that react 2 mol of water (H2O) form.
Number of moles = Mass of substance / Mass of one mole Number of moles
mass of substance = 2.4g
molar mass of calcium hydroxide is (1 ×40.078g/mol Ca) +(2 × 15.999g/mol O) + (2 × 1.008g/mol H) = 74.092 g/mol Ca (OH)2
substituting the value,
number of moles = 2.4 / 74.029
= 0.032 moles
Therefore, moles of calcium hydroxide will be 0.032 moles
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Write the equation for the reaction described: A solid metal oxide, , and hydrogen are the products of the reaction between metal and steam. (Use the lowest possible coefficients. Use the pull-down boxes to specify states such as (aq) or (s). If a box is not needed, leave it blank.)
Answer:
Pb + 2H2O --> PbO2 + 2H2
Explanation:
Products:
Solid metal; PbO2
Hydrogen; H
Reactants:
Metal; Pb
Steam; H2O
Reactants --> Products
Pb + H2O --> PbO2 + H2
Upon balancing we have;
Pb + 2H2O --> PbO2 + 2H2
of their
(c) Ethanol is an alcoholic beverage which can be brewed
from cassava. Outline the process by which ethanol can
be prepared.
[3]
(d) Ethanol is used as a fuel.construct a balanced chemical
equation for its complete combustion.
[2]
[Total:10 Marks]
s?
Answer:
the chemical equation of ethanol as fuel is
C2H5OH(l)+3 02(g)------2CO2(g)+3H2O(g)
the preparation process of ethanol from cassava is
cassava flour---liquification---saccharification---cooling---fermentation---distillation---ethanol
A solution containing a unknown ionic compound, vigorously bubbles when hydrochloric acid (HCl) is added to the solution. This might indicate that the solution contains which anion?
Answer:
CO3^2-
Explanation:
In qualitative analysis, we try to use chemical reactions to determine the composition of an unknown substance. The addition of certain reagents to the unknown solution gives certain results that show the presence or absence of certain species from the unknown sample.
When dilute HCl is added to an unknown sample and effervescence is observed, then the unknown sample must contain CO3^2- or HCO3^-. The presence of these species is confirmed if the gas evolved is passed through limewater and the gas turns limewater milky.
Which response has both answers correct? Will a precipitate form when 250 mL of 0.33 M Na 2CrO 4 are added to 250 mL of 0.12 M AgNO 3? [K sp(Ag 2CrO 4) = 1.1 × 10 –12] What is the concentration of the silver ion remaining in solution?
Answer:
A precipitate will form.
[Ag⁺] = 2.8x10⁻⁵M
Explanation:
When Ag⁺ and CrO₄²⁻ are in solution, Ag₂CrO₄(s) is produced thus:
Ag₂CrO₄(s) ⇄ 2 Ag⁺(aq) + CrO₄²⁻(aq)
Ksp is defined as:
Ksp = 1.1x10⁻¹² = [Ag⁺]² [CrO₄²⁻]
Where the concentrations [] are in equilibrium
Reaction quotient, Q, is defined as:
Q = [Ag⁺]² [CrO₄²⁻]
Where the concentrations [] are the actual concentrations
If Q < Ksp, no precipitate will form, if Q >= Ksp, a precipitate will form,
The actual concentrations are -Where 500mL is the total volume of the solution-:
[Ag⁺] = [AgNO₃] = 0.12M ₓ (250mL / 500mL) = 0.06M
[CrO₄²⁻] = [Na₂CrO₄] = 0.33M × (250mL / 500mL) = 0.165M
And Q = [0.06M]² [0.165M] = 5.94x10⁻⁴
As Q > Ksp; a precipitate will form
In equilibrium, some Ag⁺ and some CrO₄⁻ reacts decreasing its concentration until the system reaches equilibrium. Equilibrium concentrations will be:
[Ag⁺] = 0.06M - 2X
[CrO₄²⁻] = 0.165M - X
Where X is defined as the reaction coordinate
Replacing in Ksp expression:
1.1x10⁻¹² = [0.06M - 2X]² [0.165M - X]
Solving for X:
X = 0.165M → False solution. Produce negative concentrations.
X = 0.0299986M
Replacing, equilibrium concentrations are:
[Ag⁺] = 0.06M - 2(0.0299986M)
[CrO₄²⁻] = 0.165M - 0.0299986M
[Ag⁺] = 2.8x10⁻⁵M[CrO₄²⁻] = 0.135M
A spontaneous galvanic cell consists of a Pb electrode in a 1.0 M Pb(NO3)2 solution and a Cd electrode in a 1.0 M Cd(NO3)2 solution. What is the standard cell potential for this galvanic cell
Answer:
0.27 V
Explanation:
Given that the both half cells contain 1.0 molar solutions of their respective electrolytes.
E°Pb= -0.13 V
E°Cd = -0.40 V
Since it is a galvanic cell, the electrode having a more negative electrode potential will serve as the anode and the electrode having a less negative electrode potential will serve as the cathode.
Hence cadmium will serve as the anode and lead will serve as the cathode.
E°cell = E°cathode - E°anode
E°cell = -0.13 - (-0.40)
E°cell = 0.27 V
Define the following terms - you may need to consult your lecture text or other suitable resource:
a. monomer,
b. repeating unit,
c. condensation polymerization,
d. cross-linked polymer
Answer:
a) Monomers: monomers are unit molecules, that can react together with other monomers, to form a long chain molecule called a polymer. Th polymer formed can also be in a three dimensional network. The process of this conversion of monomers to polymers is called polymerization.
b) Repeating unit: A repeating unit is a unit of the polymer formed, whose repetition would produce a long complete polymer chain. A polymer is made up of these repeating links of molecules that form a long chain of molecules.
c) Condensation polymerization: This is a form of condensation reaction, that involves the combination of molecules into polymers with the loss of small molecules such as water or methanol as by products.
d) Cross-linked polymer: This is a polymer formed from a type of bonding of molecules. The bonding is usually in the form of covalent bonds or ionic bonds and the polymers can be either synthetic polymers or natural polymers. The cross-links leads to an alteration in the physical properties of the polymer.
The definition of following terms are :
a) Monomers:
The monomers are unit atoms, that can respond in conjunction with other monomers, to create a long chain molecule called a polymer.
The polymer shaped can too be in a three dimensional arrange.
b) Repeating unit:
A rehashing unit may be a unit of the polymer shaped, whose reiteration would produce a long total polymer chain.
A polymer is made up of these rehashing joins of atoms that shape a long chain of molecules.
c) Condensation polymerization:
This is often a frame of condensation response, that includes the combination of particles into polymers with the misfortune of little particles such as water or methanol as by products.
d) Cross-linked polymer:
This can be a polymer shaped from a sort of holding of particles.
The cross-links leads to an modification within the physical properties.
DefinitionsDefinition is a rhetorical style that uses various techniques to impress upon the reader the meaning of a term, idea, or concept.
Definition may be used for an entire essay but is often used as a rhetorical style within an essay that may mix rhetorical styles.
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Which solution, if either, would create the higher osmotic pressure (compared to pure water): one prepared from 1.0 g of NaCl in 10 mL of water or 1.0 g of CsBr in 10 mL of water
Answer: NaCl would give the higher pressure
Explanation:
Osmotic pressure depends only on the number of ions.
NaCl dissociates as Na+ and Cl- ; CsBr dissociates as Cs+ and Br-
But the concentration of the solutions are different.
Concentration (morality ) of NaCl = Moles /Litre = (1 g /58.44g/mol)/0.01L
Total number of ions in NaCl solution = 2 x (1 g /58.44g/mol)/0.01L ( 1 mol NaCl gives 2 moles ions, 1 mol Na+ and 1 mol Cl-)
= 1.71×2RT
Similarly total number of ions in CsBr solution = 2 x (1 g /212.80 g/mol)/0.01L
= 0.47×2RT
Therefore osmotic pressure is higher in NaCl solution.
The argon atoms are excited into an excited state before emitting the 488.0 nm laser. It is known that the energy of the first ionization energy of argon is 1520 kJ mol-1. What is the energy level of the excited state (in unit eV) lies below the vacuum energy level (0 eV)
Answer:
Explanation:
Given that:
The argon atoms are excited into an excited state before emitting the 488.0 nm laser.
the energy of the first ionization energy of argon is 1520 kJ mol-1.
SInce 1 eV = 96.49 kJ/mol
Therefore, the energy of the first ionization energy of argon in eV is = ( 1520/ 96.49) eV
= 15.75 eV
To find where the energy level of the excited state lies below the vacuum energy level, let's first determine, the energy liberated by using planck expression.
[tex]E = \dfrac{hc}{\lambda}[/tex]
[tex]E = \dfrac{6.6 \times 10^{-34} \times 3 \times 10^8}{488 \times 10^{-9}}[/tex]
[tex]E = \dfrac{1.98 \times 10^{-25}}{488 \times 10^{-9}}[/tex]
[tex]E = \dfrac{1.98 \times 10^{-25}}{488 \times 10^{-9}}[/tex]
[tex]E =4.057 \times 10^{-19} \ J[/tex]
Converting Joules (J) to eV ; we get,
[tex]E =\dfrac{4.057 \times 10^{-19}}{1.6 \times 10^{-19}}[/tex]
E = 2.53 eV
The energy levels of the first exited state = -13.223 eV
When equation for neutralization of HBr by Ca(OH)2 is correctly balanced, how many molecules of water will be formed
Answer:
When equation for neutralization of HBr by Ca(OH)₂ is correctly balanced, 1.2046*10²⁴ molecules of water will be formed
Explanation:
A neutralization reaction is one in which an acid (or acidic oxide) reacts with a base (or basic oxide). In the reaction a salt is formed and in most cases water is formed. A Salt is an ionic compound formed by the union of ions and cations through ionic bonds.
In the reactions of a strong acid (those substances that completely dissociate) with a strong base (they dissociate completely, giving up all their OH-), the complete neutralization of the species is carried out:
2 HBr (aq) + Ca(OH)₂ (s) → CaBr₂ (aq) + 2 H₂O (l)
The reaction is already balanced, complying with the law of conservation of matter. This law states that since no atom can be created or destroyed in a chemical reaction, the number of atoms that are present in the reactants must be equal to the number of atoms present in the products.
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), 2 moles of water H₂O are formed.
On the other hand, Avogadro's Number or Avogadro's Constant is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023 * 10²³ particles per mole. Avogadro's number applies to any substance.
Then you can apply the following rule of three: if 1 mole of H₂O contains 6.023*10²³ molecules, 2 moles of H₂O, how many molecules does it contain?
[tex]amount of molecules=\frac{2moles*6.023*10^{23}molecules }{1 mole}[/tex]
amount of molecules= 1.2046*10²⁴ molecules
When equation for neutralization of HBr by Ca(OH)₂ is correctly balanced, 1.2046*10²⁴ molecules of water will be formed
The compound methylamine, CH3NH2, is a weak base when dissolved in water. Write the Kb expression for the weak base equilibrium that occurs in an aqueous solution of methylamine:
Answer:
Kb = [CH₃NH₃⁺] × [OH⁻] / [CH₃NH₂]
Explanation:
According to Brönsted-Lowry acid-base theory:
An acid is a substance that donates H⁺.A base is a substance that accepts H⁺.When methylamine reacts with water, it behaves as a Brönsted-Lowry base, according to the following reaction.
CH₃NH₂(aq) + H₂O(l) ⇄ CH₃NH₃⁺(aq) + OH⁻(aq)
The basic equilibrium constant (Kb) is:
Kb = [CH₃NH₃⁺] × [OH⁻] / [CH₃NH₂]
The literature value for the Ksp of Ca(OH)2 at 25 °C is 4.68E−6. Imagine you ran the experiment and got a calculated value for Ksp which was too high. Select all of the possible circumstances which would cause this result.
A. The HCl was more concentrated than the labeled molarity (0.0500 M).
B. The Ca[OH]2 solution may have been supersaturated.
C. The HCl was less concentrated than the labeled molarity (0.0500 M).
D. The Ca[OH]2 solution may have been unsaturated.
E. The titration flask may have not been clean and had a residue of a basic solution.
F. The titration flask may have not been clean and had a residue of an acidic solution.
Answer:
D. The Ca[OH]2 solution may have been unsaturated
Explanation:
The solubility product constant Ksp of any given chemical compound is a term used to describe the equilibrium between a solid and the ions it contains solution. The value of the Ksp indicates the extent to which any compound can dissociate into ions in water. A higher the Ksp, implies more greater solubility of the compound in water.
If the Ksp is more than the value in literature, this false value must have arisen from the fact that the solution was unsaturated hence it appears to be more soluble than it should normally be when saturated.
The second-order decomposition of NO2 has a rate constant of 0.255 M-1s-1. How much NO2 decomposes in 8.00 s if the initial concentration of NO2 (1.00 L volume) is 1.33 M
Answer:
0.9718M
Explanation:
Rate constant, k = 0.255 M-1s-1
time, t = 8.00 s
Initial concentration, [A]o = 1.33 M
Final concentration, [A] = ?
These quantities are represented by the equation;
1 / [A] = 1 / [A]o + kt
1 / [A] = 1 /1.33 + (0.255 * 8)
1 / [A] = 0.7519 + 2.04
[A] = 1 / 2.7919 = 0.3582 M
How much of NO2 decomposed is obtained from the change in concentration;
Change in concentration = Initial - Final
Change = 1.33 - 0.3582 = 0.9718M
A monoprotic weak acid, HA , dissociates in water according to the reaction HA(aq)+H2O(l)↽−−⇀H3O+(aq)+A−(aq) The equilibrium concentrations of the reactants and products are [HA]=0.260 M , [H3O+]=4.00×10−4 M , and [A−]=4.00×10−4 M . Calculate the Ka value for the acid HA.
Answer:
Ka = 6.15x10⁻⁷
Explanation:
Ka is defined as dissociation constant in the equilibrium of a weak acid with water. The general reaction is:
HA(aq) + H₂O(l) ⇆ H₃O⁺(aq) + A⁻(aq)
And Ka is defined as the ratio between molar concentrations in equilibrium of products over reactants as follows:
Ka = [H₃O⁺] [A⁻] / [HA]
You don't take water in the equilibrium beacuse is a pure liquid
Replacing with the concentrations of the problem:
Ka = [H₃O⁺] [A⁻] / [HA]
Ka = [4.00x10⁻⁴] [4.00x10⁻⁴] / [0.260]
Ka = 6.15x10⁻⁷
Consider the compound hydrazine N2H4 (MW = 32.0 amu). It can react with I2 (MW = 253.8 amu) by the following reaction 2 I2 + N2H4 ------------- 4 HI + N2 (a) How many grams of I2 are needed to react with 36.7 g of N2H4? (b) How many grams of HI (MW = 127.9 amu) are produced from the reaction of 115.7 g of N2H4 with excess iodine?
Answer:Cobb
Explanation:What y'all
A balloon has an initial volume of 2.954 L containing 5.50 moles of helium. More helium is added so that the balloon expands to 4.325 L. How much helium (moles) has been added if the temperature and pressure stay constant during this process.
Answer:
8.05 moles
Explanation:
5.50 / 2.954 = x / 4.325
x = 8.05
According to ideal gas equation, if the temperature and pressure stay constant during the process 0.520 moles have been added so that the balloon expands to 4.325 L.
What is ideal gas equation?The ideal gas equation is a equation which is applicable in a hypothetical state of an ideal gas.It is a combination of Boyle's law, Charle's law,Avogadro's law and Gay-Lussac's law . It is given as, PV=nRT where R= gas constant whose value is 8.314.The law has several limitations.The law was proposed by Benoit Paul Emile Clapeyron in 1834.
In the given example if pressure and temperature are constant then V=nR substituting V=4.325 l and R=8.314 so n=V/R=4.325/8.314=0.520 moles.
Thus, 0.520 moles of helium are added if the temperature and pressure stay constant during this process.
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What would be the voltage (Ecell) of a voltaic cell comprised of Cd(s)/Cd2+(aq) and Zr(s)/Zr4+(aq) if the concentrations of the ions in solution were [Cd2+] = 0.5 M and [Zr4+] = 0.5 M at 298K?
Answer:
1.05 V
Explanation:
Since;
E°cell= E°cathode- E°anode
E°cathode= -0.40 V
E°anode= -1.45 V
E°cell= -0.40-(-1.45) = 1.05 V
Equation of the process;
2Zr(s) + 4Cd^2+(aq) ---->2Zr^4+(aq) + 4Cd(s)
n= 8 electrons transferred
From Nernst's equation;
Ecell = E°cell - 0.0592/n log Q
Ecell= E°cell - 0.0592/8 log [0.5]/[0.5]
Since log 1=0
Ecell= E°cell= 1.05 V
Acetonitrile (CH3CN) is an important industrial chemical. Among other things, it is used to make plastic moldings, which have multiple uses, from car parts to Lego bricks. Which one of the following statements about acetonitrile is not correct?a. Acetonitrile has 16 valence electrons in its Lewis structure. b. Acetonitrile has one triple bond. c. Acetonitrile has one pair of nonbonding electrons. d. All atoms satisfy the octet rule in acetonitrile. e. One carbon atom and the nitrogen atom have nonzero formal charges.
Answer:
One carbon atom and the nitrogen atom have nonzero formal charges.
Explanation:
The compound Acetonitrile has sixteen valence electrons as is easily San from its structure. It contains a carbon nitrogen triple bond with a lone pair of electrons on nitrogen. All atoms satisfy the octet rule and there is no hyper valent atom in the molecule.
The formal charge an carbon and nitrogen is calculated as follows;
No. of valence electron on atom - [non bonded electrons + no. of bonds]
Therefore, for carbon and nitrogen, we have;
formal charge on carbon = 4 - (0 + 4) = 0
formal charge on nitrogen = 5 - (2 + 3) = 0
Hence carbon and nitrogen both possess zero formal charges.
Calculate the energy required to heat 566.0mg of graphite from 5.2°C to 23.2°C. Assume the specific heat capacity of graphite under these conditions is ·0.710J·g−1K−1 . Be sure your answer has the correct number of significant digits.
Answer:
7.23 J
Explanation:
Step 1: Given data
Mass of graphite (m): 566.0 mgInitial temperature: 5.2 °CFinal temperature: 23.2 °CSpecific heat capacity of graphite (c): 0.710J·g⁻¹K⁻¹Step 2: Calculate the energy required (Q)
We will use the following expression.
Q = c × m × ΔT
Q = 0.710J·g⁻¹K⁻¹ × 0.5660 g × (23.2°C-5.2°C)
Q = 7.23 J
A. Rank the following substances in order of decreasing standard molar entropy (S∘).
Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
Br(g)
Cl2(g)
I2(g)
F2(g)
B. Rank the following substances in order of decreasing standard molar entropy (S∘).
Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
H2S(g)
H2O(g)
H2O2(g)
C. Rank the following substances in order of decreasing standard molar entropy (S∘).
Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
C(s, amorphous)
C(s, diamond)
C(s, graphite)
Answer:
A. Rank the following substances in order of decreasing standard molar entropy (S∘).
Rank the gases from largest to smallest standard molar entropy
I2(g)>Br2(g)>Cl2(g)>F2(g)
B. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
H2O2(g)>H2S(g) >H2O(g)
C. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
C(s, amorphous) >C(s, graphite)>C(s, diamond)
Explanation:
Hello,
In this case, we can apply the following principles to explain the order:
- The greater the molar mass, the larger the standard molar entropy.
- The greater the molar mass and the structural complexity, the larger the standard molar entropy.
- The greater the structural complexity, the larger the standard molar entropy.
A. Rank the following substances in order of decreasing standard molar entropy (S∘).
Rank the gases from largest to smallest standard molar entropy
I2(g)>Br2(g)>Cl2(g)>F2(g)
This is due to the fact that the greater the molar mass, the larger the standard molar entropy.
B. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
H2O2(g)>H2S(g) >H2O(g)
This is due to the fact that the greater the molar mass and the structural complexity, the larger the standard molar entropy as the hydrogen peroxide has four bonds and weights 34 g/mol as well as hydrogen sulfide that has two bonds only.
C. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.
C(s, amorphous) >C(s, graphite)>C(s, diamond)
Since the molecular complexity is greater in the amorphous carbon (messy arrangement), mid in the graphite and lower in the diamond (well organized).
Regards.
How are animals used in vaccine development?
Answer:
Animals whose certain organs closely match those of humans or have similar genetic makeup are used in vaccine tests because the results can closely resemble those same results on humans.
Explanation:
Answer:
they use them to test the effectiveness of the vaccine.
Explanation:
A 25.00 mL sample of unknown concentration of HNO3 solution requires 22.62 mL of 0.02000 M NaOH to reach the equivalence point. What is the concentration of the unknown HNO3 solution
Answer:The concentration of the unknown HNO3 solution = 0.01809 M
Explanation:
For the acid-base reaction, HNO3 + NaOH-----> NaN03 + H20
we have that
C1 V1 = C2 V2
Where ,
C1 = concentration of HNO3=?
V1 = volume of HNO3 = 25.00 mL,
V2 = volume of NaOH = 22.62 mL,
C2 = concentration of NaOH = 0.02000 M
Therefore ,
25.00 mL x C1 = 22.62 mL x 0.02000 M
= (22.62 mL / 25.00 mL) x 0.02000 M = 0.01809 M
The concentration of the unknown HNO3 solution = 0.01809 M
Human blood typically contains 1.04 kg/L of platelets. A 1.89 pints of blood would contain what mass (in grams) of platelets
A 1.89 pints of blood would contain 873 grams of platelets.
To calculate the amount of platelets present in 1.89 pints, it is first necessary to transform this unit of volume into liters:
1 pint = 473.2 mL[tex]1.89 \times 473.2 = 894.3 mL[/tex]
1000 L = 1mL
[tex]\frac{894.3}{1000}= 0.84L[/tex]
Now, just calculate the amount of platelets present in 0.84L:
[tex]\frac{1.04\times10^{3}g}{xg}=\frac{1L}{0.84L}[/tex]
x = 873 grams
So, a 1.89 pints of blood would contain 873 grams of platelets.
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The compound sodium hydroxide is a strong electrolyte. Write the transformation that occurs when solid sodium hydroxide dissolves in water. Use the pull-down boxes to specify states such as (aq) or (s).
Answer:
Solid sodium hydroxide dissolves in water to form an aqueous solution of ions.
NaOH(s) ⇌ Na+(aq) + OH–(aq) ΔH1 = ?
Solid sodium hydroxide reacts with aqueous hydrochloric acid to form water and an aqueous solution of sodium chloride.
NaOH(s) + H+(aq) + Cl–(aq) ⇌ H2O(l) + Na+(aq) + Cl–(aq) ΔH2 = ?
Solutions of aqueous sodium hydroxide and hydrochloric acid react to form water and aqueous sodium chloride.
Na+(aq) + OH–(aq) + H+(aq) + Cl–(aq) ⇌ H2O(l) + Na+(aq) + Cl–(aq) ΔH3 = ?
3. What is the mass of an object with a volume of 4 L and a density of 1.25 g/mL?
Answer:
5000g
Explanation:
mass= density × volume
Since the unit of density here is g/mL, we need to convert the volume to mL.
1L= 1000mL
4L= 4 ×1000 = 4000 mL
Mass of object
= 1.25 ×4000
= 5000g
Answer:
5,000 grams
Explanation:
The mass of an object can be found by multiplying the volume by the density.
mass= volume * density
The density is 1.25 g/mL and the volume is 4 L.
First, we must convert the volume to mL. The density is given in grams per milliliter, but the volume is given in liters.
There are 1,000 mL per L. The volume is 4 L. Therefore, we can multiply 4 and 1,000.
4 * 1,000 = 4,000
The volume is 4,000 mL.
Now, find the mass of the object.
mass= volume * density
volume = 4,000
density= 1.25
mass= 4,000 * 1.25 = 5,000
Add the appropriate units for mass, in this case, grams, or g.
mass= 5,000 g
The mass of the object is 5,000 grams.
Sulfur dioxide reacts with oxygen to form sulfur trioxide. What change in hybridization of the sulfur occurs in this reaction ? g
Answer:
PLEASE LOOK INN TO THE FILE YOU WILL GET ANSWER AND ALSO SUMMARY THANKS FOR ASKING QUESTION.
Explanation:
If 2.9g of water is heated from 23.9C to 98.9C, how much heat (in calories) was added to the water?
Answer:
Explanation:
we know that
ΔH=m C ΔT
where ΔH is the change in enthalpy (j)
m is the mass of the given substance which is water in this case
ΔT IS the change in temperature and c is the specific heat constant
we know that given mass=2.9 g
ΔT=T2-T1 =98.9 °C-23.9°C=75°C
specific heat constant for water is 4.18 j/g°C
therefore ΔH=2.9 g*4.18 j/g°C*75°C
ΔH=909.15 j
Using the data: C2H4(g), = +51.9 kJ mol-1, S° = 219.8 J mol-1 K-1 CO2(g), = ‑394 kJ mol-1, S° = 213.6 J mol-1 K-1 H2O(l), = ‑286.0 kJ mol-1, S° = 69.96 J mol-1 K-1 O2(g), = 0.00 kJ mol-1, S° = 205 J mol-1 K-1 calculate the maximum amount of work that can be obtained, at 25.0 °C, from the process: C2H4(g) + 3 O2(g) → 2 CO2(g) + 2 H2O(l)
Answer:
The correct answer is 1332 KJ.
Explanation:
Based on the given information,
ΔH°f of C2H4 is 51.9 KJ/mol, ΔH°O2 is 0.0 KJ/mol, ΔH°f of CO2 is -394 KJ/mol, and ΔH°f of H2O is -286 KJ/mol.
Now the balanced equation is:
C2H4 (g) + 3O2 (g) ⇔ 2CO2 (g) + 2H2O (l)
ΔH°rxn = 2 × ΔH°f CO2 + 2 × ΔH°fH2O - 1 × ΔH°fC2H4 - 3×ΔH°fO2
ΔH°rxn = 2 (-394) + 2(-286) - 1(51.9) - 3(0)
ΔH°rxn = -1411.9 KJ
Now, the given ΔS°f of C2H4 is 219.8 J/mol.K, ΔS°f of O2 is 205 J/mol.K, ΔS°f of CO2 is 213.6 J/mol.K, and ΔS°f of H2O is 69.96 J/mol.K.
Now based on the balanced chemical reaction,
ΔS°rxn = 2 × ΔS°fCO2 + 2 ΔS°fH2O - 1 × ΔS°f C2H4 - 3 ΔS°fO2
ΔS°rxn = 2 (213.6) + 2(69.96) - 1(219.8) -3(205)
ΔS°rxn = -267.68 J/K or -0.26768 KJ/K
T = 25 °C or 298 K
Now putting the values of ΔH, ΔS and T in the equation ΔG = ΔH-TΔS, we get
ΔG = -1411.9 - 298.0 × (-0.2677)
ΔG = -1332 KJ.
Thus, the maximum work, which can obtained is 1332 kJ.
