Answer:
a. In the nickel-aluminum galvanic cell, the cathode is nickel electrode.
b. Therefore electrons flow from the aluminium electrode to the nickel electrode.
c. The aluminium electrode loses mass, while the nickel electrode electrode gains mass.
Explanation:
Voltaic or galvanic cells are electrochemical cells in which spontaneous oxidation-reduction reactions. The two halves of the redox reaction are separate and electron transfer is required to occur through an external circuit for the redox reaction to take place. That is, one of the metals in one of the half cells is oxidized while the metal of the other half cell is reduced, producing an exchange of electrons through an external circuit. This makes it possible to take advantage of the electric current.
Given:
E ⁰N i ⁺² = − 0.23 V is the standard reduction potential for the nickel ion
E ⁰ A l ⁺³ = − 1.66 V is the standard reduction potential for the aluminum ion
The most negative potentials correspond to more reducing substances. In this case, the aluminum ion is the reducing agent, where oxidation takes place. In the anodic half cell oxidations occur, while in the cathode half cell reductions occur. So the aluminum cell acts as the anode while the nickel cell acts as the cathode.
So a. In the nickel-aluminum galvanic cell, the cathode is nickel electrode.
The metal that is oxidized gives electrons to the metal that is reduced through the outer conductor. Then the electrons flow spontaneously from the anode to the cathode.
Then b. Therefore electrons flow from the aluminium electrode to the nickel electrode.
Ni⁺², being the cathode, accepts electrons, becoming Ni (s) and depositing on the Ni electrodes.
So, c. The aluminium electrode loses mass, while the nickel electrode electrode gains mass.
An unknown gas diffuses 5 times slower than that of H2.The moleculer mass of unknown gas is??
Answer:
50.
Explanation:
We can write Graham's Law of Diffusion as:
(Rate 1)^2 = Molecular Mass 2
-------------- -------------------------
(Rate 2)^2 Molecular Mass 1
So using the Given Information:
1^2 / (1/5)^2 = Molecular Mass of unknown gas / 2, so:
25 = M/2
M = 50.
Account for the change when NO2Cl is added using the reaction quotient Qc. Match the words in the left column to the appropriate blanks in the sentences on the right.
1. decreases
2. loss
3. Increases
4. greater
A. Disturbing the equilibrium by adding NO2Cl______Qc to a value_____than Kc.
B. To reach a new state of equilibrium, Qc therefore______which means that the denominator of the expression for Qc______.
C. To accomplish this, the concentration of reagents______, and the concentration of products_______.
Answer:
A. Disturbing the equilibrium by adding NO2Cl decreases Qc to a value less than Kc.
B. To reach a new state of equilibrium, Qc therefore increases which means that the denominator of the expression for Qc decreases.
C. To accomplish this, the concentration of reagents decreases, and the concentration of products increases.
Explanation:
Hello,
In this case, for the equilibrium reaction:
[tex]NO_2Cl(g)+NO(g)\rightleftharpoons NOCl(g)+NO_2(g)[/tex]
Whose equilibrium expression is:
[tex]Kc=\frac{[NO_2][NOCl]}{[NO_2Cl][NO]}[/tex]
The proper matching is:
A. Disturbing the equilibrium by adding NO2Cl decreases Qc to a value less than Kc, since the denominator becomes greater, therefore, Qc decreases.
B. To reach a new state of equilibrium, Qc therefore increases which means that the denominator of the expression for Qc decreases, since the lower the denominator, the higher Qc as it has the concentration of reactants.
C. To accomplish this, the concentration of reagents decreases, and the concentration of products increases, since the reactants must be consumed in order to reestablish equilibrium by shifting the reaction towards the products.
Best regards.
What is the mole fraction of urea, CO(NH2)2, in a solution prepared by dissolving 4.0 g of urea in 32.0 g of methanol, CH3OH
Answer:
0.0630
Explanation:
The molar mass of urea = 60 g/mol
we all know that:
[tex]\mathtt{number \ of \ moles = \dfrac{mass }{molar \ mass}}[/tex]
Then; the number of moles of urea
= [tex]\mathtt{\dfrac{4.0 \ g}{60 \ g/mol}}[/tex]
= 0.0667 mol
Similarly; the number of moles of methanol
= [tex]\mathtt{\dfrac{32 \ g}{32.04 \ g/mol}}[/tex]
= 0.9988 mol
The total number of moles = (0.0667 + 0.9988) mol
= 1.0655 mol
Finally,the mole fraction of urea [tex]\mathtt{(X_{urea})}[/tex] = [tex]\mathtt{\dfrac{ n_{urea}}{(n_{urea}+n_{methanol})}}[/tex]
[tex]\mathtt{(X_{urea})}[/tex] = [tex]\mathtt{\dfrac{0.0667 \ mole}{1.0655 \ mole}}[/tex]
= 0.0630
g A chemist combines 59.9 mL of 0.282 M potassium bromide with 15.4 mL of 0.512 M silver nitrate. (a) How many grams of silver bromide will precipitate
Answer:
[tex]m_{AgBr}=1.48gAgBr[/tex]
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
[tex]KBr(aq)+AgNO_3(aq)\rightarrow AgBr(s)+KNO_3(aq)[/tex]
Thus, since the potassium bromide and silver nitrate are in a 1:1 mole ratio, the first step is to identify the limiting reactant, by considering the reacting volumes of reactants in order to compute the available moles of potassium bromide and the moles of potassium bromide consumed by the 15.4 mL of 0.512-M solution of silver nitrate:
[tex]n_{KBr}=0.0599L*0.282\frac{molKBr}{L} =0.0169molKBr\\\\n_{KBr}^{consumed}=0.0154L*0.512\frac{molAgNO_3}{L} *\frac{1molKBr}{1molAgNO_3}=0.00788molKBr[/tex]
In such a way, since less moles are consumed than available, we infer that silver nitrate is the limiting reactant, for which the resulting grams of silver bromide (molar mass 187.8 g/mol) result:
[tex]m_{AgBr}=0.00788molAgNO_3*\frac{1molAgBr}{1molAgNO_3} *\frac{187.8gAgBr}{1molAgBr} \\\\m_{AgBr}=1.48gAgBr[/tex]
Best regards.
Calculate the pH of a solution that is 0.210 M in nitrous acid (HNO2) and 0.290 M in potassium nitrite (KNO2). The acid dissociation constant of nitrous acid is 4.50 × 10-4.
Answer:
pH = 3.49
Explanation:
We have a buffer system formed by a weak acid (HNO₂) and its conjugate base (NO₂⁻ coming from KNO₂). We can calculate the pH of a buffer ssytem using the Henderson-Hasselbach equation.
pH = pKa + log [base] / [acid]
pH = -log Ka + log [NO₂⁻] / [HNO₂]
pH = -log 4.50 × 10⁻⁴ + log 0.290 M / 0.210 M
pH = 3.49
The pH of the solution containing 0.210 M nitrous acid (HNO₂) and 0.290 M potassium nitrite (KNO₂) is 3.49
We'll begin by calculating the the pKa of acid. This can be obtained as follow:
Acid dissociation constant (Ka) = 4.50×10¯⁴
pKa =?pKa = –Log Ka
pKa = –Log 4.50×10¯⁴
pKa = 3.35Finally, we shall determine the pH of the solution.pKa = 3.35
Concentration of HNO₂, [HNO₂] = 0.210 M
Concentration of KNO₂, [KNO₂] = 0.290 M
pH =?The pH of the solution can obtain by using the Henderson-Hasselbach equation as illustrated below:
pH = pKa + log [base] / [acid]pH = pKa+ log [NO₂⁻] / [HNO₂]
pH = 3.35 + log (0.290 / 0.210)
pH = 3.49Thus, the pH of the solution is 3.49
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How many moles of barium sulfate are produced from 0.100 mole of barium chloride?
Answer:
0.100 moles of barium sulfate are produced from 0.100 moles of barium chloride.
Explanation:
Barium chloride and sodium sulfate react according to the following balanced reaction:
BaCl₂ + Na₂SO₄ → BaSO₄ + 2 NaCl
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reagent and products participate in the reaction:
BaCl₂: 1 moleNa₂SO₄: 1 moleBaSO₄: 1 moleNaCl : 2 molesThen you can apply the following rule of three: if 1 mole of BaCl₂ produces 1 mole of BaSO₄, 0.100 mole of BaCl₂ how many moles of BaSO₄ does it produce?
[tex]amount of moles of BaSO_{4} =\frac{0.100 mole of BaCl_{2}* 1 mole of BaSO_{4} }{1 mole of BaCl_{2}}[/tex]
amount of moles of BaSO₄= 0.100
0.100 moles of barium sulfate are produced from 0.100 moles of barium chloride.
For the following reaction, predict whether the equilibrium lies predominantly to the left or to the right. Explain.
NH4+(aq) + Br-(aq) - NH3(aq) + HBr(aq)
If Ka
(NH4+) = 5.6 x 10-10
Answer:
Lies predominantly to the left.
Explanation:
In the reaction:
NH4⁺(aq) + Br-(aq) ⇄ NH3(aq) + HBr(aq)
Conjugate acid + Ion ⇄ weak base + strong acid
HBr is a strong acid whereas NH3/NH4⁺ are the weak base and its conjugate base. A strong acid as HBr dissociates completely in solution as H⁺ and Br⁻. That means in solution you will never have HBr without dissociation doing the reaction:
lies predominantly to the left.A sample is found to contain 1.29×10-11 g of salt. Express this quantity in picograms
Answer:12.9e-12g or in short 12.9pg
Explanation:as p=1e-12
Chemistry
What is a chemical reaction
Answer:
A process that involves rearrangement
Explanation:
A chemical reaction is the process that involves rearrangement of the molecular or ironic structure of a substance, as a distinct from a change in physical form or a nuclear reaction.
Answer:
Explanation:
Chemistry
The chemical reaction H2(g) + ½ O2(g) → H2O(l) describes the formation of water from its elements.
The reaction between iron and sulfur to form iron(II) sulfide is another chemical reaction, represented by the chemical equation:
8 Fe + S8 → 8 FeS
What type of chemist exclusively studies most carbon compounds?
-biochemist
-physical chemist
-organic chemist
-inorganic chemist
Answer:
Organic chemist? I do not know.
Explanation:
Thanks you.
The type of chemist exclusively studies most carbon compounds are organic chemist. Therefore, option C is correct.
What is an organic chemist ?The structure, characteristics, and reactivity of compounds containing carbon are studied by organic chemists. Additionally, they create novel organic materials with distinct features and uses.
Analytical capabilities, communication skills, and numeracy skills are three of the most important soft skills for an organic chemist.
Organic chemists often work in research and development in labs at universities, pharmaceutical, industrial, and biotechnology businesses, as well as government agencies, according to the American Chemical Society.
According to one assessment, organic chemistry is the hardest college course. According to certain statistics, almost one out of every two students in organic chemistry leave the course. The hopes of a medical career come tumbling down for those who fit this description. Organic chemistry is undoubtedly challenging.
Thus, option C is correct.
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How many grams is 5.8 moles of hydrochloric acid (HCI)?
Answer to the nearest 0.01 g.
Answer:
211.47 grams
Explanation:
We need to set up a dimensional analysis to solve this problem by converting from moles to grams.
First, find the molar mass of HCl. Since the molar mass of H (hydrogen) is 1.01 g/mol and the molar mass of Cl (chlorine) is 35.45 g/mol, then the molar mass of HCl is:
1.01 + 35.45 = 36.46 g/mol
We have 5.8 moles of HCl, so multiply by its molar mass:
(5.8 mol) * (36.46 g/mol) = 211.468 ≈ 211.47 g
The answer is thus 211.47 grams.
~ an aesthetics over
Answer:
[tex]\large\boxed{211.47}\\[/tex] grams
Explanation:
First, you need to gather the atomic masses of the elements involved in the compound - hydrogen and chlorine. Referencing a modern periodic table will give you this information.
Hydrogen has an atomic weight of 1.00784 and Chlorine has an atomic mass of 35.453.Add those two values together - 1.00784 + 35.453 = 36.46084Multiply this value by 5.8 (one mole is equivalent to the atomic mass of the compound) - 5.8 x 36.46084 = 211.472872Round to the nearest 0.01 gram - 211.47[tex]\large\boxed{211.47}[/tex] is the final answer.
Any process with a negative change in enthalpy and a positive change in entropy will be:_______.
a. spontaneous
b. nonspontaneous
c. spontaneous at high temperatures
d. spontanteous at low temperatures
Answer:
a. spontaneous
Explanation:
Hello,
In this case, since the Gibbs free energy is a metric that allows us to know whether a chemical reaction is spontaneous (Gibbs free energy less than 0) or nonspontaneous (Gibbs free energy greater than 0) we can mathematically define it as:
[tex]\Delta G=\Delta H-T\Delta S[/tex]
Thus, if the enthalpy is negative and the entropy is negative, the subtraction become always negative, for which the Gibbs free energy is negative as well, therefore, based on the aforementioned, any process with a negative change in enthalpy and a positive change in entropy will be: a. spontaneous.
Best regards.
Cite examples of how copper deposits occur. Choose one or more: A. as an agglomeration metal B. as a native metal C. in carbonate ore minerals D. in sulfide ore minerals
Answer:
A. as an agglomeration metal
B. as a native metal
D. in sulfide ore minerals
Explanation:
Copper is a metal with symbol Cu and atomic number 29. It has a pinkish-orange color and is malleable, ductile and has a high thermal and electrical conductivity. This is why it is often used in electrical appliances.
Copper exists as an agglomeration metal, as a native metal or in sulfide ore minerals such as Cu2S.
The examples of copper deposits are agglomeration metal, as a native metal or in sulfide ore minerals. Option A, B, and D are correct.
Copper is a metal with high thermal and electrical conductivity. hence, it is often used in electrical appliances.
Copper found as an agglomeration metal, as a native metal or in sulfide ore minerals such as [tex]\bold { Cu_2S.}[/tex]
Therefore, the examples of copper deposits are agglomeration metal, as a native metal or in sulfide ore minerals. Option A, B, and D are correct.
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PLEASE HELP!! 40 POINTS
Answer:
1) 6.524779402×10^(-17)
2)521.1g
3)113
Explanation:
Answer: 1) 6.524779402×10^(-17)
2)521.1g
Explanation:
Determine which set of properties correctly describes copper (Cu)?
A. Giant structure, conducts electricity, high melting point, soluble in water, malleable
B. Malleable, brittle, soluble in oil or gasoline, high melting point, simple structure
C. Ionic lattice, conducts electricity, soluble in oil or gasoline, low melting point, ductile
D. Malleable, conducts electricity, high melting point, giant structure, metallic lattice
Answer:
D. Malleable, conducts electricity, high melting point, giant structure, metallic lattice
Explanation:
Copper is a metal with an atomic number of 29. This metal is soft and reddish in color which explains why it is very malleable(beaten to form various shapes without breaking).
All metals are good conductors of electricity including copper which is also a metal. Metals generally are insoluble in water. Copper also has a high melting point which is a characteristic of metals due to their giant structure and metallic lattice which makes it difficult to be broken down.
Arrange the following substances in the order of increasing entropy at 25°C. HF(g), NaF(s), SiF 4(g), SiH 4(g), Al(s) lowest → highest
Answer:
Al(s)<NaF(s)<HF(g)<SiH4(g)<SiF4(g)
Explanation:
Hello,
In this case, we can arrange the increasing order of entropy at 25 \°C by taking into account, at first, that since solids are more molecularly organized than gases, the first we have solid sodium fluoride and solid aluminium, but in this case, as the higher the molar mass, the higher the entropy, the molar mass of aluminium is 27 g/mol and 42 g/mol for sodium fluoride, therefore, we first have:
Al(s)<NaF(s)
Afterwards, since the molar mass of hydrogen fluoride (HF), silicon fluoride (SiF4) and silane (SiH4) are 20, 104 and 32 g/mol respctively, since silicon fluoride has the greater molar mass, it also has the higher entropy. In such a way, the overall order turns out:
Al(s)<NaF(s)<HF(g)<SiH4(g)<SiF4(g)
Best regards.
Heterocyclic aromatic compounds undergo electrophilic aromatic substitution in a similar fashion to that undergone by benzene with the formation of a resonance-stabilized intermediate. Draw all of the resonance contributors expected when the above compound undergoes bromination
Answer:
See explanation
Explanation:
When we talk about electrophilic substitution, we are talking about a substitution reaction in which the attacking agent is an electrophile. The electrophile attacks an electron rich area of a compound during the reaction.
The five membered furan ring is aromatic just as benzene. This aromatic structure is maintained during electrophilic substitution reaction. The attack of the electrophile generates a resonance stabilized intermediate whose canonical structures have been shown in the image attached.
Using the following diagram, determine which of the statements below is true: The activation energy for the forward reaction is −60 J. The overall energy change for the forward reaction is −20 J. The activation energy for the reverse reaction is −80 J. The overall energy change for the reverse reaction is −40 J.
Answer:its saturated or unsaturaded
Explanation:
What is the density of a 10 kg mass that occupies 5 liters?
( pls need help)
Answer: d=2000 g/L
Explanation:
Density is mass/volume. The units are g/L. Since we are given mass and volume, we can divide them to find density. First, we need to convert kg to g.
[tex]10kg*\frac{1000g}{1kg} =10000 g[/tex]
Now that we have grams, we can divide to get density.
[tex]d=\frac{10000g}{5 L}[/tex]
d=2000g/L
A closed-end manometer was attached to a vessel containing argon. The difference in the mercury levels in the two arms of the manometer was 9.60 cm. Atmospheric pressure was 783 mm Hg. The pressure of the argon in the container was ________ mm Hg.
Answer:
96 mmHg
[tex]h=96mmHg[/tex]
Explanation:
From this question,manometer end is closedw, So we can deduced that the height of the column will not be affected by the atmospheric pressure .
The difference of height of the mercury level is given as,
h=9.60cm
h=9.60(10mm/1cm)
[tex]h=96mm[/tex]
But it is obvious that in this closed end manometer.the pressure of the gas is equal to the height
P(gas)=h
P(gas)=96mmHg
This pressure is as a result of the presence of gas.
Therefore, the pressure of the argon gas in the container is 96mmHg.
The pressure of the argon in the container was 96mmHg.
We were told that the manometer has closed ends which means that the
height will not be affected by atmospheric pressure.
The height which is the difference in mercury level is
h=9.60cm
We can convert it to millimeter by multiplying it by 10
h=9.60 × 10 = 96mm
The pressure of the closed end manometer will be equal to the height
P(gas)=h
P(gas)=96mmHg
The pressure of the argon gas in the container is 96mmHg.
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Determine the oxidation state for each of the elements below. The oxidation state of ... silver ... in ... silver oxide Ag2O ... is ... ___ . The oxidation state of sulfur in sulfur dioxide SO2 is ___ . The oxidation state of iron in iron(
Answer:
The oxidation state of silver in [tex]\rm Ag_2O[/tex] is [tex]+1[/tex].
The oxidation state of sulfur in [tex]\rm SO_2[/tex] is [tex]+4[/tex].
Explanation:
The oxidation states of atoms in a compound should add up to zero.
Ag₂OThere are two silver [tex]\rm Ag[/tex] atoms and one oxygen [tex]\rm O[/tex] atom in one formula unit of [tex]\rm Ag_2O[/tex]. Therefore:
[tex]\begin{aligned}&\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times \text{Oxidation state of $\rm O$} = 0\end{aligned}[/tex].
The oxidation state of oxygen in most compounds (with the exception of peroxides and fluorides) is [tex]-2[/tex]. Silver oxide [tex]\rm Ag_2O[/tex] isn't an exception. Therefore:
[tex]\begin{aligned}&\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times \text{Oxidation state of $\rm O$} = 0\\ &\rm 2 \times \text{Oxidation state of $\rm Ag$}+ \rm 1 \times (-2) = 0\end{aligned}[/tex].
Solve this equation for the (average) oxidation state of [tex]\rm Ag[/tex]:
[tex]\text{Oxidation state of $\rm Ag$} = 1[/tex].
SO₂Similarly, because there are one sulfur [tex]\rm S[/tex] atom and two oxygen [tex]\rm O[/tex] atoms in each [tex]\rm SO_2[/tex] molecules:
[tex]\begin{aligned}&\rm 1\times \text{Oxidation state of $\rm S$}+ \rm 2 \times \text{Oxidation state of $\rm O$} = 0\end{aligned}[/tex].
The oxidation state of [tex]\rm O[/tex] in [tex]\rm SO_2[/tex] is also [tex]-2[/tex], not an exception, either.
Therefore:
[tex]\begin{aligned}&\rm 1 \times \text{Oxidation state of $\rm S$}+ \rm 2 \times \text{Oxidation state of $\rm O$} = 0\\ &\rm 1 \times \text{Oxidation state of $\rm S$}+ \rm 2 \times (-2) = 0\end{aligned}[/tex].
Solve this equation for the oxidation state of [tex]\rm S[/tex] here:
[tex]\text{Oxidation state of $\rm S$} = 4[/tex].
Which molecule is NOT hypervalent?
Select the correct answer below:
SF
PBr3
PBr5
XeFo
Answer:
PBr3 is NOT hypervalent
Explanation:
The molecule that is not hypervalent is PBr3
A molecule can be defined as the smallest part of a substance that can exist independently.
It is formed by the chemical combination of two or more atoms.
A molecule is said to be hypervalent when more than four pairs of electrons are around the central atom.
A molecule is said to be hypovalent when less than four pairs of electrons are around the central atom.
From the question, the molecule that is hypovalent is PBr3
This is because, phosphorus can make hypervalent compounds, but in this specific example it is sharing three bonds and has one lone pair, so it has simply a full octet.
Therefore, the molecule that is not hypervalent is PBr3.
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A hypothetical metal crystallizes with the face-centered cubic unit cell. The radius of the metal atom is 198 picometers and its molar mass is 195.08 g/mol. Calculate the density of the metal in g/cm3.
Answer:
7.38 g/cm³ is the density of the metal
Explanation:
In a Face-centered cubic unit cell you have 4 atoms. Also, the edge length is √8×r (r is radius of the atom).
To solve this problem, we need first to calculate the volume of the unit cell and then, with molar mass calculate the mass of 4 atoms. As density is the ratio between mass and volume we can obtain this value.
Volume of the unit cellVolume = a³
a = √8×r
(r = 198x10⁻¹²m)
a = 5.6x10⁻¹⁰ m
Volume = 1.756x10⁻²⁸ m³
1m = 100cm → 1m³ = (100cm)³:
1.756x10⁻²⁸ m³× ((100cm)³ / 1m³) =
1.756x10⁻²² cm³ → Volume of the unit cell in cm³Mass of the unit cell:There are 4 atoms of gold:
4 atoms × (1mol / 6.022x10²³ atoms) = 6.64x10⁻²⁴ moles of gold
As 1 mole weighs 195.08g:
6.64x10⁻²⁴ moles of gold × (195.08g / mol) =
1.296x10⁻²¹g is the mass of the unit cellDensity of the metal:1.296x10⁻²¹g / 1.756x10⁻²² cm³ =
7.38 g/cm³ is the density of the metalThe density of the metal is 7.40 g/cm³
In cubic crystal system, face-centered cubic FFC is the name given to sort of atom arrangement observed in which structure is made up of atoms organized in a cube with a portion of an atom in each corner and six extra atoms in the center of each cube face.
It is expressed by using the formula:
[tex]\mathbf{\rho = \dfrac{Z \times M}{N_A\times a^}}[/tex]
where;
[tex]\rho[/tex] = density of the metalZ = atoms coordination no = 4 (for FCC)Molar mass (M) = 195.8 g/molAvogadro's constant (NA) = 6.022 × 10²³ /mola = edge lengthFor face-centered cubic FFC;
The edge length [tex]\mathbf{a =2 \sqrt{2}\times r }[/tex]
[tex]\mathbf{a =2 \sqrt{2}\times 198 \ pm }[/tex]
[tex]\mathbf{a =560.0285 \ pm }[/tex]
a = 5.60 × 10⁻⁸ cm
Replacing it into the previous equation, we have:
[tex]\mathbf{\rho = \dfrac{4 \times 195.8}{6.022 \times 10^{23} \times( 5.60 \times 10^{-8} )^3}}[/tex]
[tex]\mathbf{\rho = 7.40\ g/cm^3 }[/tex]
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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.
A 10.00-mL aliquot of vinegar requires 16.95 mL of the 0.4874 M standardized NaOH solution to reach the end point of the titration. Demonstrate how to calculate the molarity of the vinegar solution (HC2H3O2). Show complete work below. Answer: 0.8261 M.
Answer:
0.8261 M.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
HC2H3O2 + NaOH —> NaC2H3O2 + H2O
From the balanced equation above, we obtained the following:
Mole ratio of the acid, HC2H3O2 (nA) = 1
Mole ratio of the base, NaOH (nB) = 1
Data obtained from the question include the following:
Volume of acid, HC2H3O2 (Va) = 10 mL
Molarity of acid, HC2H3O2 (Ma) =..?
Volume of base, NaOH (Vb) = 16.95 mL Molarity of base, NaOH (Mb) = 0.4874 M
Finally, we shall determine the molarity of the acid solution, as follow:
MaVa/MbVb = nA/nB
Ma x 10 / 0.4874 x 16.95 = 1
Cross multiply
Ma x 10 = 0.4874 x 16.95
Divide both side by 10
Ma = (0.4874 x 16.95) /10
Ma = 0.8261 M.
Therefore, the molarity of the vinegar solution (HC2H3O2) is 0.8261 M.
Complete the unit conversion by entering the correct numbers
A=
B=
C=
Answer: A=1, B=3, C=12
Explanation:
For this problem, you will need to know your unit conversions. There are 3 ft in 1 yard. Knowing this, we can find A, B, C.
For A and B, we know that we want to cancel out ft so the answer can be in yards. To do so, we need to put B=3 and A=1.
Now that we know the unit conversion, we can directly solve.
36 ft×(1 yd/3 ft)=12 yd
Our final answer is A=1, B=3, C=12.
Answer:
A=1,000, B=1, C=5,400
Explanation:
the question was 5.4L x AmL / BmL = CmL
If we represent the equilibrium as:...N2O4(g) 2 NO2(g) We can conclude that: 1. This reaction is: A. Exothermic B. Endothermic C. Neutral D. More information is needed to answer this question. 2. When the temperature is increased the equilibrium constant, K: A. Increases B. Decreases C. Remains the same D. More information is needed to answer this question. 3. When the temperature is increased the equilibrium concentration of NO2: A. Increases B. Decreases C. Remains the same D. More information is needed to answer this question.
Answer:
1. This reaction is: B. Endothermic.
2. When the temperature is increased the equilibrium constant, K: A. Increases.
3. When the temperature is increased the equilibrium concentration of NO2: A. Increases.
Explanation:
Hello,
In this case, considering the images, we can state that the red color at high temperature is due to the presence of nitrogen dioxide (product) and the lower coloring is due to the presence of dinitrogen tetroxide (reactant) at low temperature.
With the aforementioned, we can conclude that the chemical reaction:
[tex]N_2O_4(g) \rightleftharpoons 2 NO_2(g)[/tex]
Is endothermic since high temperatures favor the formation of the product and the low temperatures favor the consumption of the the reactant. thereby:
1. This reaction is: B. Endothermic.
2. When the temperature is increased the equilibrium constant, K: A. Increases. In this particular case, since the dinitrogen tetroxide has 1 molecule and nitrogen dioxide two molecules in the chemical reaction, the entropy change should be positive, therefore, by increasing the T, the Gibbs free energy of reaction becomes more negative:
[tex]G=H-TS[/tex]
As Gibbs free energy becomes more negative, the equilibrium constant becomes bigger given their relationship:
[tex]K=exp(-\frac{\Delta G}{RT} )[/tex]
3. When the temperature is increased the equilibrium concentration of NO2: A. Increases.
Regards.
1. What volume in milliliters of 0.100 M HClO₃ is required to neutralize 40.0 mL of 0.140 M KOH? 2. A 25.0 mL solution of HNO₃ is neutralized with 15.7 mL of 0.250 M Ba(OH)₂. What is the concentration of the original HNO₃ solution?
Answer:
The correct answer is 1) 56 ml and 2) 0.314 M
Explanation:
1. The reaction taking place in the given case is,
HClO₃ + KOH ⇒ KClO₃ + H2O, the molarity of HClO₃ given is 0.100 M, the molarity of KOH given is 0.140 M and the volume of KOH given is 40 ml, there is a need to find the volume of HClO₃.
Therefore, the mole of HClO₃ = mole of KOH
= MHClO₃ × VHClO₃ = MKOH × VKOH
= 0.100 M × VHClO₃ = 0.140 M × 40 ml
VHClO₃ = 0.140 M × 40 ml/0.100 M
VHClO₃ = 56 ml.
2. The reaction taking place is,
2HNO₃ + Ba(OH)₂ ⇒ Ba(NO₃)₂ + 2H₂O
The volume of HNO₃ given is 25 ml, the molarity of Ba(OH)2 is 0.250 M, the volume of Ba(OH)2 is 15.7 ml, the n or the number of moles of HNO₃ is 2, and the n of Ba(OH)2 is 1, the concentration or M of HNO₃ is,
M₁V₁/n₁ = M₂V₂/n₂
M₁ × 25/ 2 = 0.25 × 15.7/1
M₁ or molarity of HNO₃ = 0.314 M
1. The volume of HClO₃ required to neutralize the KOH is 56.0 mL
2. The concentration of the original HNO₃ solution is 0.314 M
1.
First, we will write a balanced chemical equation for the reaction
The balanced chemical equation for the reaction is
HClO₃ + KOH → KClO₃ + H₂O
This means,
1 mole of HClO₃ is required to neutralize 1 mole of KOH
From the titration formula
[tex]\frac{C_{A}V_{A} }{C_{B}V_{B}} = \frac{n_{A}}{n_{B}}[/tex]
Where
[tex]C_{A}[/tex] is the concentration of acid
[tex]C_{B}[/tex] is the concentration of base
[tex]V_{A}[/tex] is the volume of acid
[tex]V_{B}[/tex] is the volume of base
[tex]n_{A}[/tex] is the mole ratio of acid
[tex]n_{B}[/tex] is the mole ratio of base
From the given information,
[tex]C_{A} = 0.100 \ M[/tex]
[tex]C_{B} = 0.140 \ M[/tex]
[tex]V_{B} = 40.0 \ mL[/tex]
From the balanced chemical equation
[tex]n_{A} = 1[/tex]
[tex]n_{B} =1[/tex]
Putting the values into the formula, we get
[tex]\frac{0.100 \times V_{A} }{0.140 \times 40.0} = \frac{1}{1}[/tex]
∴ [tex]0.100 \times V_{A} = 0.140 \times 40.0[/tex]
[tex]V_{A}=\frac{0.140\times 40.0}{0.100}[/tex]
[tex]V_{A}=\frac{5.60}{0.100}[/tex]
[tex]V_{A}=56.0 \ mL[/tex]
Hence, the volume of HClO₃ required to neutralize the KOH is 56.0 mL
2.
First, we will write the balanced chemical equation for the reaction
The balanced chemical equation for the reaction is
2HNO₃ + Ba(OH)₂ → Ba(NO₃)₂ + 2H₂O
This means, 2 mole of HNO₃ is required to neutralize 1 mole Ba(OH)₂
From the given information,
[tex]V_{A} = 25.0\ mL[/tex]
[tex]C_{B} = 0.250 \ M[/tex]
[tex]V_{B} = 15.7 \ mL[/tex]
From the balanced chemical equation
[tex]n_{A} = 2[/tex]
[tex]n_{B} =1[/tex]
Also, Using the titration formula
[tex]\frac{C_{A}V_{A} }{C_{B}V_{B}} = \frac{n_{A}}{n_{B}}[/tex]
We get
[tex]\frac{C_{A} \times 25.0 }{0.250 \times 15.7} = \frac{2}{1}[/tex]
Then,
[tex]C_{A} = \frac{2\times 0.250 \times 15.7} {1 \times 25.0}[/tex]
[tex]C_{A} =\frac{7.85}{25.0}[/tex]
[tex]C_{A} =0.314 \ M[/tex]
Hence, the concentration of the original HNO₃ solution is 0.314 M
Learn more here: https://brainly.com/question/16870772
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 valuePlease tell the answer
Answer:
see the photo
Explanation:
it was the answer
