Answer:
HBr, CH3C(OH)2 and CH3COOH, Br-
Explanation:
The conjugate acid-base pairs acid reacts with base to form a conjugate acid and conjugate base.
Conjugate acid is formed when a bases receives a proton (H+) and a conjugate base is formed when an acid losses a proton (H+).
From the given equation:
HBr, CH3C(OH)2 and CH3COOH, Br- are conjugate acid-base pair, where HBr is an acid and CH3C(OH)2 is a conjugate acid while CH3COOH and Br- is the conjugate base.
Atoms are indivisible spheres. 1.plum pudding model 2.Dalton model 3.Bohr model
Answer: 2. Dalton Model
Explanation:
John Dalton proposed that atoms are indivisible spheres. Although his model of an atom was not entirely new to the scientific world since the ancient Greeks has made a similar statement in the past ( all matter are made up of small indivisible particle called atom).
As of when Dalton proposed his model of an atom, electrons and nucleus where yet to be discovered.
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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g What is the molarity of hydrochloric acid if 40.95 mL of HCl is required to neutralize 0.550 g of sodium oxalate, Na2C2O4
Answer:
0.0002 M
Explanation:
The molarity of the HCl required would be 0.0002 M.
First, let us consider the balanced equation of the reaction:
[tex]Na_2C_2O_4 + 2HCl = 2NaCl + H_2 + 2CO_2[/tex]
Stoichiometrically, 1 mole of [tex]Na_2C_2O_4[/tex] reacts with 2 moles of [tex]HCl[/tex] for a complete neutralization reaction.
Recall that: mole = [tex]\frac{mass}{molar mass}[/tex]
Mole of 0.550 g sodium oxalate = 0.550/134 = 0.0041 mole
If 1 mole [tex]Na_2C_2O_4[/tex] requires 2 moles HCl, then 0.0041 mole will require:
0.0041 x 2 = 0.0082 mole HCl
Volume of the HCl = 40.95 L
Molarity = mole/volume
Hence, molarity of the HCl = 0.0082/40.95 = 0.0002 M
Why can gasses change volume?
A. The forces holding the gas particles together are
stronger than gravity.
B. The gas particles have no mass, so they can change volume.
C. Gravity has no effect on gas particles, so they can float away.
O D. There are no forces holding the gas particles together.
Answer:
There are no forces holding the gas particles together.
Explanation:
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
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
When scientists are ready to publish the result of their experiments why is it important for them to include a description of the procedure they used
Answer: So other scientist can replicate the experiment and see if they get the same results in other words, test reliability.
Explanation:
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.
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.
2. In what part of an atom can protons be found?
a. Inside the electrons
b. Inside the neutrons
C. Inside the atomic nucleus
d. Inside the electron shells
Answer:
c
Explanation:
it's found inside the atomic nucleus
2) What is the concentration (M) of CH3OH in a solution prepared by dissolving 11.7 g of CH3OH in sufficient water to give exactly 230 mL of solution?
A small amount of solid calcium hydroxide is shaken vigorously in a test tube almost full of water until no further change occurs and most of the solid settles out. The resulting solution is:______.
Answer:
Lime water, [tex]Ca(OH)_{2}_({aq} )[/tex] is formed.
Explanation:
Lime-water is a clear and colourless dilute solution of aqueous calcium hydroxide salt.
Small amounts of calcium hydroxide salt, [tex]Ca(OH)_{2}_(s)[/tex] is sparsely soluble at room temperature when dispersed vigorously. if in excess, a white suspension called 'milk of lime'is formed.
I hope this explanation is helpful.
An aqueous solution of potassium bromide, KBr, contains 4.34 grams of potassium bromide and 17.4 grams of water. The percentage by mass of potassium bromide in the solution is 20 %.
Answer:
True
Explanation:
The percentage by mass of a substance in a solution can be calculated by dividing the mass of the substance dissolved in the solution by the total mass of the solution. This can be expressed mathematically as:
Percentage by mass = mass of substance in solution/mass of solution x 100
In this case;
mass of KBr = 4.34 grams
mass of water = 17.4 grams
mass of solution = mass of KBr + mass of water = 4.34 + 17.4 = 21.74
Percentage by mass of KBr = 4.34/21.74 x 100
= 19.96 %
19.96 is approximately 20%.
Hence, the statement is true.
1. Explain what the police siren sounds like to Jane:
2. Explain what the police siren sounds like to John:
3. Explain why the police siren sounds different between Jane and John:
Answer:
1. the siren has a lower pitch to Jane
2. the siren has a higher pitch to John
3. sound different due to moving away from Jane making the sound wave lengths longer and moving toward John making the wave lengths shorter
Explanation:
The Doppler effect expresses that sound is comparative with the spectator or observer. This is demonstrated valid by the model given with Jane and John. To one individual it could sound low and to someone else it could sound high, in light of where they are tuning in from. To John, the police alarm playing is a higher pitch. Be that as it may, to Jane this equivalent alarm is a totally extraordinary pitch and is heard lower than in comparison to the john.
This is a prime case of the Doppler Effect. They sound distinctive on the grounds that the sound is moving far from Jane making the sound frequencies longer and it is advancing toward John making the frequencies shorter. This impacts how the sound is heard by the human ear.
Explain why, when the guanidino group of arginine is protonated, the double-bonded nitrogen is the nitrogen that accepts the proton. There is a scheme of a reversible reaction, where one equivalent of the reactant reacts with two equivalents of H plus. The reactant is H2NCNHCH2CH2CH2CHCO minus, with an NH group, with a lone pair at the N atom, double-bonded to the first (from left to right) carbon, an NH2 group attached to the fifth carbon, an O atom double-bonded to the sixth carbon and a lone pair of electrons at the first and the second N atoms of the chain. The product has the same structure as the reactant, except that not an NH group with a lone pair, but an NH2 plus group is double-bonded to the first carbon. In addition, an NH3 plus group is attached to the fifth carbon instead of the NH2 group.
Answer:
Due to the resonance structures
Explanation:
In the question:
"Explain why, when the guanidino group of arginine is protonated, the double-bonded nitrogen is the nitrogen that accepts the proton. There is a scheme of a reversible reaction, where one equivalent of the reactant reacts with two equivalents of H plus"
We have to take into account the structure of the amino acid arginine. In which, we have the amino and the carboxylic groups in the right and the guanidine group in the left.
In this group, we have a central carbon with three nitrogen atoms around and a double bond with the nitrogen on the top. This nitrogen on the top will accept the proton because the structure produced will have a positive charge on this nitrogen. Then, the double bond with the carbon can be delocalized into the nitrogen producing a positive charge in the carbon.
In this structure (the carbocation), we can have several resonance structures. In the blue option, we can produce a double bond with the nitrogen on the right. In the purple option, we can produce a double bond with the nitrogen on the left.
In conclusion, if the nitrogen in the top on the guanidine group accepts an hydrogen atom and we will have several resonance structures that can stabilize the molecule. Due to this, the nitrogen in the top its the best option to accept hydrogens.
See figure 1
I hope it helps!
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
Using only sodium carbonate, Na2CO3, sodium bicarbonate, NaHCO3, and distilled water determine how you could prepare 50.0 mL of a 0.20 M solution that is buffered to a pH of 10.3. The total molarity of the ions should be 0.20 M. The Ka of the hydrogen carbonate ion, HCO3 - , is 4.7 x 10-11 .
Answer:
Weight 0.4326 g of sodium bicarbonate and 0.5141 g of sodium carbonate, dissolve it in distilled water and then bring the solution to a final volume of 50.0 mL using distilled water.
Explanation:
The pH of a buffered solution can be calculated using the Henderson-Hasselbalch equation:
[tex] pH = pKa + log(\frac{[Na_{2}CO_{3}]}{[NaHCO_{3}]}) [/tex]
We have that pH = 10.3 and the Ka is 4.7x10⁻¹¹, so:
[tex] 10.3 = -log(4.7 \cdot 10^{-11}) + log(\frac{[Na_{2}CO_{3}]}{[NaHCO_{3}]}) [/tex]
[tex] \frac{[Na_{2}CO_{3}]}{[NaHCO_{3}]} = 0.94 [/tex] (1)
Also, we know that:
[tex] [Na_{2}CO_{3}] + [NaHCO_{3}] = 0.20 M [/tex] (2)
From equation (2) we have:
[tex] [Na_{2}CO_{3}] = 0.20 - [NaHCO_{3}] [/tex] (3)
By entering (3) into (1):
[tex] \frac{0.20 - [NaHCO_{3}]}{[NaHCO_{3}]} = 0.94 [/tex]
[tex] 0.94*[NaHCO_{3}] + [NaHCO_{3}] = 0.20 [/tex]
[tex] [NaHCO_{3}] = 0.103 M [/tex]
Hence, the [Na_{2}CO_{3}] is:
[tex] [Na_{2}CO_{3}] = 0.20 - [NaHCO_{3}] = 0.20 M - 0.103 M = 0.097 M [/tex]
Now, having the concentrations and knowing the volume of the buffer solution we can find the mass of the sodium carbonate and the sodium bicarbonate, as follows:
[tex]m_{Na_{2}CO_{3}} = C*V*M = 0.097 mol/L*0.050 L*105.99 g/mol = 0.5141 g[/tex]
[tex]m_{NaHCO_{3}} = C*V*M = 0.103 mol/L*0.050 L*84.007 g/mol = 0.4326 g[/tex]
Therefore, to prepare 50.0 mL of a 0.20 M solution that is buffered to a pH of 10.3 we need to weight 0.4326 g of sodium bicarbonate and 0.5141 g of sodium carbonate, dissolve it in distilled water and then bring the solution to a final volume of 50.0 mL using distilled water.
I hope it helps you!
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.
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
What happens to the rate of dissolution as the temperature is increased in a gas solution?
A.
The rate stays the same.
B.
The rate decreases.
C.
The rate increases.
D.
There is no way to tell.
Answer:
The rate decreases
Explanation:
When we dissolve a gas in a water, the process is exothermic. This implies that heat is evolved upon dissolution of a gas in water.
Recall from Le Chateliers principle that for exothermic reactions, an increase in temperature favours the reverse reaction. The implication of these is that when the temperature of the gas is increased, less gas will dissolve in water.
Hence increase in temperature decreases the rate of solubility of a gas in water.
Answer:
B.
The rate decreases.
Explanation:
The heat of vaporization of water is 40.66 kJ/mol. How much heat is absorbed when 3.11 g of water boils at atmospheric pressure?
Answer:
The amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure is 7.026 kJ.
Explanation:
A molar heat of vaporization of 40.66 kJ / mol means that 40.66 kJ of heat needs to be supplied to boil 1 mol of water at its normal boiling point.
To know the amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure, the number of moles represented by 3.11 g of water is necessary. Being:
H: 1 g/moleO: 16 g/molethe molar mass of water is:
H₂O= 2* 1 g/mole + 16 g/mole= 18 g/mole
So: if 18 grams of water are contained in 1 mole, 3.11 grams of water in how many moles are present?
[tex]moles of water=\frac{3.11 grams*1 mole}{18 gramos}[/tex]
moles of water= 0.1728
Finally, the following rule of three can be applied: if to boil 1 mole of water at its boiling point it is necessary to supply 40.66 kJ of heat, to boil 0.1728 moles of water, how much heat is necessary to supply?
[tex]heat=\frac{0.1728 moles*40.66 kJ}{1 mole}[/tex]
heat= 7.026 kJ
The amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure is 7.026 kJ.
In a buffer solution made of acetic acid and sodium acetate, if a small amount of acid is added, the added acid will react with whome?
Answer:
The acid reacts with the conjugate base producing more weak acid.
Explanation:
A buffer solution is defined as the mixture of a weak acid and its conjugate base or a weak base with its conjugate acid.
The acetic buffer, CH₃COOH/CH₃COO⁻, is in equilibrium with water as follows:
CH₃COOH(aq) + H₂O(l) ⇄ CH₃COO⁻(aq) + H₃O⁺
When an acid HX (Source of H₃O⁺) is added to the buffer, the reaction that occurs is:
CH₃COO⁻ + HX → CH₃COOH
The acid reacts with the conjugate base producing more weak acid.In fact, this is the principle of the buffer:
An acid reacts with the conjugate base producing weak acid. And the weak acid reacts with a base producing conjugate base
If a small amount of acid is added to an acetic acid-sodium acetate buffer, the acid will react with the acetate ion from sodium acetate.
We have a buffer formed by acetic acid and sodium acetate.
What is a buffer?A buffer is a solution used to resist abrupt changes in pH when an acid or a base is added.
How are buffers formed?They can be formed in 1 of 2 ways:
By a weak acid and its conjugate base.By a weak base and its conjugate acid.Our buffer is formed by a weak acid (acetic acid) and its conjugate base (acetate ion from sodium acetate).
When an acid (HX) is added, it is neutralized by the basic component of the buffer. The generic net ionic equation is:
H⁺ + CH₃COO⁻ ⇄ CH₃COOH
If a small amount of acid is added to an acetic acid-sodium acetate buffer, the acid will react with the acetate ion from sodium acetate.
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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.
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 valueWrite 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
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
By heating a 93% pure kclo3 sample, what percentage of its mass is reduced?
2KCLO3---->2KCL+3O2
Explanation:
free your mind drink water and go outside take fresh air you will get answers
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
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
How much work (in Joules) is required to expand the volume of a pump from 0.00 L to 2.50 L against an external pressure of 1.10 atm
Answer:
W = 278.64375 Joules
Explanation:
The information given in this problem are;
Initial volume = 0L
Final volume = 2.50L
ΔV = 2.50 - 0 = 2.50 L
External pressure, P = 1.10 atm
Work = ?
These parameters are related by the equation;
w = - P ΔV
W = - (1.10 )(2.50)
W = 2.75 L atm
Upon conversion to joules;
1 liter atmosphere is equal to 101.325 joule
W = 278.64375 Joules
