Answer:
The concentrations are: [K⁺] = 1.2 M, [OH⁻] = 0.016 M, [CO₃²⁻] = 1.18 M and [H⁺] = 6.25x10⁻¹³ M.
Explanation:
The dissociation equation of K₂CO₃ in water is:
K₂CO₃(aq) ⇄ K⁺(aq) + CO₃²⁻(aq) (1)
Also, the CO₃²⁻ will react with water as follows:
CO₃²⁻(aq) + H₂O(l) ⇄ HCO₃⁻(aq) + OH⁻(aq) (2)
The constant of the reaction (2) is:
[tex] Kb = \frac{[OH^{-}][HCO_{3}^{-}]}{[CO_{3}^{-2}]} = 2.08 \cdot 10^{-4} [/tex]
The solution of K₂CO₃ is 1.2 M, and since the mole ratio of K₂CO₃ with K⁺ and CO₃²⁻ is 1:1, then we have:
[tex] [K_{2}CO_{3}] = [K^{+}] = [CO_{3}^{-2}] = 1.2 M [/tex]
Now, from equation (2) we have:
CO₃²⁻(aq) + H₂O(l) ⇄ HCO₃⁻(aq) + OH⁻(aq) (3)
1.2 - x x x
[tex] 2.08 \cdot 10^{-4} = \frac{[OH^{-}][HCO_{3}^{-}]}{[CO_{3}^{-2}]} [/tex]
[tex] 2.08 \cdot 10^{-4} = \frac{x^{2}}{1.2 - x} [/tex]
[tex] 2.08 \cdot 10^{-4}*(1.2 - x) - x^{2} = 0 [/tex] (4)
By solving equation (4) for x we have:
x = 0.016 M = [HCO₃⁻] = [OH⁻]
Hence, the CO₃²⁻ concentration is:
[CO₃²⁻] = 1.2 M - 0.016 M = 1.18 M
Finally, the concentration of [H⁺] is:
[tex] [H^{+}][OH^{-}] = 10^{-14} [/tex]
[tex][H^{+}] = \frac{10^{-14}}{[OH^{-}]} = \frac{10^{-14}}{0.016} = 6.25 \cdot 10^{-13} M[/tex]
Therefore, the concentrations are: [K⁺] = 1.2 M, [OH⁻] = 0.016 M, [CO₃²⁻] = 1.18 M and [H⁺] = 6.25x10⁻¹³ M.
I hope it helps you!
Which, if any, of the two molecular geometries is likely to be polar if the outer atoms are identical and there is a difference in electronegativity between the central atom and the outer atoms
Answer:
Polar molecules are not symmetrical
Explanation:
Even though the structures of the molecules involved were not shown in the question, but I will proceed to give a general explanation of the conditions that describe a polar molecule.
First of all, symmetrical molecules are non-polar and asymmetrical molecules are polar. This is the reason why CF4 will be a nonpolar molecule but H2O will be a polar molecule. Some symmetrical molecules may may posses polar bonds or dipoles but these dipoles eventually cancel out since the molecule is symmetrical in nature.
Summarily, if a molecule possess the same type of atoms attached to the central atom with some symmetry axes, like C3, C4 etc., we will end up with a non polar molecule but if we have a nonplanar molecule, then we will end up finding it to be polar.
A spinning turbine can generate electricity only in the form of a/an _______ current.
Of all the alternative energy technologies presented in this section, only solar panels produce a/an _______ current.
Answer:
The correct answer is - alternating and direct, in order.
Explanation:
Alternating current is is type of electric current that is characterized by the direction of the flow of electrons in continuously switches its directs in opposite manner at regular cycles. While direct current or DC is flow of the electrons that move from starting to end in one direction.
Spinning turbines always leads to the alternating electric current while only solar energy produces the direct current with the help of the solar panels.
Thus, the correct answer is - alternating and direct, in order.
Answer:
1. alternating
2. direct
3. The sun heats up the atmosphere as Earth spins, creating areas of high and low temperature. This temperature difference causes wind to start moving through convection, which can then drive a wind turbine to produce electricity.
Explanation:
From Penn
An energy of 13.6 eV is needed to ionize an electron from the ground state of a hydrogen atom. Selecting the longest wavelength that will work from the those given below, what wavelength is needed if a photon accomplishes this task
Answer:
The wavelength is [tex]\lambda = 91nm[/tex]
Explanation:
From the question we are told that
The energy required is [tex]E = 13.6 \ eV[/tex]
This energy needed in form of a photon can be mathematically represented as
[tex]E = \frac{h * c }{\lambda }[/tex]
where h is the Planck constant with a value [tex]h = 4.1357 * 10{-15} eV \cdot sec[/tex]
and c is the speed of light which is
substituting values
[tex]13.6 eV = \frac{ 4.1357 * 10^{-15} eV * 3.0*10^8 }{\lambda }[/tex]
= > [tex]\lambda = 9.1*10^{-8}[/tex]
[tex]\lambda = 91nm[/tex]
How many grams of dipotassium succinate trihydrate (K2C4H4O4·3H2O, MW = 248.32 g/mol) must be added to 660.0 mL of a 0.0577 M succinic acid solution to produce a pH of 5.869 ? Succinic acid has pKa values of 4.207 (pKa1) and 5.636 (pKa2).
Answer:
the amount of grams of dipotassium succinate trihydrate K₂C₄H₄O₄·3H₂O = 41.798g
Explanation:
Given that:
The volume of [tex]K_2C_4H_4O_4.3H_2O[/tex] = 660.mL
The molarity of succinic acid = 0.0577 M
The pH of the solution = 5.869
The pKa₁ = 4.207
The pKa₂ = 5.636
The pKa₂ is required to be used for the determination of given that succinic acid is dissociated twice.
Using Henderson-HasselBalch Equation,
[tex]PH =pKa + log\dfrac{[Salt]}{[Acid]}[/tex]
we know that :
The volume of [tex]K_2C_4H_4O_4[/tex] = 660.mL
The molarity of succinic acid = 0.0577 M
number of moles = Molarity [tex]\times[/tex] Volume (liters)
number of moles = 0.0577 [tex]\times[/tex] 0.660 L
number of moles = 0.038082 mol
The balanced chemical reaction for this equation can be expressed as follows:
[tex]K_2C_4H_4O_4+C_4H_4O_4^{2-} \longleftrightarrow 2KC_4H_4O_4^-[/tex]
Here, the number of moles of [tex]K_2C_4H_4O_4= C_4H_4O_4^{2-}[/tex]
number of moles of [tex]2KC_4H_4O_4^-[/tex] = 2 × 0.038082 mol
number of moles of [tex]2KC_4H_4O_4^-[/tex] = 0.076164 mol
From the Henderson-HasselBalch Equation,
[tex]pH =pKa + log\dfrac{[Salt]}{[Acid]}[/tex]
[tex]pH =pKa_2 + log\dfrac{[K_2C_4H_4O_4]}{[KC_4H_4O_4^-]}[/tex]
[tex]pH =pKa_2 + log\dfrac{n_{K_2C_4H_4O_4}}{n_{KC_4H_4O_4^-}}[/tex]
[tex]5.869 =5.636+ log \dfrac{n_{K_2C_4H_4O_4}}{0.076164 }[/tex]
[tex]0.233= log \dfrac{n_{K_2C_4H_4O_4}}{0.076164 }[/tex]
[tex]10^{0.233}= \dfrac{n_{K_2C_4H_4O_4}}{0.076164 }[/tex]
[tex]1.710015315= \dfrac{n_{K_2C_4H_4O_4}}{0.076164 }[/tex]
[tex]1.710015315 \times 0.076164= n_{K_2C_4H_4O_4[/tex]
[tex]n_{K_2C_4H_4O_4}= 0.1302416[/tex]
SInce; the number of moles of [tex]K_2C_4H_4O_4= C_4H_4O_4^{2-}[/tex]
[tex]n_{C_4H_4O_4^{2-}} =[/tex]( 0.1302416 + 0.038082) mol
[tex]n_{C_4H_4O_4^{2-}} =[/tex] 0.1683236 mol
The grams of dipotassium succinate trihydrate = numbers of moles of dipotassium succinate trihydrate × Molar mass
The grams of dipotassium succinate trihydrate = 0.1683236 mol × 248.32 g/mol
The grams of dipotassium succinate trihydrate = 41.798g
Thus , the amount of grams of dipotassium succinate trihydrate K₂C₄H₄O₄·3H₂O = 41.798g
Question 4
1 points
Save Answer
A solution is prepared at 25°C that is initially 0.42 M in an "X" base, a weak base with Kb=7.4X10^-4, and 0.42 M in conjugate acid "HX". Calculate the pH of the
solution. Round your answer to 2 decimal places.
(Hint: pkw=pka + pkb)
A Moving to another question will save this response.
< Question 4 of 12>
Answer:
pH = 10.87
Explanation:
You can find pH of a buffer (Mixture of a weak base, X, and its conjugate acid, HX) by using H-H equation:
pH = pKa + log [X] / [HX]
pKb is -log Kb = 3.13
Using Hint pKw = pKa + pKb (pKw = 14)
pKw = pKa + pKb
14 - 3.13 = pKa
10.87 = pKa
Replacing in H-H equation:
pH = 10.87 + log [0.42M] / [0.42]
pH = 10.87 + log 1
pH = 10.87A civil engineer designs mostly:
A. building structures.
B. computer parts.
C. new foods.
D. technology that flies.
One of the nuclides in spent nuclear fuel is U-235, an alpha emitter with a half-life of 703 million years. How long will it take for an amount of U-235 to reach 24.0% of its initial amount? Express your answer in years to three significant figures. View Available Hint(s)
Answer:
1447584654 years or 1.44 billion years
Explanation:
From the formula;
0.693/t1/2=2.303/t log No/N
Where;
t1/2 = half life of the U-235 nuclides
No= amount of U-235 initially present
N= amount of U-235 present after a time t
t= time taken for N amount of U-235 to remain
Since N= 0.24No
Substituting values
0.693/703×10^6 = 2.303/t log No/0.24No
9.8578×10^-10 = 2.303/t log 1/0.24
9.8578×10^-10 = 1.427/t
t= 1.427/9.8578×10^-10
t= 1447584654 years
In movies about space, there is frequently a space battle scene where ships
explode in big fireballs. Why are these scenes unscientific?
O
A. The explosion would not be a fireball, but would point towards the closest gravity
source, a planet or star. They are only fireballs on Earth because we are already on a
gravity source.
B. None of these
C. There is no oxygen in space, so there can be no combustion.
OD. Space is very cold, there would not be enough heat energy for an explosion to occur.
-
The correct answer is C. There is no oxygen in space, so there can be no combustion.
Explanation:
Fire and flames are the result of a chemical process known as combustion. Moreover, for combustion to occur there are two essential elements. The first one is a fuel or a substance that releases energy and ignites, and the second one is an oxidant, which accepts electrons. This mix and reaction causes high temperatures and release of heat in the form of fire and flames.
This implies, that for fireballs or any other form of fire to exist there must be oxygen or any substance that replaces it. This does not occur in space because the levels of oxygen are extremely low, this means, at least oxygen is added fireballs are not possible in this context as there is no oxygen, and therefore no combustion (Option C).
Answer:
C.) There is no oxygen in space, so there can be no combustion.
Explanation:
I got it correct on founders edtell
True or False
1. Density is considered a chemical (i.e., not a physical) property. TRUE FALSE
2. When naming an ionic compound containing a transition element such as iron (Fe), the name must include a Roman numeral to indicate the charge of the metal ion. TRUE FALSE
3. The neutron was discovered about 20 years after the electron and proton because it has no charge (in order for it to be detected). TRUE FALSE
4. When we balance a chemical equation, we are observing the law of conservation of mass as well as the part of Dalton’s theory that atoms are neither created or destroyed in a chemical reaction TRUE FALSE
5. When a gas is heated up in a closed container, the kinetic energy of the molecules or atoms of the gas increase, which leads to a decrease in the pressure of the gas. TRUE FALSE
6. The amount of enthalpy (heat energy) for a reaction is directly proportional to the amount (number of moles or grams) of the reactants. TRUE FALSE
7. The combined gas law works for any gas (i.e., you do not need to know the chemical formula). TRUE FALSE
8. A balloon with 10.0 g of CO2 gas will have more molecules than a 10.0 g sample of NO gas. TRUE FALSE
9. Unless a sample is at absolute zero (kelvins), the particles in the sample will have kinetic energy and have some kind of motion. TRUE FALSE
Answer:
1. False
2. True
3. True
4. True
5. True
6. True
7. True
8. False
9. True
Explanation:
Density is a physical property since its measurement does not involve any chemical process.
Since transition elements exhibit variable oxidation states, the actual oxidation state of the transition element must be specified in the compound.
Due to the fact that neutron has no charge, it was discovered by Chadwick long after the electron and proton were discovered.
The balancing of chemical reaction equations is a demonstration that atoms are neither created no destroyed. It also shows that mass is neither created nor destroyed in chemical reactions.
When a gas is heated, it expands. Its volume and its kinetic energy increases. Since volume and pressure are inversely proportional (Boyle's law) the pressure decreases.
Enthalpy is said to be an extensive property. This implies that the magnitude of change in enthalpy is known to depend on the amount of reactants that is actually reacted.
The combined gas law is applicable to all ideal gas systems irrespective of their individual chemical formulas.
10g of CO2 contains 0.227 moles of CO2 while 10g of NO contains 0.33 moles of NO hence 10.0 g of NO will contain more molecules than 10.0g of CO2.
If a sample is not at absolute zero, the particles are known to possess kinetic energy which decreases continuously until absolute zero is attained.
AB2AB2 has a molar solubility of 3.72×10−4 M3.72×10−4 M. What is the value of the solubility product constant for AB2AB2? Express your answer numerically.
Answer:
Ksp = 2.06x10⁻¹⁰
Explanation:
For AB₂. solubility product constant, Ksp, is written as follows:
AB₂(s) ⇄ A²⁺ + 2Br⁻
Ksp = [A²⁺] [Br⁻]²
Molar solubility represents how many moles of AB₂ are soluble per liter of solution. and is obtained from Ksp:
AB₂(s) ⇄ A²⁺ + 2Br⁻
AB₂(s) ⇄ X + 2X
where X are moles that are soluble (Molar solubility)
Ksp = [X] [2X]²
Ksp = 4X³As molar solubility of the salt is 3.72x10⁻⁴M:
Ksp = 4X³
Ksp = 4(3.72x10⁻⁴)³
Ksp = 2.06x10⁻¹⁰Balance the following skeleton reaction and identify the oxidizing and reducing agents: Include the states of all reactants and products in your balanced equation. You do not need to include the states with the identities of the oxidizing and reducing agents.
CrO_4^2- (aq) + N_2O(g) rightarrow Cr^3+ (aq) + NO(g) [acidic]
The oxidizing agent is:_______.
The reducing agent is:_______.
Answer:
Oxidizing agent - CrO4^2-
Reducing agent- N2O
Explanation:
Let us look at the equation closely;
CrO4^2- (aq) + 3N2O(g) ------------> Cr^3+ (aq) + 3NO(g) [acidic]
The reduction half equation is;
CrO4^2- (aq) + 3e -------->Cr^3+ (aq)
Oxidation half equation is;
3N2O(g) ------>3 NO(g) +3 e
Note that the oxidizing agent participates in the reduction half equation while the reducing agent participates in the oxidation half equation as seen above.
A sample of drinking water contains 668 ppm of lead. How many grams of lead are there in 100.0 g of this water?
Answer:
0.0668g
Explanation:
Step 1: Given data
Concentration of lead: 668 ppm (mg/kg)
Mass of water: 100.0 g
Step 2: Convert the mass of water to kilograms
We will use the relationship 1 kg = 1,000 g.
[tex]100.0g \times \frac{1kg}{1,000g} = 0.1000kg[/tex]
Step 3: Calculate the mass of lead in 0.1000 kg of water
There are 668 mg of Pb in 1 kg of water.
[tex]0.1000kgWater \times \frac{668mgPb}{1kgWater} = 66.8mgPb[/tex]
Step 4: Convert the mass of Pb to grams
We will use the relationship 1 g = 1,000 mg.
[tex]66.8mg \times \frac{1g}{1,000mg} = 0.0668g[/tex]
Solid sodium oxide and gaseous water are formed by the decomposition of solid sodium hydroxide (NaOH) .
Write a balanced chemical equation for this reaction.
Answer:
2NaOH(s) → Na₂O(s) + H₂O(g)
Hope that helps.
Why are antiparallel beta sheets more stable than parallel beta sheets?
Answer:
The side chains of the amino acids alternate above and below the sheet
Explanation:
Hydrogen bonds are formed between the amine and carbonyl groups across strands. ... Antiparallel ß sheets are slightly more stable than parallel ß sheets because the hydrogen bonding pattern is more optimal.
a) During the workup of the reaction, an aqueous solution of sodium bicarbonate was added to the cooled reaction mixture. Why was this done
Answer:
The purpose of adding an aqueous solution of sodium bicarbonate is either to extract a certain compound or to remove/neutralize acidic compounds present in the reaction mixture.
Explanation:
Upon adding sodium bicarbonate, carbon dioxide is released (gaseous state at room temperature), which helps build up pressure that is able to push out the unwanted gas/liquid.
What products result from mixing aqueous solutions of Cr(NO3)2(aq) and NaOH(aq)? Question 10 options: Cr(OH)2(s), Na+(aq), and NO3−(aq) Cr(OH)2(s) and NaNO3(s) Cr2(OH)2(aq) and NaNO3(aq) Cr(OH)2(aq) and NaNO3(s) Cr(OH)2(s), N2(g), and H2O(l)
Answer:
Cr(OH)2(s), Na+(aq), and NO3−(aq)
Explanation:
Let is consider the molecular equation;
2NaOH(aq) + Cr(NO3)2(aq) -----> 2NaNO3(aq) + Cr(OH)2(s)
This is a double displacement or double replacement reaction. The reacting species exchange their partners. We can see here that both the sodium ion and chromium II ion both exchanged partners and picked up each others partners in the product.
Sodium ions and nitrate ions now remain in the solution while chromium II hydroxide which is insoluble is precipitated out of the solution as a solid hence the answer.
what is the name of this molecule?
Draw the structural formula of the principal organic product formed when ethyl benzoate is treated with C6H5MgBr (two equivalents), then HCl/H2O.
Answer:
Ph3OH
Explanation:
The reaction is between ethyl benzoate (PhCOOC2H5) and a Grignard reagent PhMgBr(C6H5MgBr).
The first step in the reaction mechanism is that the ethyl benzoate is converted to PhCOPh by the first molecule of Grignard reagent.
The second molecule of Grignard reagent now converts PhCOPh to Ph3O^-. In the presence of acid, Ph3O^- is now protonated to yield Ph3OH which is the major organic product of the reaction.
See image attached for more details.
A chemist adds of a M sodium carbonate solution to a reaction flask. Calculate the millimoles of sodium carbonate the chemist has added to the flask. Round your answer to significant digits.
Answer:
256 milimoles of Sodium Carbonate are in the flask
Explanation:
The chemist adds 180mL of a 1.42M sodium carbonate solution.
Molarity is an unit of concentration in chemistry defined as the ratio between moles of solute (Sodium Carbonate, in this case) per liter of solution.
A 1.42M solution contains 1.42 moles of Sodium carbonate in 1L of the solution.
As you have 180mL = 0.180L, moles of sodium carbonate you have are:
0.180L × (1.42 moles Sodium Carbonate / 1L) = 0.2556 moles of sodium carbonate
1000milimoles are equal to 1mole. 0.2556 moles are:
0.2556 moles × (1000 milimoles / 1 mole) =
256 milimoles of Sodium Carbonate are in the flaskIn the laboratory you are asked to make a 0.694 m copper(II) iodide solution using 455 grams of water. How many grams of copper(II) iodide should you add
Answer:
100.2g of CuI₂ you must add
Explanation:
Molality, m, is defined as the ratio between moles of solute and kg of solvent.
In the problem, you have a 0.694m of copper (II) iodide -CuI₂, molar mass: 317.35 g/mol-. That means there are 0.694 moles of CuI₂ per kg of water.
As you have 455g = 0.455kg of water -solvent-, moles of CuI₂ are:
0.455kg ₓ (0.694 moles CuI₂ / kg) = 0.316 moles of CuI₂
Using molar mass, grams of CuI₂ in the solution are:
0.316moles CuI₂ ₓ (317.35g / mol) =
100.2g of CuI₂ you must addA certain radioactive nuclide has a half life of 1.00 hour(s). Calculate the rate constant for this nuclide. s-1 Calculate the decay rate for 1.000 mole of this nuclide. decays s-1
Answer:
k= 1.925×10^-4 s^-1
1.2 ×10^20 atoms/s
Explanation:
From the information provided;
t1/2=Half life= 1.00 hour or 3600 seconds
Then;
t1/2= 0.693/k
Where k= rate constant
k= 0.693/t1/2 = 0.693/3600
k= 1.925×10^-4 s^-1
Since 1 mole of the nuclide contains 6.02×10^23 atoms
Rate of decay= rate constant × number of atoms
Rate of decay = 1.925×10^-4 s^-1 ×6.02×10^23 atoms
Rate of decay= 1.2 ×10^20 atoms/s
Arrange the following oxides in order of increasing acidity.
Rank from least acidic to most acidic. To rank items as equivalent,overlap them.
CaO
P2O5
SO3
SiO2
Al2O3
CO2
Answer:
Based on the Modern Periodic table, there is an increase in the electropositivity of the atom down the group as well as increases across a period. On comparing the electropositivities of the mentioned oxides central atom, it is seen that Ca is most electropositive followed by Al, Si, C, P, and S is the least electropositive.
With the decrease in the electropositivity, there is an increase in the acidity of the oxides. Thus, the increasing order of the oxides from the least acidic to the most acidic is:
CaO > Al2O3 > SiO2 > CO2 > P2O5 > SO3. Hence, CaO is the least acidic and SO3 is the most acidic.
Since acidity increases across a period from left to right, and decreases down a group, the oxides can be ranked from the least acidic to the most acidic as follows:
[tex]\mathbf{CaO < Al_2O_3 < SiO_2 < CO_2 < P_2O_5 <SO_3}[/tex]
The least acidic is CaOThe most acidic is [tex]SO_3[/tex]Note the following:
Acidity of an oxide depends on its electronegativity.Non-metals are more electronegative, while metals are less electronegative.Acidity of oxides increases across a period as you move from left to the right side of a periodic table.Acidity of oxides decreases down a group (column) in a periodic table.Using the periodic table diagram given in the attachment below, we can rank the given oxides according to their increasing acidity.
CaO, is the least, because it is an oxide of the metal, Calcium, which is at the far left in group 2 in the periodic table.The next is, [tex]Al_2O_3[/tex]. Aluminum is a metal from group 3.[tex]SiO_2[/tex] is an oxide of Silicon, also in group 4 but below Carbon.[tex]CO_2[/tex] is an oxide of Carbon, from group 4.
[tex]P_2O_5[/tex] is an oxide of the non-metal, Phosphorus, a group 5 element
[tex]SO_3[/tex] is an oxide of the non-metal, Sulphur, a group 6 element.
Therefore, since acidity increases across a period from left to right, and decreases down a group, the oxides can be ranked from the least acidic to the most acidic as follows:
[tex]\mathbf{CaO < Al_2O_3 < SiO_2 < CO_2 < P_2O_5 <SO_3}[/tex]
The least acidic is CaOThe most acidic is [tex]SO_3[/tex]Learn more here:
https://brainly.com/question/12200588
Calculate the amount of heat energy, in kJ, needed to increase the temperature of 200g of water by 300 C.
Answer:
Q = 246 kJ
Explanation:
It is given that,
Mass of water, m = 200 g
Let initial temperature, [tex]T_i=5^{\circ}[/tex]
Final temperature of water, [tex]T_f=300^{\circ} C[/tex]
We know that the specific heat capacity of water, [tex]c=4.18\ J/g-^{\circ} C[/tex]
So, the heat energy needed to raise the temperature is given by :
[tex]Q=mc\Delta T\\\\Q=200\times 4.18\times (300-5)\\\\Q=246620\ J[/tex]
or
Q = 246 kJ
So, the heat energy of 246 kJ is needed.
The height of a column of mercury in a closed-end manometer is 13.2 cm. What is the pressure of the gas in torr
Answer:
[tex]P_{gas}=131.96torr[/tex]
Explanation:
Helo,
In this case, the pressure must be computed as follows:
[tex]P_{gas}=gh\rho _{Hg}[/tex]
Which is using the density of mercury (13.6 g/mL) and its height, thus, we obtain (using the proper units):
[tex]P_{gas}=9.8\frac{m}{s^2}*(13.2cm *\frac{1m}{100cm} )*(13.6\frac{g}{cm^3}*\frac{1kg}{1000g}*\frac{1x10^6cm^3}{1m^3} )\\\\P_{gas}=17592.96Pa*\frac{760torr}{101325Pa} \\\\P_{gas}=131.96torr[/tex]
Best regards.
Zn + 2 HCl --> H2 + ZnCl2 If 1.70 g of Zn are reacted, how many grams of ZnCl2 can be created? Show work and process and I will give brainliest
Explanation:
first find the the number of moles of of zinc .
as the number of moles of zinc and ZnCl2 is same we can calculate the mass of ZnCl2.
One kilogram of butane (C4H10) is burned with 25 kg of air that is at 308C and 90 kPa. Assuming that the combustion is complete and the pressure of the products is 90 kPa, determine (a) the percentage of theoretical air used and (b) the dew-point temperature of the products.
Explanation:
Combustion of butane gives following reaction
C4H10 + 13/2 O2 → 5H20 + 4CO2
Therefore, One Kg of Butane consists of 1000/58 moles . That is 17.24 moles.
17.24 moles of butane reacts with (6.5×17.24) moles of O2 = 112.06 moles = 3.586 Kg O2.
Weight % of O2 in Air = 20 %
Mass of theoretical air used is 3.586×5 = 17.93 Kg = 71.72 %
The dew point temperature of H2O is :
D = (237.3×B)/(1-B)
B = ln(E/6.108)/17.27
E is vapor pressure at given temp.
At T = 30° C , E = 31.8 mm of Hg = 41.84 Milli Bars
B = 0.1114
D = 29.74°C
Which of the following expressions is the correct equilibrium-constant expression for the equilibrium between dinitrogen tetroxide and nitrogen dioxide?
N2O4(g) ⇌ 2NO2(g)
a. [NO]^2 [N2O4]
b. [NO2]/ [N2O4]^2
c. [NO2]/[N2O4]
d. [NO2]^2/[N2O4]
Answer:
D. [NO₂]²/[N₂O₄]
Explanation:
The equilibrium constant expression for a reaction is products over reactants. Since NO₂ has a coefficient of 2, it will become an exponent.
So, it would be:
[NO₂]²/[N₂O₄]
Hope that helps.
The correct equilibrium-constant expression for the equilibrium between dinitrogen tetroxide (N₂O₄) and nitrogen dioxide (NO₂) is option d. [NO₂]²/[N₂O₄].
In the balanced equation N₂O₄(g) ⇌ 2NO₂(g), the stoichiometric coefficients indicate that two moles of NO₂ are produced for every one mole of N₂O₄ consumed.
The equilibrium constant expression is derived from the molar concentrations of the species involved in the equilibrium. In this case, the expression is written as [NO₂]²/[N₂O₄], where [NO₂] represents the molar concentration of NO₂ and [N₂O₄] represents the molar concentration of N₂O₄.
By squaring the concentration of NO₂ and dividing it by the concentration of N₂O₄, the equilibrium-constant expression correctly accounts for the stoichiometry of the reaction and the relative concentrations of the species involved.
Hence, the correct option is option d.
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A cannonball made of iron has a volume of 121 cm3. What is the mass, in kilograms, of the cannonball? Express your answer to three significant figures and include the appropriate units.
Answer:the mass is 35456
Explanation:
because it works
The density has been the division of the mass to the volume of the substance. The mass of an iron cannonball of density 7.86 grams/cm³ is estimated to be 0.951 kilograms.
What is the relationship between mass and density?The mass (m) has been calculated by multiplying the density (D or ρ) and volume (V) of the iron cannonball. The measurement of the amount is called mass whereas the measurement of the space occupied by the substance volume is given as density.
The amount of the substance contained in a volume (milliliters, liters, and cubic centimeters) along with the density of the substance has been used to give mass. The mass can be estimated in kilograms, grams, milligrams, etc.
Given,
Density of iron cannonball (D) = 7.86 grams/cm³
Volume of cannonball (V) = 121 cm³
The formula to calculate mass is given as:
Density = mass ÷ volume
Mass = density × volume
Substituting values in the above formula, the mass of the iron cannonball in grams is calculated as:
mass = 7.86 grams/cm³ × 121 cm³
mass = 951.06 grams
Therefore, the iron cannonball with a density of 7.86 grams/cm³ has a mass of 0.951 kilograms.
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Your question is incomplete, but most probably your full question was, A cannonball made of iron has a volume of 121 cm³. if the iron has a density of 7.86 g/cm³, what is the mass, in kilograms, of the cannonball?
Which of the choices is not a statement or direct application of the second law of thermodynamics? There are no 100% efficient heat engines. The change in internal energy of a system can be found by combining the heat energy added to a system minus the work done by the system. Heat energy naturally transfers from a high temperature substance to a low temperature substance. All natural processes tend toward states of increased entropy.
Answer:
The change in internal energy of a system can be found by combining the heat energy added to a system minus the work done by the system is not a direct application of second law of thermodynamics.
Explanation:
Second law of thermodynamics states that heat can be transfer spontaneously from high temperature to low temperature only.
The change in internal energy of a system can be found by combining the heat energy added to a system minus the work done by the system is not a direct application of second law of thermodynamics because according to the second law of thermodynamics it is impossible in any system for heat transfer from a source to completely convert to work done in a cyclical process( bring the system to its original stage after each cycle) in which the system then return to it's original stage.
Determine the cell notation for the redox reaction given below.
Sn(s) + 2H+(aq) ⟶ Sn2+(aq) + H2(g)
a. H+(aq) | H2(g) | Pt ∥ Sn(s) | Sn2+(aq)
b. H2(g) | H+(aq) | Pt ∥ Sn2+(aq) | Sn(s)
c. Sn2+(aq) | Sn(s) ∥ H2(g) | H+(aq) | Pt
d. Sn(s) | Sn2+(aq) ∥ H+(aq) | H2(g) | Pt
e. Sn(s) | H2(g) ∥ Sn2+(aq) | H+(aq) | Pt
Answer:
The correct answer is d. Sn(s) | Sn²⁺(aq) ∥ H⁺(aq) | H₂(g) | Pt
Explanation:
The half reactions are:
2H⁺(aq) + 2 e- ⟶ H₂(g) (reduction)
Sn(s) ⟶ Sn²⁺(aq) + 2 e- (oxidation)
In the cell notation, there are two electrodes in which are separated the reduction reaction from the oxidation reaction. In the left electrode occurs the oxidation reaction (anode) while in the right electrode occurs the reduction reaction (cathode). The general form of the cell notation is the following:
anode reaction∥ cathode reaction
where the two bars ( ∥ ) represent the physical barrier between the electrodes. A single bar ( | ) is used to represent a phase separation.
In this redox reaction, the half reaction of the anode is Sn(s) ⟶ Sn²⁺(aq) + 2 e-; whereas the half reaction of the cathode is 2H⁺(aq) + 2 e- ⟶ H₂(g).
The componens are written in order according to the half reaction. Since Sn²⁺ and H⁺ ions are in solution, a platinum electrode is used and represented as Pt. Thus, the cell notation is:
Sn(s) | Sn²⁺(aq) ∥ H⁺(aq) | H₂(g) | Pt