Answer:
(a) [tex]m=2.69m[/tex]
(b) [tex]x_{LiBr}=0.099[/tex]
(c) [tex]\% LiBr=18.9\%[/tex]
Explanation:
Hello,
In this case, given the molality in mol/L, we can compute the required units of concentration assuming a 1-L solution of acetonitrile and lithium bromide that has 1.80 moles of lithium bromide:
(a) For the molality, we first compute the grams of lithium bromide in 1.80 moles by using its molar mass:
[tex]m_{LiBr}=1.80mol*\frac{86.845 g}{1mol}=156.32g[/tex]
Next, we compute the mass of the solution:
[tex]m_{solution}=1L*0.826\frac{g}{mL}*\frac{1000mL}{1L}=826g[/tex]
Then, the mass of the solvent (acetonitrile) in kg:
[tex]m_{solvent}=(826g-156.32g)*\frac{1kg}{1000g}=0.670kg[/tex]
Finally, the molality:
[tex]m=\frac{1.80mol}{0.670kg} \\\\m=2.69m[/tex]
(b) For the mole fraction, we first compute the moles of solvent (acetonitrile):
[tex]n_{solvent}=669.68g*\frac{1mol}{41.05 g} =16.31mol[/tex]
Then, the mole fraction of lithium bromide:
[tex]x_{LiBr}=\frac{1.80mol}{1.80mol+16.31mol}\\ \\x_{LiBr}=0.099[/tex]
(c) Finally, the mass percentage with the previously computed masses:
[tex]\% LiBr=\frac{156.32g}{826g}*100\%\\ \\\% LiBr=18.9\%[/tex]
Regards.
Interpret the following equation for a chemical reaction using the coefficients given: CO(g) Cl2(g) COCl2(g) On the particulate level: _________ of CO(g) reacts with _________ of Cl2(g) to form _________ of COCl2(g). On the molar level: _________ of CO(g) reacts with _________ of Cl2(g) to form _________ of COCl2(g).
Answer:
On the particulate level: 6.02 * 10²³ particles of CO(g) reacts with 6.02 * 10²³ particles of Cl₂(g) to form 6.02 * 10²³ particles of COCl2(g).
On the molar level: 1 mole of CO(g) reacts with 1 mole of Cl2(g) to form 1 mole of COCl₂(g).
Explanation:
The particulate level refers to the microscopic or atomic level of substances. It also involves the ions, protons, neutrons and molecules present in substances.
The molar level refers to the quantitative measure of substances in terms of the mole, where a mole represents the amount of substances containing the Avogadro number of particles which is equal to 6.02 * 10³ particles.
Equation of the reaction: CO(g) + Cl₂(g) ----> COCl₂(g)
From the equation above, I mole of CO gas reacts with 1 mole of Cl₂ gas to produce 1 mole of COCl₂ gas.
Since 1 mole of a substance contains 6.02 * 10²³ particles, on a particulate level, 6.02 * 10²³ particles of CO gas reacts with 6.02 * 10²³ particles of Cl₂ gas to produce 6.02 * 10²³ particles of COCl₂ gas.
Identify whether each species functions as a Brønsted-Lowry acid or a Brønsted-Lowry base in this net ionic equation. HF (aq) + SO32- ⇌ F- + HSO3- Brønsted-Lowry _____ Brønsted-Lowry _____ Brønsted-Lowry _____ Brønsted-Lowry _____ In this reaction: The formula for the conjugate _____ of HF is The formula for the conjugate _____ of SO32- is
Explanation:
A Bronsted-Lowry base is a substance that accepts a proton in the form of a hydrogen (H) atom.
On the other hand;
Bronsted-Lowry acid is the substance that donates the proton.
HF (aq) + SO32- ⇌ F- + HSO3-
In the forward reaction;
Bronsted-Lowry acid : HF
Bronsted-Lowry base: SO32-
In the backward reaction;
Bronsted-Lowry acid : HSO3-
Bronsted-Lowry base: F-
The conjugate base of HF is F-
The conjugate acid of SO32- is HSO3-
AHP for the formation of rust (Fe2O3) is -826 kJ/mol. How much energy is
involved in the formation of 5.00 grams of rust?
A 25.9 kJ
B 25.9 J
C 66.0 kJ
D 66.0)
Answer:
A- 25.9 kJ
Explanation:
ΔH of formation is defined as the amount of energy that is involved in the formation of 1 mole of substance.
ΔH of rust is -826kJ/mol, that means when 1 mole of rust is formed, there are released -826kJ.
Moles of 5.00g of Fe₂O₃ (Molar mass: 159.69g/mol) are:
5.00g ₓ (1 mole / 159.69g) = 0.0313 moles of Fe₂O₃.
If 1 mole release -826kJ, 0.0313 moles release:
0.0313 moles ₓ (-826kJ / 1 mole) = -25.9kJ
Thus, heat involved is:
A- 25.9 kJIdentify the Lewis acids and Lewis bases in the following reactions:
1. H+ + OH- <-> H2O Lewis acid: Lewis base:
2. Cl- + BCl3 <-> BCl4- Lewis acid: Lewis base:
3. K+ + 6H2O <-> K(H2O)6+ Lewis acid: Lewis base:
Answer: 1. [tex]H^++OH^-\rightarrow H_2O[/tex] Lewis acid : [tex]H^+[/tex], Lewis base : [tex]OH^-[/tex]
2. [tex]Cl^-+BCl_3\rightarrow BCl_4^-[/tex] Lewis acid : [tex]BCl_3[/tex], Lewis base : [tex]Cl^-[/tex]
3. [tex]K^++6H_2O\rightarrow K(H_2O)_6[/tex] Lewis acid : [tex]K^+[/tex], Lewis base : [tex]H_2O[/tex]
Explanation:
According to the Lewis concept, an acid is defined as a substance that accepts electron pairs and base is defined as a substance which donates electron pairs.
1. [tex]H^++OH^-\rightarrow H_2O[/tex]
As [tex]H^+[/tex] gained electrons to complete its octet. Thus it acts as lewis acid.[tex]OH^-[/tex] acts as lewis base as it donates lone pair of electrons to electron deficient specie [tex]H^+[/tex].
2. [tex]Cl^-+BCl_3\rightarrow BCl_4^-[/tex]
As [tex]BCl_3[/tex] is short of two electrons to complete its octet. Thus it acts as lewis acid. [tex]Cl^-[/tex] acts as lewis base as it donates lone pair of electrons to electron deficient specie [tex]BCl_3[/tex].
3. [tex]K^++6H_2O\rightarrow K(H_2O)_6[/tex]
As [tex]K^+[/tex] is short of electrons to complete its octet. Thus it acts as lewis acid. [tex]H_2O[/tex] acts as lewis base as it donates lone pair of electrons to electron deficient specie [tex]K^+[/tex].
Now construct a different electrochemical cell. You put a zinc metal anode in contact with a 0.052 M solution of zinc nitrate and a silver cathode in contact with a 0.0042 M solution of silver(I) nitrate. What is the value of the electric potential at the moment the reaction begins
Answer:
[tex]1.66~V[/tex]
Explanation:
We have to start with the half-reactions for both ions:
[tex]Zn^+^2~+2e^-~->Zn[/tex] V= -0.76
[tex]Ag^+~e^-~->~Ag[/tex] V= +0.80
If we want a spontaneous reaction (galvanic cell) we have to flip the first reaction, so:
[tex]Zn~->~Zn^+^2~+2e^-~[/tex] V= +0.76
[tex]Ag^+~+~e^-~->~Ag[/tex] V= +0.80
If we want to calculate ºE we have to add the two values, so:
ºE=0.76+0.80 = 1.56 V
Now, we have different concentrations. So, if we want to calculate E we have to use the nerts equation:
[tex]E=ºE~+~\frac{0.059}{n}LogQ[/tex]
On this case, Q is equal to:
[tex]Q=\frac{[Zn^+^2]}{[Ag^+]^2}[/tex]
Because the total reaction is:
[tex]Zn~+~2Ag^+~->~Zn^+^2~+~2Ag[/tex]
So, the value of "Q" is:
[tex]Q=\frac{[0.052 M]}{[0.0042]^2}=2947.84[/tex]
Now, we can plug all the values in the equation (n=2, because the amount of electrons transferred is 2). So:
[tex]E=1.56~V~+~\frac{0.059}{2}Log(2947.84)=1.66~V[/tex]
I hope it helps!
How many moles of CO2 are produced when 84 0 mol O2 completely react?
Answer:
Explanation:boom
What are plastic bottles made of?
Polyethylene
halogen
silicon
Alkyl groups
Answer:
polyethylenes
Explanation:
the plastic bottles used to hold potable water and other drinks are made from polyethylene because, the material is both strong and light.
hope this helped!
Answer: Polyethylenes
Explanation: I got 100% on the test :)
At 25 °C, what is the hydroxide ion concentration, [OH−] , in an aqueous solution with a hydrogen ion concentration of [H+]=1.9×10−9 M?
Answer:
The correct answer will be "5.26 × 10⁻⁶".
Explanation:
The given values is:
[tex][H^{+}]=1.9\times 10^{-9} M[/tex]
As we know,
⇒ [tex]pH+pOH=14[/tex]
On taking log, we get
⇒ [tex]-log[H^{+}] + -log[OH^{-}] = 14[/tex]
Now,
Taking "log" as common, we get
⇒ [tex]log[H^{+}][OH^{-}]= -14[/tex]
⇒ [tex][H^{+}][OH^{-}]= 10^{-14}[/tex]
⇒ [tex][OH^{-}]=\frac{10^{-14}}{[H^{+}]}[/tex]
On putting the estimated value of "[tex][H^{+}][/tex]", we get
⇒ [tex]=\frac{10^{-14}}{1.9\times 10^{-9}}[/tex]
⇒ [tex]=5.26\times 10^{-6}[/tex]
Fishing trawlers in a certain bay catch a large variety of marketable fish along with a species of eel that is toxic. They normally kill the eels and throw them back into the sea. What term is used to refer to the eel? The eel species is called a(n) ______ of the fishing operation.
Answer:
1. Non-target
2. Bycatch
Explanation:
In the fishing industry, the main aim of the industry is to capture fishes that can be used or eaten and sell. A variety of fishes are captured for this purpose and since they are used therefore are known as Target catch.
But there are some species which has to be discarded because they are toxic and not useful. These non-useful species like eel which gets captured in the net while capturing other fishes are known as Non-target fish.
The eel fish which gets captured is known as bycatch fishes in the fishing operation.
Thus, Non-target and Bycatch are the correct answer.
Answer:
Its just bycatch
Explanation:
The eel species is called a(n) bycatch of the fishing operation.
When you look at an ant up close, using a convex lens, what do you see?
Answer:
You would be able to see the ants clearly with the unique body parts.
Explanation:
Convex lens is also known as Converging lens. It helps in converging rays of light to become a principal focus which is usually very clear and visible to the eyes. The converging lens when used to view the ant makes the ant appear very visible and the individual is able to see all the unique body parts of the ant. This type of lens is used in individuals who are longsighted.
For the iodine trichloride molecule: a. Determine the number of valence electrons for each atom in the molecule b. Draw the Lewis Dot structure c. Describe why the molecule is drawn this way (i.e. any extra rules/steps needed?) d. Show the polarity of each bond and for the molecule by drawing in the dipole +à
Answer:
Explanation:
a. Determine the number of valence electrons for each atom in the molecule
In this case we both atoms are halogens. Therefore we will have 7 electrons for each atom.
b. Draw the Lewis Dot structure
In this case, the formula is [tex]ICl_3[/tex], so the central atom would be "I" and the "Cl" atoms would be placed around "I". See figure 1
c. Describe why the molecule is drawn this way (i.e. any extra rules/steps needed?)
In this specific case, the "I" atom don't follow the octet rule. We will have an expanded octet for iodine (more than 8 electrons).
d. Show the polarity of each bond and for the molecule by drawing in the dipole +d
The negative dipole would be placed in the atom with higher electronegativity, in this case "Cl". The positive dipole would be placed in the atom with low electronegativity, in this case "I".
I hope it helps!
Determine the amount of heat (in kJ) associated with the production of 5.71 × 104 g of ammonia according to the following equation. N2(g) + 3H2(g) 2NH3ΔH°rxn = −92.6 kJ Assume that the reaction takes place under standard-state conditions at 25°C.
Answer:
[tex]Q=-3.11x10^5kJ[/tex]
Explanation:
Hello,
In this case, for the given reaction, we are given the standard enthalpy of reaction per mole of ammonia that is -92.6 kJ, it means, that forming one mole of ammonia will release 92.6 kJ of energy. In such a way, for the formation of 5.71x10⁴ g of ammonia, the following amount of heat will be released:
[tex]Q=5.71x10^4gNH_3*\frac{1molNH_3}{17gNH_3}*-92.6\frac{kJ}{molNH_3}\\ \\Q=-3.11x10^5kJ[/tex]
Best regards.
The amount of the heat associated with the production of 5.71×10⁴ g of ammonia, NH₃ is –311026.732 KJ
We'll begin by calculating the number of mole in 5.71×10⁴ g of NH₃
Mass of NH₃ = 5.71×10⁴ g
Molar mass of NH₃ = 14 + (3×1) = 17 g/mol
Mole of NH₃ =?Mole = mass / molar mass
Mole of NH₃ = 5.71×10⁴ / 17
Mole of NH₃ = 3358.82 moles Finally, we shall determine the heat required to produce 3358.82 moles (i.e 5.71×10⁴ g) of NH₃. This can be obtained as follow:N₂(g) + 3H₂(g) —> 2NH₃(g) ΔH°rxn = −92.6 kJ
Since reaction took place at standard conditions, it means:
1 moles of NH₃ required −92.6 kJ
Therefore,
3358.82 moles of NH₃ will require = 3358.82 × –92.6 = –311026.732 KJ
Thus, the amount of the heat associated with the production of 5.71×10⁴ g of ammonia, NH₃ is –311026.732 KJ
Learn more: https://brainly.com/question/17332795
how many moles of helium gas occupy 22.4 L at 0 degreeC at 1 atm pressure
Answer:
1 mole of the gas occupies 22. 4L at O0C at 1atm pressure. Hence, the correct option is C.
Explanation:
Calcium chloride reacts with sodium hydroxide to form solid calcium hydroxide, Ca(OH)2. The balanced net ionic equation is
Group of answer choices
Ca+2(aq) + 2OH-(aq) + 2 Na+(aq) + 2 Cl-(aq) → Ca(OH)2(s).
Ca+2(aq) + 2OH-(aq) + Na+(aq) + Cl-(aq) → Ca(OH)2(s).
Ca+2(aq) + 2OH-(aq) → Ca(OH)2(s).
CaCl2(aq) + 2NaOH(aq) → CaOH(s) + NaCl(aq).
CaCl2(aq) + 2NaOH(aq) → Ca(OH)2(s) + 2NaCl(aq).
Answer:
Ca²⁺ + 2 OH⁻ → Ca(OH)₂(s)
Explanation:
In chemistry, the net ionic equation is a way to write a chemical reaction whereas you write only the ions that are involved in the reaction.
When calcium chloride, CaCl₂ reacts with sodium hydroxide, NaOH to produce Ca(OH)₂ the only ions involved in the reaction are Ca²⁺ and OH⁻, thus, the balanced net ionic equation is:
Ca²⁺ + 2 OH⁻ → Ca(OH)₂(s)
Cl⁻ and Na⁺ are not involved in the reaction and you don't have to write them.
The balanced net ionic equation for the reaction between calcium chloride and sodium hydroxide is
Ca²⁺(aq) + 2OH⁻(aq) —> Ca(OH)₂(s)The ionic equation for the reaction between calcium chloride and sodium hydroxide can be written as follow:
Calcium chloride => CaCl₂
Sodium hydroxide => NaOH
In solution,
CaCl₂(aq) —> Ca²⁺(aq) + 2Cl⁻(aq)
NaOH(aq) —> Na⁺(aq) + OH⁻(aq)
CaCl₂(aq) + NaOH(aq) —>
Ca²⁺(aq) + 2Cl⁻(aq) + Na⁺(aq) + OH⁻(aq) —> Ca(OH)₂(s) + 2Cl⁻(aq) + Na⁺(aq)
Cancel the spectator ions (i.e Cl⁻ and Na⁺) and write 2 before OH⁻ to obtain the net ionic equation as shown below:
Ca²⁺(aq) + 2OH⁻(aq) —> Ca(OH)₂(s)Learn more: https://brainly.com/question/9489859
A 25.0 mLsample of an acetic acid solution is titrated with a 0.175 M NaOH solution. The equivalence point is reached when 37.5 mL of the base is added. The concentration of acetic acid in the sample was:_______,
A) 0.263
B) 0.365
C) 0.175
D) 1.83×10−4
E) 0.119
Answer:
0.263M of CH₃COOH is the concentration of the solution.
Explanation:
The reaction of acetic acid (CH₃COOH) with NaOH is:
CH₃COOH + NaOH → CH₃COO⁻Na⁺ + H₂O
1 mole of acetic acid reacts per mole of NaOH to produce sodium acetate and water.
In the equivalence point, moles of acetic acid are equal to moles of NaOH and moles of NaOH are:
0.0375L × (0.175 moles / L) = 6.56x10⁻³ moles of NaOH = moles of CH₃COOH.
As the sample of acetic acid had a volume of 25.0mL = 0.025L:
6.56x10⁻³ moles of CH₃COOH / 0.0250L =
0.263M of CH₃COOH is the concentration of the solution
What is the Arrhenius definition of an acid? A substance that increases H3O+ concentration when it is dissolved in water. A substance that increases OH– concentration when it is dissolved in water. A compound that donates protons. A compound that accepts protons.
Answer:
A substance that increases H3O+ concentration when it is dissolved in water.
Explanation:
Note that H3O+ and H+ are used quite interchangeably in chemistry.
An acid makes the H+ content higher, thereby decreasing the pH.
Answer:
a
A substance that increases H3O+ concentration when it is dissolved in water.
Explanation:
Consider a solution containing 0.100 M fluoride ions and 0.126 M hydrogen fluoride. The concentration of fluoride ions after the addition of 5.00 mL of 0.0100 M HCl to 25.0 mL of this solution is __________ M.
a. 0.0980
b. 0.0817
c. 0.0167
d. 0.0850
e. 0.00253
Answer:
The answer is "Option b"
Explanation:
In this question first we calculates the moles in F-, HF, and in HCL, which can be defined as follows:
Formula:
[tex]\ Number \ of \ moles\ = \ Molarity \times \ Volume \ in \ litter[/tex]
[tex]\ moles \ in\ F- = 0.100 \ M \times 0.0250 L\\\\[/tex]
[tex]=\ 0.0025 \ moles[/tex]
[tex]\ moles \ in \ HF \ = 0.126M \times 0.0250 L[/tex]
[tex]= 0.00315 \ moles[/tex]
[tex]\ moles \ in \ HCl = 0.0100M \times 0.00500 L[/tex]
[tex]= 0.00005 \ moles[/tex]
[tex]\ Reaction: \\\\F - + H+ \rightarrow HF[/tex]
[tex]\Rightarrow \ moles \ in \ F- = 0.0025 \\\\\Rightarrow \ moles \ in \ H+ = 0.00005 \\\\ \Rightarrow \ moles \ in \ HF = 0.00315\\\\ \ total \ moles = 0.00250 -0.0000500 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.00315 + 0.00005\\\\\ total \ moles =0.00245 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.00245[/tex]
[tex]\ total \ volume \ in \ the \ solution = \ V = \ 0.0300 L\\\\ after \ addition \ of \ HCl \ the \ concentration \ of \ F- \ = 0.00245\ moles \div V[/tex]
[tex]=\frac{ 0.00245 \ moles }{0.0300L}\\\\= \frac{245 \times 10^4}{300 \times 10^5} \\\\= \frac{245}{3000} \\\\ = 0.0817 M[/tex]
given a k value of 0.43 for the following aqueous equilibrium suppose sample z is placed into water such that its original concentration is 0.033M assume there was zero initial concentration of either A(aq) or B(ag) once equilibrium has occured what will be the equilibrium concentration of z? K=0.43
Answer:
Less than 0.033 M:
[tex][Z]_{eq}=2.4x10^{-3}M[/tex]
Explanation:
Hello,
In this case, the described equilibrium is:
[tex]2A+B\rightarrow 2Z[/tex]
Thus, the law of mass action is:
[tex]K=\frac{[Z]^2}{[A]^2[B]}=0.43[/tex]
Nevertheless, given the initial concentration of Z that is 0.033 M, we should invert the equilibrium since the reaction will move leftwards:
[tex]\frac{1}{K}=\frac{[A]^2[B]}{[Z]^2}=\frac{1}{0.43}=2.33[/tex]
Know, by introducing the change [tex]x[/tex] due to the reaction extent, we can write:
[tex]2.33=\frac{(2x)^2*x}{(0.033-2x)^2}[/tex]
Which has the following solution:
[tex]x_1=2.29M\\x_2=0.0181M\\x_3=0.0153M[/tex]
But the correct solution is [tex]x_3=0.0152M[/tex] since the other solutions make the equilibrium concentration of Z negative which is not possible. In such a way, its concentration at equilibrium is:
[tex][Z]_{eq}=0.033M-2(0.0153M)[/tex]
[tex][Z]_{eq}=2.4x10^{-3}M[/tex]
Which is clearly less than 0.033 M since the addition of a product shift the reaction leftwards in order to reestablish equilibrium (Le Chatelier's principle).
Regards.
Round off the following measurement to three significant digits: 29.950g
Answer:
30.0 g.
Explanation:
Hello,
In this case, for us to round the given number off to three significant figures, we firstly realize it has initially five significant figures. Thus, cutting at the third digit, which is the second nine, we will have 29.9 g, nonetheless, as a five is after such nine, we should round the nine to ten, so the result is 30.0 g.
Best regards.
A compound is known to be Na2CO3, Na2SO4, NaOH, NaCl, NaC2H3O2, or NaNO3. When a barium nitrate solution is added to a solution containing the unknown a white precipitate forms. No precipitate is observed when a magnesium nitrate solution is added to a solution containing the unknown. What is the identity of the unknown compound
Answer:
Na₂SO₄
Explanation:
Barium nitrate, Ba(NO₃)₂ produce precipitate with SO₄²⁻, CO₃²⁻. That means the precipitate could be obtained from Na₂SO₄ and Na₂CO₃.
Also, magnesium nitrate, Mg(NO₃)₂, produce precipitate just with CO₃²⁻. As the unknown solution produce no precipitate, the unknown compound is:
Na₂SO₄
The height of the sun in the sky at noontime is called
Answer:
It is called 'solar noon'
Explanation:
It's when the sun is at its highest
And also the moment the sun crosses the meridian.
what is the color of benzene and bromine
Explanation:
Benzene is colorless, with a sweet odour.
Color of Bromine is reddish brown .
Hope this helps.
A compound has an empirical formula of CHN. What is the molecular formula, if it’s molar mass is 135.13 g/mol? (C=12.01 amu, H=1.008 amu, N= 14.01)
Answer:
well the MF is 224.78 g/mol
Explanation:
just times them all by the molor mass and divide it by 3
Which diagram represents this molecule?
Answer:
C
Explanation:
The molecule has 8 carbon atoms joined by 7 C-C bonds.
The first two diagrams show 6 carbon atoms, not 8.
The last two diagrams show line segments representing C-C bonds. Only choice C shows 7 such segments.
The appropriate choice is C.
Answer:
C.
Explanation:
The reaction of hydrogen and iodine to produce hydrogen iodide has a Kc of 54.3 at 703 K. Given the initial concentrations of H2 and I2 are 0.453 M, what will the concentration of HI be at equilibrium
Answer:
[HI] = 0.7126 M
Explanation:
Step 1: Data given
Kc = 54.3
Temperature = 703 K
Initial concentration of H2 and I2 = 0.453 M
Step 2: the balanced equation
H2 + I2 ⇆ 2HI
Step 3: The initial concentration
[H2] = 0.453 M
[I2] = 0.453 M
[HI] = 0 M
Step 4: The concentration at equilibrium
[H2] = 0.453 - X
[I2] = 0.453 - X
[HI] = 2X
Step 5: Calculate Kc
Kc = [Hi]² / [H2][I2]
54.3 = 4x² / (0.453 - X(0.453-X)
X = 0.3563
[H2] = 0.453 - 0.3563 = 0.0967 M
[I2] = 0.453 - 0.3563 = 0.0967 M
[HI] = 2X = 2*0.3563 = 0.7126 M
If a pork roast must absorb 1500 kJkJ to fully cook, and if only 14% of the heat produced by the barbeque is actually absorbed by the roast, what mass of CO2CO2 is emitted into the atmosphere during the grilling of the pork roast?
You need to know the amount of heat generated by the combustion reaction.
Assuming propane as fuel, you can use thiis data:
C3H8(g)+5O2(g)---3CO2(g)+4H2O(g) ΔH= -2217 KJ
So when 3 moles of CO2 is emmitted 2217 kJ of heat is produced.
The molar wegiht of CO2 is 12 g/mol + 2 * 16 g/mol = 44 g/mol.
Then 3 mol * 44 g / mol = 132 g of CO2 are produced with 2217 kJ of heat.
Now you have to calculate how much energy you need to produce if only 12% is abosrbed by the pork
Energy absorbed by the pork = 12% * total energy =>
total energy = energy absorbed by the pork / 0.12 = 1700 kJ / 0.12 = 14,166.67 kJ.
Now, state the proportion:
132 g CO2 / 2217 kJ = x / 14,166.7 kJ =>
x = 14,166.67 * 132 / 2217 = 843.48 g CO2.
Answer: 843 g of CO2
what is the balanced equation for calcium sulfate?
Answer:
CaSO4
Explanation:
Calcium sulfate (or calcium sulphate) is the inorganic compound with the formula CaSO4 and related hydrates.
Give two examples (i.e. list 2 elements that are examples) of: a. an atom with a half-filled subshell b. an atom with a completely filled outer shell c. an atom with its outer electrons occupying a half-filled subshell and a filled subshell
Answer:
an atom with a half-filled subshell - hydrogen
an atom with a completely filled outer shell - argon
an atom with its outer electrons occupying a half-filled subshell and a filled subshell- copper
Explanation:
The outermost shell or the valence shell of the atom is the last shell in the atom. Chemical reactions occur at this outer most shell. The number of electrons on the outermost shell of an atom determines the group to which it belongs in the periodic table as well as its chemical properties.
Hydrogen has a half filled 1s sublevel. Only one electron is present in this sublevel.
Let us consider argon
1s2 2s2 2p6 3s2 3p6
The outermost ns and np levels are completely filled. Thus the outermost shell is completely filled.
In the last case; let us look at the electronic configuration of nitrogen;
1s2 2s2 2p3
The outermost 2p subshell is exactly half filled while the 2s sublevel is fully filled. The outermost shell of nitrogen is made up of 2s2 and 2p3 sublevels.
What are extensive properties of Oxygen?
What is the limiting reactant if 8 g of Ba reacts with 2.8 g of Al2(SO4)3?
Answer:
Al2(SO4)3 is the limiting reactant
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
3Ba + Al2(SO4)3 → 2Al + 3BaSO4
Next, we shall determine the mass of Ba and the mass of Al2(SO4)3 that reacted from the balanced equation. This is illustrated below:
Molar mass of Ba = 137g/mol
Mass of Ba from the balanced equation = 3 x 137 = 411g
Molar mass of Al2(SO4)3 = 2x27 + 3[32 + (16x4)]
= 54 + 3[32 + 64]
= 54 + 3[96]
= 54 + 288 = 342g/mol
Mass of Al2(SO4)3 from the balanced equation = 1 x 342 = 342g
Summary:
From the balanced equation above,
411g of Ba reacted with 342g of Al2(SO4)3.
Finally, we shall determine the limiting reactant as follow:
From the balanced equation above,
411g of Ba reacted with 342g of Al2(SO4)3.
Therefore, 8g of Ba will react with
= (8 x 342/411 = 6.66g of Al2(SO4)3.
From the calculations made above, we can see that it will take a higher mass of Al2(SO4)3 i.e 6.66g than what was given i.e 2.8g to react completely with 8g of Ba.
Therefore, Al2(SO4)3 is the limiting reactant and Ba is the excess reactant.