Answer the following question: In a space shuttle, the CO2 that the crew exhales is removed from the air by a reaction within canisters of lithium hydroxide. Let's assume that one astronaut exhales about 825. L of CO2 daily. What mass of water will be produced when this amount reacts with LiOH? The other product of the reaction is Li2CO3. When answering this question include the following:
Have both the unbalanced and balanced chemical equations.
Explain how to find the molar mass of the compounds.
Explain how the balanced chemical equation is used to find the ratio of moles (hint: step 3 in the video).
Explain how many significant figures your answer needs to have.
The numerical answer
Your Answer:
Answer:
See Explanation
Explanation:
The unbalanced reaction equation is;
CO2(g) + LiOH(aq) + Li2CO3(aq) + H2O(l)
The balanced chemical reaction equation;
CO2(g) + 2LiOH(aq) + Li2CO3(aq) + H2O(l)
The molar mass of each compound is the sum of the relative atomic masses of the atoms of the elements that compose the compounds.
For CO2
12 + 16(2) = 12 + 32 = 44 g/mol
For LiOH
7 + 16 + 1 = 24 g/mol
For Li2CO3
2(7) + 12 + 3(16) = 74 g/mol
For H2O
2(1) + 16 = 18 g/mol
From the balanced chemical reaction equation, the mole ratio is 1:2:1:1. This is obtained from the stoichiometric coefficient of each reactant in the balanced chemical reaction equation.
From the question;
1 mole of CO2 occupies 22.4 L
x moles of CO2 occupies 825 L
x = 1 mole * 825 L/22.4 L
x = 36.8 moles
From the reaction equation;
1 mole of CO2 produces 1 mole of water
36.8 moles of CO2 produces 36.8 moles of water
Mass of 36.8 moles of water = 36.8 moles * 18 g/mol
= 662 g of water
The answer must have three significant figures because that is the same number of significant figures to which values were given in the question.
Zn(s), Zn(NO3)2 (0.3 M) || Cu(s), CuCl2 (0.5 M)
(Need help solving these questions, so I can solve the rest of the questions like this in my lab. It is greatly appreciated. Thank you)
2. Write the half reactions that occur at the anode and cathode in this electrochemical cell. Annotate which half reaction occurs at the anode and the cathode.
3. Write the overall balance reaction of this electrochemical cell.
4. Calculate E°cell of this electrochemical cell. (include units)
5. Calculate the reaction quotient (Q) of this reaction.
6. Calculate the expected Ecell for this reaction
Answer:
See Explanation
Explanation:
At the anode;
Zn(s) -----> Zn^2+(aq) + 2e
At the cathode;
Cu^2+(aq) + 2e ------> Cu(s)
Overall electrochemical reaction;
Zn(s) + Cu^2+(aq) ------> Zn^2+(aq) + Cu(s)
E°cell = E°cathode - E°anode
E°cell = 0.34 - (-0.76)
E°cell = 1.1 V
Q = [0.3 M]/[0.5 M]
Q = 0.6
From Nernst equation;
Ecell = E°cell - 0.0592/n log Q
Ecell = 1.1 - 0.0592/2 log (0.6)
Ecell = 1.1 - 0.0296 log (0.6)
Ecell = 1.11 V
If 5.25 mL of HCl requires 4.96 mL of 0.9845 M NaOH to reach the equivalence point,
what is the concentration of the HCI?
[tex]M_{A}V_{A}=M_{B}V_{B}\\(5.25)M_{A}=(4.96)(0.9845)\\M_{A}=\frac{(4.96)(0.9845)}{5.25} \approx \boxed{0.93 \text{ M}}[/tex]
Which of the statements below about an acid-base buffer solution is/are true?
I. It can be prepared by combining a strong acid with a salt of its conjugate base.
II. It can be prepared by combining a weak acid with a salt of its conjugate base.
III. It can be prepared by combining a weak base with its conjugate acid.
IV. The pH of a buffer solution does not change when the solution is diluted.
V. A buffer solution resists changes in its pH when an acid or base is added to it.
A. I, II, and IV.
B. II, III, and V.
C. II, III, IV, and V.
D. I, II, IV, and V.
E. II, III, and IV.
Answer:
C. II, III, IV, and V.
Explanation:
Acid buffer is generally formed by the combination of a weak acid as well as the salt of the conjugate base.
Basic buffer is formed the combination of a weak base and also the salt of the conjugate acid.
On dilution the ration of the concentration terms of the salt and weak acid/base does not change. Hence the pH of the buffer solution does not change.
When acid or base is added to buffer, it resists changes in the pH.
Therefore, option (C) is correct.
ASAP PLEASE AND THANK YOU
What is the molar mass of a pure gas that has the density of 1.40 g/L at STP?
Answer:
bro what Is this like I dont even kno
Answer:
O2 is the answer I believe
what are super fuels? Describe both types with examples. In your view, what makes super fuels?
[tex]\bcancel{\huge\pink{\fbox{\tt{࿐αɴѕωєя࿐}}}}[/tex]
Super fuel or high performance fuel is petrol with a higher octane rating. Most standard brands of fuel have a 95 octane rating. But super fuel typically has a higher rating - around 98 - which can make the engine work more efficiently and improve performance.Why would a doctor most likely restrict a patient's contact with other people while the patient receives internal
radiation?
The patient's stress and anxiety would be eliminated.
High levels of radiation can diffuse through the patient's skin.
Social contact would increase the effect of the radiation treatment.
Radioactive material can leave the patient's body through saliva, sweat, and urine.
D. Radioactive material can leave the patient’s body through saliva, sweat, and urine.
I took the test n got it right ¯\_(ツ)_/¯
A sample of Ne(g) has a volume 250 mL at 752 mm Hg. What is the
new volume if the temperature and amount of gas held constant, the
pressure is;
a) lowered to 385 mm Hg.
b) Increased to 3.68 atm.
Answer: a) 525 ml
b) 67.2 ml
Explanation:
Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.
[tex]P\propto \frac{1}{V}[/tex] (At constant temperature and number of moles)
[tex]P_1V_1=P_2V_2[/tex]
where,
a) [tex]P_1[/tex] = initial pressure of gas = 752 mm Hg
[tex]P_2[/tex] = final pressure of gas = 385 mm Hg
[tex]V_1[/tex] = initial volume of gas = 250 ml
[tex]V_2[/tex] = final volume of gas = ?
[tex]752\times 250=385\times V_2[/tex]
[tex]V_2=525ml[/tex]
Therefore, the volume at 385 mm Hg is 525 ml.
b) [tex]P_1[/tex] = initial pressure of gas = 752 mm Hg
[tex]P_2[/tex] = final pressure of gas = 3.68 atm = 2796.8 mm Hg (760mmHg=1atm)
[tex]V_1[/tex] = initial volume of gas = 250 ml
[tex]V_2[/tex] = final volume of gas = ?
[tex]752\times 250=2796.8\times V_2[/tex]
[tex]V_2=67.2ml[/tex]
Therefore, the volume at 3.68 atm is 67.2 ml.
Need help with this question please.
Answer:
12.8
Explanation:
14 = pOH + pH
pH = 14 - pOH
pH = 14 - 1.2
pH = 12.8
What is the formula mass of ZrF4?
Answer:
167.217g/mol
Explanation:
Formula mass is defined as the mass in grams that a mole of a molecule weighs. To solve the formula mass of ZrF₄ we require the molar mass of Zr and of F (Molar mass Zr: 91.225g/mol; F: 18.998g/mol)
In this molecule, there is 1 mole of Zr and 4 moles of F. The formula mass is:
Zr = 1*91.225g/mol = 91.225g/mol
F = 4*18.998g/mol = 75.992g/mol
Formula mass: 91.225g/mol + 75.992g/mol
167.217g/molWhich is a correct comparison between the modern quantum model and John Dalton’s model of the atom?
Answer:
No comparison => John Dalton did NOT postulate an atomic structure.
Explanation:
Dalton's postulates are based upon generalizations of observable phenomena. Micro structure was not proposed. The atomic structure that's accepted in the scientific community evolved from conclusions of ...
Thompson's raisen pudding model => electrons suspended in a positive matrix
Rutherford's shell model => dense positive nucleus surrounded by an electron cloud. (No fine structure postulated)
Bohr Concentric Ring Model => electrons orbiting a positive nucleus in discrete energy levels much like planets orbiting the sun.
Schrodinger - Dirac Quantum Models => electrons occupying specific orbital energy levels as based upon statistical wave mechanics.
D = 22.1 g/cm3, M = 523.1 g, V = ? mL
Answer:
The volume is 22.66 mL
Explanation:
D = 22.1 g/cm³, M = 523.1 g, V = ? mL
For Volume
Density = Mass ÷ Volume
Volume = Mass ÷ Density
V = M ÷ D
V = 523.1 g ÷ 22.1 g/cm³
V = 23.66 cm³
Now, 1 cm³ = 1 mL
So,
V = 22.66 mL
Thus, The volume is 22.66 mL
-TheUnknownScientist
Draw a structural formula for the organic product formed by treating butanal with the following reagent: NaBH4 in CH3OH/H2O You do not have to consider stereochemistry. You do not have to explicitly draw H atoms. Do not include lone pairs in your answer. They will not be considered in the grading. Include counter-ions, e.g., Na , I-, in your submission, but draw them in their own separate sketcher. Do not draw organic or inorganic by-products.
Answer:
Please find the solution in the attachment file.
Explanation:
Which product is often derived from the natural environment?
Answer:
coal, cotton, chinese tradictional medicine
a layer of paint can be used to prevent iron rusting true or false
Answer:
true
Explanation:
This layer will prevent moisture from reaching the metal and therefore prevent rust. oil paint especially
What was the purpose of letting the transformed cells sit in LB for a few minutes before spreading them onto the plates?
A. This allows time for the cells to express the antibiotic resistance gene
B. This allows the cells to take up the plasmid after the heat shock procedure
C. This allows time for the cells to warm up before plating
D. This allows cells time to start glowing green
Answer: D
Explanation:
When a lead acid car battery is recharged by the alternator, it acts essentially as an electrolytic cell in which solid lead(II) sulfate is reduced to lead at the cathode and oxidized to solid lead(II) oxide at the anode. Suppose a current of is fed into a car battery for seconds. Calculate the mass of lead deposited on the cathode of the battery. Be sure your answer has a unit symbol and the correct number of significant digits.
The question is incomplete, the complete question is;
When a lead acid car battery is recharged by the alternator, it acts essentially as an electrolytic cell in which solid lead(II) sulfate PbSO₄ is reduced to lead at the cathode and oxidized to solid lead(II) oxide PbO at the anode.
Suppose a current of 96.0 A is fed into a car battery for 37.0 seconds. Calculate the mass of lead deposited on the cathode of the battery. Round your answer to 3 significant digits. Also, be sure your answer contains a unit symbol.
Answer:
3.81 g of lead
Explanation:
The equation of the reaction is;
Pb^2+(aq) + 2e ---->Pb(s)
Quantity of charge = 96.0 A * 37.0 seconds = 3552 C
Now we have that 1F = 96500 C so;
207 g of lead is deposited by 2 * 96500 C
x g of lead is deposited by 3552 C
x = 207 * 3552/2 * 96500
x = 735264/193000
x = 3.81 g of lead
i need help can someone help me
Answer:
Option D. The number of oxygen atom is the same before and after the reaction.
Explanation:
From the question given above, the following were obtained:
Robin's equation:
H₂ + O₂ —> H₂O
Alex's equation
2H₂ + O₂ —> 2H₂O
To know which equation better represents the reaction, we shall determine which of the equation is balanced.
For Robin:
H₂ + O₂ —> H₂O
Element >>> Reactant >>> Product
H >>>>>>>>> 2 >>>>>>>>>> 2
O >>>>>>>>> 2 >>>>>>>>>> 1
Robin's equation is not balanced because the number of atoms of each element in the reactant and product are not equal.
For Alex:
2H₂ + O₂ —> 2H₂O
Element >>> Reactant >>> Product
H >>>>>>>>> 4 >>>>>>>>>> 4
O >>>>>>>>> 2 >>>>>>>>>> 2
Alex' equation is balanced because the number of atoms of each element in the reactant and product are equal.
Thus, option D gives the right answer to the question.
Calculate the density, in grams per liter, of a gas at STP if 3.56 L of the gas at 36.7 °C and 758.5 mmHg weighs 0.433 g.
density:? g/L
Answer:
the density of the Gas at STP is 0.227 g/L .
Explanation:
This question involves the combined gas law . The equation for the combined gas law
If aluminum has a mass of 22.3 g, how many liters of oxygen gas are required at STP?
Answer:
27.8
Explanation:
Which of the following is an example of chemistry in forensic science?
Estimating the trajectory and impact of bullet
Calculating the force needed for a body to have fallen at a specific location Using blood spatter patterns to establish how a body was stabbed
Extending DNA strands to match samples
Answer:
Extending DNA strands to match samples
If you dilute 18.8 mL of a 3.5 M solution to make 296.6 mL of solution, what is the molarity of the dilute solution?
Answer:
0.22M
Explanation:
We will be using the law of dilutions. We are simply increasing the amount of solvent to create a larger volume of solution.
So: moles before dilution = moles after dilution & [tex]moles_{concentrated} = moles_{dilute}[/tex]. And M = moles/liter of solution, so if we express this as moles = M x [tex]L_{soln}[/tex].
That is how we derive the formula we will be using: [tex]M_{concentrated} * Vol_{conc} = M_{dilute} * Vol_{dilute}[/tex]
or
[tex]M_{1} * Vol_{1} = M_{2} * Vol_{2}[/tex]
Applying this formula to our problem, we can substitute the variables with the given values to find the molarity of the dilute solution.
M1 = 3.5M
V1 = 18.8mL
M2 = ?
V2 = 296.6mL
Equation: (3.5M)(18.8mL) = (296.6mL)(M2)
==> 65.8M*mL = 296.6mL * M2
==> M2 = (65.8 M*mL)/296.6mL
==> M2 = 0.22M
Can someone help me please!!
Answer:
1. AgNO₃ (aq) + NaCl (aq) ----> NaNO₃ (aq) + AgCl (s)
2. Li₂SO₄ (aq) + BaCl₂ (aq) ----> 2 LiCl (aq) + BaSO₄ (s)
3. 2 NaOH (aq) + MgCl₂ (aq) ----> 2 NaCl (aq) + Mg(OH)₂ (s)
Explanation:
The reaction involving the mixing of two soluble solutions to produce a precipitate is known as a precipitation reaction.
A precipitation reaction is double-replacement reaction (a reaction that exchanges the cations or the anions of two ionic compounds) in which one product is a solid precipitate.
Precipitation reactions at useful in the identification of various ions present in a solution. In order to predict the reactions that will produce a precipitate, solubility rules as given in the solubility table below can be used.
From the tables, the reactions that will produce a precipitate, as well as their balanced molecular equations are as follows:
1. AgNO₃ (aq) + NaCl (aq) ----> NaNO₃ (aq) + AgCl (s)
2. Li₂SO₄ (aq) + BaCl₂ (aq) ----> 2 LiCl (aq) + BaSO₄ (s)
3. 2 NaOH (aq) + MgCl₂ (aq) ----> 2 NaCl (aq) + Mg(OH)₂ (s)
Express your answer as a balanced half-reaction. Identify all of the phases in your answer.
(acidic) Cr2O7 2−(aq)⟶Cr 3+(aq)
(acidic) CrO4 2−(aq)⟶Cr(OH)4 −(aq)
(acidic) Bi 3+(aq)⟶BiO3 −(aq)
(acidic) CIO −(aq)⟶Cl −(aq)
(^for CIO - that is an i not an L)
Answer:
1. Cr₂O₇²−(aq) + 14H+ (aq) + 6e- ---->⟶2 Cr³+(aq) + 7H₂O (l)
2. CrO₄²− (aq)⟶+ 4H+ (aq) + 3e- ---> Cr(OH)₄ −(aq)
3. Bi³+ (aq) + 3H₂O (l) ---> BiO₃− (aq) + 6 H+ (aq) + 2 e-
4. CIO −(aq)⟶+ 2H+ (aq) + 2e- ---> Cl −(aq) + H₂O
Explanation:
The given equations are redox reaction equations expressed as as half reactions.
The first step is to identify whether the half-reaction is oxidation reduction.
Then the number of electrons gained or lost are added on the right side of the equation.
Appropriate H+ ions and water molecules are added where necessary since the reaction takes place in acidic environment
The atoms of elements involved in the reaction are balanced by adding the correct coefficients.
1. (acidic) Cr2O7 2−(aq)⟶Cr 3+(aq)
The half-reaction is reduction as the oxidation number of chromium changes from +6 to +3. Two Cr⁶+ ions accepts 3 electrons each to form Cr³+ ions
Cr₂O₇²−(aq) + 6e- ---->⟶2 Cr³+(aq)
Cr₂O₇²−(aq) + 14H+ (aq) + 6e- ---->⟶2 Cr³+(aq) + 7H₂O (l)
2. (acidic) CrO₄²− (aq)⟶---> Cr(OH)₄ −(aq)
The half-reaction is a reduction. One Cr⁶+ accepts 3 electrons to become Cr³+
CrO₄²− (aq)⟶+ 3e- ---> Cr(OH)₄ −(aq)
CrO₄²− (aq)⟶+ 4H+ (aq) + 3e- ---> Cr(OH)₄ −(aq)
3, (acidic) Bi³+ (aq)⟶---> BiO₃− (aq)
The half-reaction is an oxidation. One Bi³+ ion gives up two electrons to become Bi⁵+
Bi³+ (aq)⟶---> BiO₃− (aq) + 2e-
Bi³+ (aq) + 3H₂O (l) ---> BiO₃− (aq) + 6 H+ (aq) + 2 e-
4. (acidic) CIO −(aq)⟶---> Cl −(aq)
The half-reaction is a reduction. One Cl+ ion accepts two electrons to become Cl- ion.
CIO −(aq) + 2e-⟶---> Cl −(aq)
CIO −(aq)⟶+ 2H+ (aq) + 2e- ---> Cl −(aq) + H₂O
Chemical formula for Aluminum Oxide
Answer: Al₂O₃
Explanation:
How many moles of argon, Ar, are in 1.31×1024 Ar atoms?
Answer:
2.18 mol Ar.
Explanation:
Hello there!
In this case, according to the definition of mole as the amount of particles of a given substance, it is possible to introduce the Avogadro's number to assert that 1 mole of any element contains 6.022x10²³ atoms; thus, the moles of Ar in the given amount of atoms turns out to be:
[tex]1.31x10^{24}Ar atoms*\frac{1molAr}{6.022x10^{23}atoms}\\\\=2.18molAr[/tex]
Best regards!
A que se denomina función química
En química, el grupo de algunas sustancias compuestas que poseen propiedades químicas semejantes, denominadas propiedades funcionales, recibe el nombre de función química. ... Además están divididas en ácidos, bases, sales y óxidos; y funciones orgánicas que son las relativas a los compuestos orgánicos.
BRAINLIST
13. According to Arrhenius definition which of the following is an acid *
1 point
NaCl
КСІ
Ο Ο Ο
HCI
Al(OH)3
Answer: hcl
Explanation:
Caffeine is a compound found in some natural coffees and teas and in some colas. a. Determine the empirical formula for caffeine, using the following composition of a 100.00-g sample. 49.47 grams of carbon, 28.85 grams of nitrogen, 16.48 grams of oxygen, and 5.20 grams of hydrogen b. If the molar mass of caffeine is 194.19 g/mol, calculate its molecular formula.
Answer: The molecular formula will be [tex]C_8N_4H_{10}O_2[/tex]
Explanation:
Mass of C= 49.47 g
Mass of N = 28.85 g
Mass of O = 16.48 g
Mass of H = 5.20 g
Step 1 : convert given masses into moles.
Moles of C =[tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{49.47g}{12g/mole}=4.12moles[/tex]
Moles of N =[tex]\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{28.85g}{14g/mole}=2.06moles[/tex]
Moles of O =[tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{16.48g}{16g/mole}=1.03moles[/tex]
Moles of H =[tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{5.20g}{1g/mole}=5.20moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = [tex]\frac{4.12}{1.03}=4[/tex]
For N = [tex]\frac{2.06}{1.03}=2[/tex]
For O =[tex]\frac{1.03}{1.03}=1[/tex]
For H = [tex]\frac{5.20}{1.03}=5[/tex]
The ratio of C : N: O: H = 4: 2: 1: 5
Hence the empirical formula is [tex]C_4N_2OH_5[/tex]
The empirical weight of [tex]C_4N_2OH_5[/tex] = 4(12)+2(14)+1(16)+5(1)= 97 g.
The molecular weight = 194.19 g/mole
Now we have to calculate the molecular formula.
[tex]n=\frac{\text{Molecular weight}}{\text{Equivalent weight}}=\frac{194.19}{97}=2[/tex]
The molecular formula will be = [tex]2\times C_4N_2H_5O=C_8N_4H_{10}O_2[/tex]
An analytical chemist is titrating 111.0 mL of a 0.3700 M solution of aniline (C6H5NH2) with a 0.3500 M solution of HNO3. The pK_b of aniline is 9.37. Calculate the pH of the base solution after the chemist has added 79.1 mL of the HNO_3 solution to it.
Answer:
The answer is "4.31"
Explanation:
aniline millimoles [tex]= 111 \times 0.37 = 41.07[/tex]
added [tex]HNO_3[/tex] millimoles [tex]= 79.1 \times 0.35 = 27.685[/tex]
[tex]\to 41.07 - 27.685 = 13.385[/tex] millimoles aniline left
[tex]\to 27.685[/tex] millimoles salt formed
total volume[tex]= 111 + 79.1 = 190.1\ mL\\\\[/tex]
[tex]\to [aniline] = \frac{13.385}{190.1} = 0.07 \ M\\\\\to [salt] =\frac{ 27.685}{ 190.1} = 0.146\ M\\\\\to pOH = pKb + \frac{\log [salt]}{ [base]}\\\\\to pOH = 9.37 + \frac{\log [0.146]}{[0.07]}\\\\\to pOH = 9.69\\\\\to pH = 14 - 9.69\\\\\to pH = 4.31\\[/tex]