Answer:
1.8x10¹⁷ molecules of CO are in each breath we take
Explanation:
Parts per million, ppm, is an unit of concentration in chemistry used for very diluted solutions.
A 9ppm of X in a solution means in 1 million of molecules (1x10⁶) you have only 9 molecules of X.
In a breath we have 2x10²² molecules and 9 ppm are CO. Thus, CO molecules in each breath are:
2x10²² molecules × (9 molecules CO / 1x10⁶ molecules) =
1.8x10¹⁷ molecules of CO are in each breath we take[tex]1.8\times 10^{17}[/tex] molecules of CO are in each breath we take
The calculation is as follows:A 9ppm of X in a solution represent in 1 million of molecules[tex](1\times10^6)[/tex]you have only 9 molecules of X.
Now CO molecules in each breath is
[tex]= 2\times 10^{22}\ vmolecules \times (9\ molecules\ CO \div 1\times 10^6 molecules) \\\\= 1.8\times 10^{17}[/tex]
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A 635 mL NaCl solution is diluted to a volume of 1.13 L and a concentration of 5.00 M . What was the initial concentration C1?
Answer:
8.90 M
Explanation:
Step 1: Given data
Initial concentration (C₁): ?Initial volume (V₁): 635 mL = 0.635 LFinal concentration (C₂): 5.00 MFinal volume (V₂): 1.13 LStep 2: Calculate the initial concentration
We have a concentrated NaCl solution and we want to prepare a diluted one. We will use the dilution rule.
C₁ × V₁ = C₂ × V₂
C₁ = C₂ × V₂ / V₁
C₁ = 5.00 M × 1.13 L / 0.635 L
C₁ = 8.90 M
Answer:
[tex]\large \boxed{\text{8.90 mol/L}}[/tex]
Explanation:
We can use the dilution formula to calculate the concentration of the original solution.
[tex]\begin{array}{rcl}V_{1}c_{1} & = & V_{2}c_{2}\\\text{635 mL }\times c_{1} & = & \text{1130 mL} \times \text{5.00 mol/L}\\635 c_{1}&=& \text{5650 mol/L}\\c_{1}& = & \dfrac{5650}{635}\text{ mol/L}\\\\& = & \textbf{8.90 mol/L}\\\end{array}\\\text{The initial concentration was $\large \boxed{\textbf{8.90 mol/L }}$}[/tex]
Read the article. Use your understanding to answer the questions that follow. What type of source is this article? primary or secondary and how do you know
Answer: C
Explanation:
The article was sourced from the Oak National Laboratory
Which reasons did you include in your response? Check all of the boxes that apply.
1. The article does not present original research.
and
3. The article has references to primary sources.
Answer:
C
Explanation:
Which reasons did you include in your response? Check all of the boxes that apply.
The article does not present original research.
The article summarizes other research.
The article has references to primary sources.
NEED HELP ASAP
In 1988, three gray whales were trapped in Arctic ice. Television crews captured the frantic
attempts of hundreds of people to save the whales. Eventually, a Soviet icebreaker and U.S.
National Guard helicopters arrived to help free the whales. The cost of the rescue mission
exceeded $5 million.
i. Write a scientific question related to the whale story. (1 point)
A substance, W has a concentration of 0.02mol when its molar mass was found to be
74.0 gmol-1 . Another substance V contains 1.00x1023 atoms and has molar mass of
40.0gmol-1. Which of the two substances has the greater mass (in grams)? [L =
6.02x1023]
ii) A 250 cm3
solution contains 14.63g of sodium chloride (NaCl). Calculate the
concentration of the solution in moldm-3
[Na= 23, Cl = 35.5]
Answer:
Explanation:
mass of W in gram = mole x molecular weight
= .02 x 74 = 1.48 gm
mass of V in gram
first of all we shall calculate the no of moles of V
1 mole = 6.0 x 10²³ atoms
1 x 10²³ atoms = 1 / 6 moles
mass of V in grams
= 40 / 6
= 6.67 grams .
So V has greater mass .
ii )
molecular weight of sodium chloride
= 58.5 gm
14.63 gram of sodium chloride
= 14.63 / 58.5 = .25 moles
250 cm³ = 250 x 10⁻³ dm³
So 250 x 10⁻³ dm³ of solution contains .25 moles of salt
1 dm³ of solution will contain .25 / 250 x 10⁻³ mole
= 1 mole
so concentration of solution is 1 mole per dm³
Given that the Ksp value for Ca3(PO4)2 is 8.6×10−19, if the concentration of Ca2+ in solution is 4.9×10−5 M, the concentration of PO3−4 must exceed _____ to generate a precipitate.
Answer:
.0027 M
Explanation:
We must calculate the threshold concentration of PO3−4 using Ksp and the given concentration of Ca2+:
Ca3(PO4)2(s)⇌3Ca2+(aq)+2PO3−4(aq)
Ksp=8.6×10−19=[Ca2+]3[PO3−4]2=(4.9×10−5M)3[PO3−4]2
[PO3−4]=0.0027 M
Aqueous potassium nitrate (KNO3) and solid silver bromide are formed by the reaction of aqueous potassium bromide and aqueous silver nitrate (AgNO3). Write a balanced chemical equation for this reaction
Answer:
For the mentioned reaction, the balanced chemical equation is:
KBr (aq) + AgNO3 (s) ⇒ KNO3 (aq) + AgBr (s)
The number written in front of the ion, atoms, and molecules in a chemical reaction so that each of the elements on both the sides of reactants and products of the equation gets balanced is known as the stoichiometric coefficient.
From the mentioned balanced equation, the stoichiometric coefficient before KBr is 1, AgNO3 is 1, KNO3 is 1, as well as before AgBr is also 1. Thus, it is clear that 1 mole of potassium bromide reacts with 1 mole of silver nitrate to produce 1 mole of potassium nitrate and 1 mole of silver bromide.
How many dozen (dz) eggs are needed to make 12 muffins? What about 15.5
muffins? (hint cross out units first) *
Answer:
I think its 1.2 cause I divided 15.5 with 12 and got 1.2 as an answer
Why don't siblings look exactly alike
Answer:
Your genes play a big role in making you who you are. ... But brothers and sisters don't look exactly alike because everyone (including parents) actually has two copies of most of their genes. And these copies can be different. Parents pass one of their two copies of each of their genes to their kids.
For which of the following elements (in their normal, stable, forms) would it be correct to describe the bonding as involving "a sea of electrons"?
a. hydrogen
b. nellum
c. sulfur
d. Iodine
e. Ethium
Answer:
e. Lithium
Explanation:
Correct list of options!
a. hydrogen b. Helium c. sulfur d. Iodine e. Lithium
Sea of electrons generally refers to metal atoms. This is because of the delocalized nature of the electrons compared to non metals where the electrons are localized (fixed to a specific atoms).
Among all the elements in the options, the metal is option e. Lithium
What can you learn about the pH of a substance with the conductivity test? hint: gives you no info on concentration.
Answer:
See explanation
Explanation:
So, I'm gonna take a shot at this one and say this:
With a strongly acidic/basic solution, you'll get a high conductivity when preforming a conductivity test.
The more acidic or basic a substance is, the higher the electrical conductivity.
Based on how high or low the conductivity is, it will give you an idea of the substance's pH.
Hope that made since or gave you an idea of what you're looking for. Good luck :)
What is the pH of a solution made by mixing 15.00 mL of 0.10 M acetic acid with 15.00 mL of 0.10 M KOH? Assume that the volumes of the solutions are additive. K a = 1.8 ×× 10-5 for CH3CO2H.
Answer:
pH = 8.72
Explanation:
This is like a titration of a weak acid and a strong base, in this case, we are at the equivalence point plus we have the same mmoles of acid and base. We have completely neutralized the acid.
CH₃COOH + OH⁻ ⇄ CH₃COO⁻ + H₂O
0.1M . 15 mL 0.1M . 15 mL
We only have (0.1M . 15 mL) mmoles of acetate ion. → 1.5 mmoles
As this compound acts like a base, we propose this equilibrium:
CH₃COO⁻ + H₂O ⇄ CH₃COOH + OH⁻ Kb
We need to work with Kb and we know, that Kw = Ka. Kb so, Kb = Kw/Ka
Kb = 1×10⁻¹⁴ /1×10 ⁻⁵ = 5.55×10⁻¹⁰
Concentration of CH₃COO⁻ → 1.5 mmol / 30mL (volumes of the solutions are additive) = 0.05M
So: [CH₃COOH] . [OH⁻] / [CH₃COO⁻] = Kb
x²/ 0.05-x = 5.55×10⁻¹⁰
We can avoid the quadractic equation because Kb is so small
[OH⁻] = √(5.55×10⁻¹⁰ . 0.05) = 5.27×10⁻⁶
pOH = - log [OH⁻] → 5.28
pH = 14 - pOH = 8.72
The pH of a solution made by mixing 15.00 mL of 0.10 M acetic acid should be 8.72.
Calculation of the pH of the solution:Since the following equation should be used.
CH₃COOH + OH⁻ ⇄ CH₃COO⁻ + H₂O
0.1M . 15 mL 0.1M . 15 mL
Now
(0.1M . 15 mL) mmoles of acetate ion. → 1.5 mmoles
So,
CH₃COO⁻ + H₂O ⇄ CH₃COOH + OH⁻ Kb
Now
Kw = Ka. Kb
Kb = Kw/Ka
And,
Kb = 1×10⁻¹⁴ /1×10 ⁻⁵
= 5.55×10⁻¹⁰
Now
[CH₃COOH] . [OH⁻] / [CH₃COO⁻] = Kb
x²/ 0.05-x = 5.55×10⁻¹⁰
Now
[OH⁻] = √(5.55×10⁻¹⁰ . 0.05) = 5.27×10⁻⁶
pOH = - log [OH⁻] → 5.28
pH = 14 - pOH
= 8.72
Hence, The pH of a solution made by mixing 15.00 mL of 0.10 M acetic acid should be 8.72.
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A student wants to examine a substance by altering the bonds within its molecules. Which of the following properties of the substance should the student examine?
Question is incomplete, the complete question is as follows:
A student wants to examine a substance by altering the bonds within its molecules. Which of the following properties of the substance should the student examine?
A. Toxicity, because it can be observed by altering the state of the substance
B. Boiling point, because it can be observed by altering the state of the substance
C. Toxicity, because it can be observed by replacing the atoms of the substance with new atoms
D. Boiling point, because it can be observed by replacing the atoms of the substance with new atoms
Answer:
B.
Explanation:
A student can examine a substance without altering the bonds within the molecules by examining its boiling point.
The boiling point is the property of a substance, at which the substance changes its state, which is from solid to liquid, liquid to gas and others. So, examining the boiling point will alter the bonds within the molecules as the state of substance will change.
Hence, the correct answer is "B".
Which of the following processes have a ΔS < 0? Which of the following processes have a ΔS < 0? carbon dioxide(g) → carbon dioxide(s) water freezes propanol (g, at 555 K) → propanol (g, at 400 K) methyl alcohol condenses All of the above processes have a ΔS < 0.
Answer:
All of the above processes have a ΔS < 0.
Explanation:
ΔS represents change in entropy of a system. Entropy refers to the degree of disorderliness of a system.
The question requests us to identify the process that has a negative change of entropy.
carbon dioxide(g) → carbon dioxide(s)
There is a change in state from gas to solid. Solid particles are more ordered than gas particles so this is a negative change in entropy.
water freezes
There is a change in state from liquid to solid. Solid particles are more ordered than liquid particles so this is a negative change in entropy.
propanol (g, at 555 K) → propanol (g, at 400 K)
Temperature is directly proportional to entropy, this means higher temperature leads t higher entropy.
This reaction highlights a drop in temperature which means a negative change in entropy.
methyl alcohol condenses
Condensation is the change in state from gas to liquid. Liquid particles are more ordered than gas particles so this is a negative change in entropy.
How many liters of CH₃OH gas are formed when 3.20 L of H₂ gas are completely reacted at STP according to the following chemical reaction? Remember 1 mol of an ideal gas has a volume of 22.4 L at STP.CO(g)+ H2(g) → CH3OH
Answer:
The correct answer is 1.60 Liters.
Explanation:
The given reaction:
CO (g) + H₂(g) ⇔ CH₃OH (g)
Based on the given reaction, two moles of H₂ reacts with one mole of CO and produce one mole of CH₃OH.
It is mentioned that 3.20 L of H₂ is reacted, therefore, there is a need to convert it into moles.
As 22.4 L at standard temperature and pressure is equivalent to 1 mole.
Therefore, 1 L at STP will be, 1/22.4 mole
Now 3.20 L at STP will be,
= 1/22.4 × 3.20
= 0.1428 mole
And as mentioned in the reaction that 2 moles of H₂ gives 1 mole of CH₃OH, therefore, 1 mole of H₂ will give 1/2 mole of CH₃OH
Now, 0.1428 mole of H₂ will give,
= 0.1428/2 = 0.071 mole of CH₃OH
= 0.071 × 22.4 = 1.60 L
The volume, in liters, of CH₃OH gas formed is 1.60 L
From the question,
We are to determine the volume of CH₃OH formed
The given chemical equation for the reaction is
CO(g)+ H₂(g) → CH₃OH
The balanced chemical equation for the reaction is
CO(g)+ 2H₂(g) → CH₃OH
This means
1 mole of CO reacts with 2 moles of H₂ to produce 1 mole of CH₃OH
Now, we will determine the number of moles of H₂ present in the 3.20 L H₂ at STP
1 mol of an ideal gas has a volume of 22.4 L at STP
Then,
x mole of the H₂ gas will have a volume of 3.20 L at STP
x = [tex]\frac{3.20 \times 1}{22.4}[/tex]
x = 0.142857 mole
∴ The number of mole of H₂ present is 0.142857 mole
Since
2 moles of H₂ reacts to produce 1 mole of CH₃OH
Then,
0.142857 mole of H₂ will react to produce [tex]\frac{0.142857}{2}[/tex] mole of CH₃OH
[tex]\frac{0.142857}{2} = 0.0714285[/tex]
∴ The number of moles of CH₃OH produced = 0.0714285 mole
Now, for the volume of CH₃OH formed
Since
1 mol of an ideal gas has a volume of 22.4 L at STP
Then,
0.0714285 mol of CH₃OH will have a volume of 22.4 × 0.0714285 at STP
22.4 × 0.0714285 = 1.5999984 L ≅ 1.60 L
Hence, the volume of CH₃OH gas formed is 1.60 L
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When HCl is added to water, the [H3O+] = 0.6 M. What is the [OH-]?
What is the pH of the solution?
Answer:
pH=0.22.
[tex][OH^-]=1.66x10^{-14}M[/tex]
Explanation:
Hello,
In this case, since the pH is directly computed from the given concentration of hydronium ions:
[tex]pH=-log([H_3O^+])=-log(0.6M)=0.22[/tex]
It is widely known that the pH and POH are directly related via:
[tex]pH+pOH=14[/tex]
Therefore, the pOH is:
[tex]pOH=14-pH\\\\pOH=14-0.22=13.78[/tex]
Thus, the concentration of hydroxyl ions are computed from the pOH:
[tex]pOH=-log([OH^-]}\\\\[/tex]
[tex][OH^-]=10^{-pOH]=10^{-13.78}[/tex]
[tex][OH^-]=1.66x10^{-14}M[/tex]
Regards.
How does the spontaneity of the process below depend on temperature? PCl5(g)+H2O(g)→POCl3(g)+2HCl(g) ΔH=−126 kJ mol−1, ΔS=146 J K−1mol−
The given question is incomplete, the complete question is:
How does the spontaneity of the process below depend on temperature? PCI5(9)+H2O(g)POCI3(g) +2HCI(g) -126 kJ mol1, AS = 146 J K-'mol1 ΔΗ Select the correct answer below: nonspontaneous at all temperatures spontaneous at all temperatures spontaneous at high temperatures and nonspontaneous at low temperatures spontaneous at low temperatures and nonspontaneous at high temperatures
Answer:
The correct answer is spontaneous at all the temperatures.
Explanation:
Gibbs Free energy is an essential relation that determines the spontaneity of any reaction, that is, ΔG = ΔH - TΔS
When ΔG is less than zero, that is, negative, the reaction is considered to be in spontaneous state. Based on the given information, ΔH = -126 kJ/mol
= -126000 J/mol, it is negative
ΔS = 146 J/K/mol, it is positive
Now, ΔG = ΔH-TΔS
= (-ve) - T (+ve), Thus, when ΔH, is -ve, ΔS is +ve, -TΔS is -ve, the ΔG will be -ve. Therefore, reaction will be spontaneous at all the temperatures.
A saturated sodium carbonate solution at 0°C contains 7.1 g of dissolved sodium carbonate per 100. mL of solution. The solubility product constant for sodium carbonate at this temperature is
Answer:
[tex]Ksp=1.2[/tex]
Explanation:
Hello,
In this case, as the saturated solution has 7.1 grams of sodium carbonate, the solubility product is computed by firstly computing the molar solubility by using its molar mass (106 g/mol):
[tex]Molar \ solubility=\frac{7.1gNa_2CO_3}{0.1L}*\frac{1molNa_2CO_3}{106gNa_2CO_3}=0.67M[/tex]
Next, as its dissociation reaction is:
[tex]Na_2CO_3(s)\rightleftharpoons 2Na^+(aq)+CO_3^{2-}(aq)[/tex]
The equilibrium expression is:
[tex]Ksp=[Na^+]^2[CO_3^{2-}][/tex]
And the concentrations are related with the molar solubility (2:1 mole ratio between ionic species):
[tex]Ksp=(2*0.67)^2*(0.67)\\\\Ksp=1.2[/tex]
Best regards.
A compound (C_9H_9BrO_2) gives the following NMR data. Draw the structure of the compound.
'1^H-NMR: 1.39 ppm, t(3H); 4.38 ppm, q(2H); 7.57 ppm, d(2H); 7.90 ppm, d(2H)
13^C-NMR: 165.73; 131.56; 131.01; 129.84; 127.81; 61.18; 14.18
You do not have to consider stereo chemistry.
You do not have to explicitly draw H atoms.
Do not include lone pairs in your answer.
Answer:
ethyl 4-bromobenzoate
Explanation:
In this question, we can start with the Index of Hydrogen Deficiency (I.H.D):
[tex]I.H.D=\frac{2C+2+N-H-X}{2}=\frac{(2*9)+2+0-9-1}{2}~=~5[/tex]
This indicates, that we can have a benzene ring (I.H.D = 4) and a carbonyl group (I.H.D = 1), for a total of 5.
Additionally, in the 1H-NMR info, we have a triplet 1.39 (3H) followed by a doublet 4.38 (2H), this indicates the presence of an ethyl group ([tex]CH_3-CH_2-[/tex]). Also, in the formula, we have 2 oxygens if we have carbonyl group with 2 oxygens we have a high probability to have an ester group.
[tex]O=C-O-CH_2-CH_3[/tex]
Now, if we add this to the benzene ring and the "Br" atom that we have in the formula, we will have ethyl 4-bromobenzoate.
See figures 1 and 2 to further explanations.
I hope it helps!
You have a saturated solution of BaSO4, a slightly soluble ionic compound. What happens if you add Ba(OH)2, NaNO3, and CuSO4 to this solution
Answer:
- Addition of Ba(OH)2: favors the formation of a precipitate.
- Undergo a chemical reaction forming soluble species.
- Addition of CuSO4 : favors the formation of a precipitate.
Explanation:
Hello,
In this case, since the dissociation reaction of barium sulfate is:
[tex]BaSO_4(s)\rightleftharpoons Ba^{2+}(aq)+SO_4^{2-}(aq)[/tex]
We must analyze the effect of the common ion:
- By adding barium hydroxide, more barium ions will be added to the equilibrium system so the formation of solid barium sulfate will be favored (reaction shifts leftwards towards reactants).
- By adding sodium nitrate, the following reaction will undergo:
[tex]BaSO_4(s)+NaNO_3(aq)\rightarrow Ba(NO_3)_2(aq)+Na_2SO_4(aq)[/tex]
So the precipitate will turn into other soluble species.
- By adding copper (II) sulfate, more sulfate ions will be added to the equilibrium system so the formation of solid barium sulfate will be favored (reaction shifts leftwards towards reactants).
All of this is supported by the Le Chatelier's principle.
Best regards.
0.22 L of HNO3 is titrated to equivalence using 0.18 L of 0.2 MNaOH. What is the concentration of the HNO3?
Answer:
0.16 M
Explanation:
Data provided as per the question is below:-
Volume of [tex]HNO_3[/tex] = 0.22 L
The Volume of NaOH = 0.18 L
Morality of NaOH = 0.2
According to the given situation, the calculation of the concentration of the [tex]HNO_3[/tex] is shown below:-
For equivalence,
Number of the equivalent of [tex]HNO_3[/tex] = Number of equivalents of NaOH
[tex]= \frac{0.18\times0.2}{0.22}[/tex]
[tex]= \frac{0.036}{0.22}[/tex]
= 0.16363 M
or
= 0.16 M
Compounds A and B (both C10H14) show prominent peaks in their mass spectrum at m/z 134 and 119. Compound B also shows a less prominent peak at m/z 91. On vigorous oxidation with chromic acid, compound A is nonreactive while compound B yielded terephthalic acid.
Required:
From this information, deduce the structures of both compounds, and then draw the structure of B.
why are(±)-glucose and (-)-glucose both classified as D sugar
Answer:
See explanation
Explanation:
We must remember that the D / L nomenclature refers to the orientation of the hydroxyl group on carbon 5. If the "OH" is on the right we will have a D configuration. Yes, the "OH" is on the left we will have an L configuration. (See figure 1)
Now, the orientation of this "OH" has nothing to do with the ability of the molecule to deflect polarized light. If the molecule deflects light to the left we will have the symbol "(-)" (levorotation) if the molecule deflects light to the right we will have the symbol "(+)" (dextrorotation).
So in the "D" configuration, we can have both a right (+) and a left (-) deviation.
I hope it helps!
Hey can anyone help me please?
Answer:
D
Explanation:
D is the answer
Answer:
D is the correct option. All of the above.Explanation;Hope it helps you....thank you...
The K sp for silver(I) phosphate is 1.8 × 10 –18. Determine the silver ion concentration in a saturated solution of silver(I) phosphate.
Answer:
[tex][Ag^+]=4.82x10^{-5}M[/tex]
Explanation:
Hello,
In this case, the dissociation reaction for silver phosphate is:
[tex]Ag_3PO_4(s)\rightleftharpoons 3Ag^+(aq)+PO_4^{3-}(aq)[/tex]
Therefore, the equilibrium expression is:
[tex]Ksp=[Ag^+]^3[PO_4^{3-}][/tex]
And in terms of the reaction extent [tex]x[/tex] is:
[tex]Ksp=1.8x10^{-18}=(3x)^3(x)[/tex]
Thus, [tex]x[/tex] turns out:
[tex]1.8x10^{-18}=27x^4\\\\x=\sqrt[4]{\frac{1.8x10^{-18}}{27} } \\\\x=1.61x10^{-5}M[/tex]
In such a way, the concentration of the silver ion is:
[tex][Ag^+]=3x=3*1.61x10^{-5}M=4.82x10^{-5}M[/tex]
Best regards.
An element has an atomic number of 36, what element is it? Question 4 options: Kr K Se Es
Answer:
[tex]\Huge \boxed{\mathrm{Kr}}[/tex]
Explanation:
Krypton is an element in the periodic table with an atomic number of 36.
The symbol for Krypton is Kr.
Answer:
KR.
Explanation:
Use the periodic table for reference:
In which ONE of the following compounds would the bonding be expected to have the highest percentage of ionic character? A) LiBr B) CsCl C) BaBr2 D) NaCl E) KI
Answer:
B) CsCl
Explanation:
The ionic character is formed between two kinds of atoms having a large electronegativity differences e.g metals (like those in groups IA and IIA) and nonmetals (like those in groups VIA and VIIA). The formation of an ionic character involves a transfer of electrons from the less electronegative atom(metal) to the more electronegative atom (non-metal) such that the two kinds of atoms now have completely filled outer shell like the noble gases.
In CsCl, electrons are being transferred from Cs⁺ to Cl⁻ . As a result of this transfer , the atom of the metal becomes positively charged (cation) while that of the non-metal becomes negatively charged (anion).
The highest percentage of ionic character will occur as a result of smaller negatively charged (anion) and larger positively charged (cation). From the options given, CsCl have the highest percentage of ionic character.
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
Which of the following ionic lattices would have the highest melting point?
A. Potassium oxide
w
B. Boron nitride
C. Beryllium oxide
D. Lithium chloride
Answer:
I think, berryllium oxide, is answer.Explanation:
Hope it helps you....The ionic lattices would have the highest melting point Potassium oxide. option A is correct.
what is ionic lattice?An ionic compound is a giant structure of ions. The ions have a regular, repeating arrangement called an ionic lattice. The lattice is formed because the ions attract each other and form a regular pattern with oppositely charged ions next to each other.
Ionic compounds are held together by electrostatic forces between oppositely charged ions.
These forces are usually referred to as the ionic lattice contains such a large number of ions, that a lot of energy is needed to overcome this ionic bonding so ionic compounds have high melting and boiling points.
therefore, sodium oxide has the highest melting point. option A is correct
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Bomb calorimetry is a poor choice to determine the number of nutritional Calories in food; it consistently overestimates the Caloric content because options: A) dietary fiber isn't used by the body. B) carbohydrates don't burn to completion. C) proteins don't burn. D) water has Calories and isn't burnable.
Answer:
A) dietary fiber isn't used by the body.
Explanation:
The food we eat contains certain nutritional contents that provides energy, measured in calories (CAL) to the body. A procedure called BOMB CALORIMETRY can be used to determine the energy contents of these foods. The energy-supplying macromolecules contained in food substances we eat are carbohydrate, protein, fats etc.
Bomb calorimetry uses the method of burning the food substance in a device called bomb calorimeter, and measure the caloric content of the burnt food. Bomb calorimetry measures all the present calories in a food substance, which can include dietary fibers. Due to this reason, it is considered a poor choice in determining the number of nutritional calories in a food substance.
Dietary fibers are indigestible carbohydrates that cannot be broken down and used by the body. They pass along the alimentary canal until they are egested. Hence, they are no source of nutrients to the body. Since bomb calorimetry measures all calories including dietary fibers, it is said to overestimate the caloric content of food substances.
2NO + 2H2 ⟶N2 + 2H2O What would the rate law be if the mechanism for this reaction were: 2NO + H2 ⟶N2 + H2O2 (slow) H2O2 + H2 ⟶2H2O (fast)
Answer:
rate = [NO]²[H₂]
Explanation:
2NO + H2 ⟶N2 + H2O2 (slow)
H2O2 + H2 ⟶2H2O (fast)
From the question, we are given two equations.
In chemical kinetics; that is the study of rate reactions and changes in concentration. The rate law is obtained from the slowest reaction.
This means that our focus would be on the slow reaction. Generally the rate law is obtained from the concentrations of reactants in a reaction.
This means our rate law is;
rate = [NO]²[H₂]
