Explanation:
using boyles law
p1v1=p2v2
735 x 1.76 = 700 x V2
1293.6 = 700 x V2
V2 = 1293.6/700
V2 = 1.85L
A laboratory technician combines 35.9 mL of 0.258 M chromium(II) chloride with 35.8 mL 0.338 M potassium hydroxide. How many grams of chromium(II) hydroxide can precipitate
Answer:
0.52 g of chromium(II) hydroxide, Cr(OH)2.
Explanation:
We'll begin by calculating the number of mole of chromium (ii) chloride, CrCl2 in 35.9 mL of 0.258 M chromium(II) chloride solution.
This can be obtained as follow:
Molarity of CrCl2 = 0.258 M
Volume = 35.9 mL = 35.9/1000 = 0.0359 L
Mole of CrCl2 =?
Molarity = mole /Volume
0.258 = mole of CrCl2 /0.0359
Cross multiply
Mole of CrCl2 = 0.258 x 0.0359
Mole of CrCl2 = 0.0093 mole
Next, we shall determine the number of mole of potassium hydroxide, KOH in 35.8 mL 0.338 M potassium hydroxide solution.
This can be obtained as follow:
Molarity of KOH = 0.338 M
Volume = 35.8 mL = 35.8/1000 = 0.0358 L
Mole of KOH =.?
Molarity = mole /Volume
0.338 = mole of KOH /0.0358
Cross multiply
Mole of KOH = 0.338 x 0.0358
Mole of KOH = 0.0121 mole.
Next, we shall write the balanced equation for the reaction. This is given below:
2KOH + CrCl2 → Cr(OH)2 + 2KCl
From the balanced equation above,
2 mole of KOH reacted with 1 mole of CrCl2 to produce 1 mole of Cr(OH)2.
Next, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
2 mole of KOH reacted with 1 mole of CrCl2.
Therefore, 0.0121 mole of KOH will react with = (0.0121 x 1)/2 = 0.00605 mole of CrCl2.
From the calculations made above, we can see that only 0.00605 mole out of 0.0093 mole of CrCl2 is needed to react completely with 0.0121 mole of KOH.
Therefore, KOH is the limiting reactant.
Next, we shall determine the number of mole of Cr(OH)2 produced from the reaction.
In this case, we shall be using the limiting reactant because it will give the maximum yield of Cr(OH)2.
The limiting reactant is KOH and the number of mole of Cr(OH)2 produced can be obtained as illustrated below:
From the balanced equation above,
2 mole of KOH reacted to produce 1 mole of Cr(OH)2.
Therefore, Therefore, 0.0121 mole of KOH will react with = (0.0121 x 1)/2 = 0.00605 mole of Cr(OH)2.
Finally, we shall convert 0.00605 mole of Cr(OH)2 to grams.
This is illustrated below:
Mole of Cr(OH)2 = 0.00605 mole
Molar mass of Cr(OH)2 = 52 + 2(16 + 1) = 52 + 2(17) = 86 g/mol
Mass of Cr(OH)2 =..?
Mole = mass /Molar mass
0.00605 = mass of Cr(OH)2/86
Cross multiply
Mass of Cr(OH)2 = 0.00605 x 86
Mass of Cr(OH)2 = 0.52 g
Therefore, 0.52 g of chromium(II) hydroxide, Cr(OH)2 was produced.
A compound is found to contain 27.29 % carbon and 72.71 % oxygen by weight. To answer the questions, enter the elements in the order presented above. 1. What is the empirical formula for this compound
Answer:
CO2
Explanation:
Mass of carbon = 27.29
Mass of oxygen = 72.21
Step 1:
We have to first convert these masses to moles
Carbon = 2.29/2.01 = 2.274
Oxygen = 72.71/16 = 4.544
Step 2:
We have to divide the mols by the smallest mol to get simplest ratio of whole number.
The smallest mol is 2.274. we have to divide the mols by this.
2.274/2.274 = 1
4.544/2.274 = 2
Our empirical formula is therefore CO2
Given the following reaction and data, A + B → Products
Experiment A (M) B (M) Rate (M/s)
1 1.50 1.50 0.320
2 1.50 2.50 0.320
3 3.00 1.50 0.640
Required:
a. What is the rate law of the reaction?
b. What is the rate constant?
Answer:
a. Rate = k×[A]
b. k = 0.213s⁻¹
Explanation:
a. When you are studying the kinetics of a reaction such as:
A + B → Products.
General rate law must be like:
Rate = k×[A]ᵃ[B]ᵇ
You must make experiments change initial concentrations of A and B trying to find k, a and b parameters.
If you see experiments 1 and 3, concentration of A is doubled and the Rate of the reaction is doubled to. That means a = 1
Rate = k×[A]¹[B]ᵇ
In experiment 1 and to the concentration of B change from 1.50M to 2.50M but rate maintains the same. That is only possible if b = 0. (The kinetics of the reaction is indepent to [B]
Rate = k×[A][B]⁰
Rate = k×[A]b. Replacing with values of experiment 1 (You can do the same with experiment 3 obtaining the same) k is:
Rate = k×[A]
0.320M/s = k×[1.50M]
k = 0.213s⁻¹An oxidation-reduction reaction in which 3 electrons are Transferred has ∆G° = +18.55 kJ at 25°C? What is the value of E°?
Answer:
The correct answer is: 0.064 V
Explanation:
In a oxidation-reduction reaction, the relation between Gibbs free energy (ΔGº) and cell potential (Eºcell) is given by:
[tex]\Delta G^{0} = -nFE^{0} _{cell}[/tex]
where n is the number of electrons that are transferred in the reaction and F is the Faraday constant (96,500 C/mol e-). Given: ∆G° = +18.55 kJ and n= 3 mol e-, we calculate Eºcell as follows:
+18.55 kJ = (-3 mol e-) x (96500 C/mol e-) x Eºcell
Eºcell= (+18.55 kJ)/(-3 mol e-) x (96500 C/mol e-)
Eºcell= (18550 J)/ (289500 C)
Eºcell= 0.064 J/C
Since 1 Volt= 1 Joule/1 Coulomb, thus:
Eºcell= 0.064 J/C = 0.064 V
Which of the following is a half-reaction? A. Zn+CuSO4−> B. 2Cl−−>Cl2+2e− C. H2+1/2O2−>H2O D. −>Cu+ZnSO4
Answer:
2Cl——>Cl2+2e-
Explanation:
It shows an electron loss or gain
A star is observed from two positions of Earth in its orbit, in summer and winter. Which of these is the best method to calculate the approximate distance of the star from Earth? measure the parallax and use it in calculations measure the red shift of emitted light and use it in calculations use doppler effect to calculate the shift in light traveling from star to Earth in winter use doppler effect to calculate the shift in light traveling from star to Earth in summer
Answer:
measure the parallax and use it in calculations
Explanation:
got it right on test
There are many more stars at different distances from the earth. The distance to the stars calculated in light years and it is measured using parallax method.Thus option a is correct.
What is parallax method?Parallax method is used to measure the approximate distance of stars from earth. It uses the position of nearby star from two points opposite to earth and the small angular displacement observed from the remote stars are noted.
The orbit radius of earth and distance to the stars can be calculated from the parallactic angle p, that is one second of arc. Thus the distance is described in the units parsec.
The distance to the stars are usually calculated in light years. One parsec equals 3.26 light years. The nearest star to earth is named as proxima century having the distance parallax 0.76813'' which equals 4.24 light years. Thus, parallax is inversely proportional to the distance.
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Consider these metal ion/metal standard reduction potentials Cd2+(aq)|Cd(s) Zn2+(aq)|Zn(s) Ni2+(aq)|Ni(s) Cu2+(aq)|Cu(s) Ag+(aq)|Ag(s) -0.40 V -0.76 V ‑0.25 V +0.34 V +0.80 V Based on the data above, which species is the best reducing agent?
Answer:
The best reducing agent is Zn(s)
Explanation:
A reducing agent must to be able to reduce another compound, by oxidizing itself. Consequently, the oxidation potential must be high. The oxidation potential of a compound is the reduction potential of the same compound with the opposite charge. Given the reduction potentials, the best reducing agent will be the compound with the most negative reduction potential. Among the following reduction potentials:
Cd₂⁺(aq)|Cd(s) ⇒ -0.40 V
Zn²⁺(aq)|Zn(s) ⇒ -0.76 V
Ni²⁺(aq)|Ni(s) ⇒‑0.25 V
Cu²⁺(aq)|Cu(s) ⇒ +0.34 V
Ag⁺(aq)|Ag(s) ⇒ +0.80 V
The most negative is Zn²⁺(aq)|Zn(s) ⇒ -0.76 V
From this, the most reducing agent is Zn. Zn(s) is oxidized to Zn²⁺ ions with the highest oxidation potential (0.76 V).
the pain reliever codeine is a weak base with a kb equal to 1.6 x 10^-6. what is the ph of a 0.05 m aqueous codeine solution
Answer:
[tex]pH=10.45[/tex]
Explanation:
Hello,
In this case, for the dissociation of the given base, we have:
[tex]base\rightleftharpoons OH^-+CA[/tex]
Whereas CA accounts for conjugated acid and OH⁻ for the conjugated base. In such a way, equilibrium expression is:
[tex]Kb=\frac{[OH^-][CA^+]}{[base]}[/tex]
And in terms of the reaction extent [tex]x[/tex] we can write:
[tex]1.6x10^{-6}=\frac{x*x}{0.05M-x}[/tex]
For which the roots are:
[tex]x_1=-0.000284M\\x_2=0.000282M[/tex]
For which clearly the result is the positive root which also equals the concentration of hydroxyl ions and we can compute the pOH:
[tex]pOH=-log([OH^-])=-log(0.000282)\\\\pOH=3.55[/tex]
And the pH:
[tex]pH=14-pOH=14-3.55\\\\pH=10.45[/tex]
Regards.
The pH of the solution is 10.45.
Let us represent codeine with the generic formula BH. We can set up the ICE table as follows;
:B(aq) + H2O(l) ⇄ BH(aq) + OH^-(aq)
I 0.05 0 0
C -x +x +x
E 0.05 - x x x
We know that the Kb of codeine is 1.6 x 10^-6, Hence;
1.6 x 10^-6 = x^2/0.05 - x
1.6 x 10^-6 (0.05 - x ) = x^2
8 x 10^-8 - 1.6 x 10^-6x = x^2
x^2 + 1.6 x 10^-6x - 8 x 10^-8 = 0
x = 0.00028 M
The concentration of hydroxide ions = 0.00028 M
Given that pOH = - log[0.00028 M]
pOH = 3.55
pH + pOH = 14
pH = 14 - 3.55
pH = 10.45
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15. Calculate the critical angle of glass and water combination. Show your calculation. 16. What is the critical angle for the interface between Mystery A and glass
Answer:
15. Critical angle of glass and water combination, θ = 62.45°
16. Critical angle for the interface between Mystery A and glass, θ = 37.93°
Note; The question is incomplete. The complete question is as follows:
Medium Air Water Glass Mystery A Mystery B Table-2 Speed (m/s) 1.00 C 0.75 c 0.67 0.41 c 0.71 c n 1.00 1.33 1.50 Index of Refraction n of a given medium is defined as the ratio of speed of light in vacuum, c to the speed of light in a medium, v. n = c/v
Table-4: Incident Angle (degrees) Reflected Angle Refracted angle (degrees) (degrees) % Intensity of reflected ray 0 10 20 30 40 50 N/A N/A N/A 30 40 50 0 11.3 22.7 34.2 46.3 59.5 N/A N/A N/A 0.67 1.22 3.08 % Intensity of refracted ray 100 100 100 99.33 98.78 96.92
When rays travel from a denser medium to a less dense medium, we can define a critical angle of incidence θ such that refracted angle θ₂ = 90°. Applying Snell's law: Critical angle θ = sin-1(n₂/n₁).
When the angle of incidence is greater than the critical angle, 100% of the light intensity is reflected. This is called total internal reflection because all the light is reflected.
15. Calculate the critical angle of glass and water combination. Show your calculation.
16. What is the critical angle for the interface between Mystery A and glass?
Explanation:
15. Applying Snell's law; Critical angle θ = sin-1(n₂/n₁).
where n₂,refractive index of water = 1.33, n₁, refractive index of glass = 1.50 since glass is denser than water
θ = sin-1(1.33/1.50)
θ = 62.45°
Critical angle of glass and water combination, θ = 62.45°
16. Refractive index of mystery A , n = c/v
where v = 0.41 c
therefore, n = c / 0.41 c = 2.44
Critical angle for the interface between Mystery A and glass, θ = sin-1(n₂/n₁).
where n₂,refractive index of glass = 1.50, n₁, refractive index of mystery A = 2.44 since mystery A is denser than glass as seen from its refractive index
θ = sin-1(1.50/2.44)
θ = 37.93°
Critical angle for the interface between Mystery A and glass, θ = 37.93°
alculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer. Express your answer using two decimal places.
A 1.0-L buffer solution contains 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2. The value of Ka for HC2H3O2 is 1.8×10−5.
Calculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer.
Answer:
The pH of this solution = 5.06
Explanation:
Given that:
number of moles of CH3COOH = 0.100 mol
volume of the buffer solution = 1.0 L
number of moles of NaC2H3O2 = 0.100 mol
The objective is to Calculate the pH of the solution, upon addition of 0.035 mol of NaOH to the original buffer.
we know that concentration in mole = Molarity/volume
Then concentration of [CH3COOH] = [tex]\mathtt{ \dfrac{0.100 \ mol}{ 1.0 \ L }}[/tex] = 0.10 M
The chemical equation for this reaction is :
[tex]\mathtt{CH_3COOH + OH^- \to CH_3COO^- + H_2O}[/tex]
The conjugate base is CH3COO⁻
The concentration of the conjugate base [CH3COO⁻] is = [tex]\mathtt{ \dfrac{0.100 \ mol}{ 1.0 \ L }}[/tex]
= 0.10 M
where the pka (acid dissociation constant)for CH3COOH = 4.74
If 0.035 mol of NaOH is added to the original buffer, the concentration of NaOH added will be = [tex]\mathtt{ \dfrac{0.035 \ mol}{ 1.0 \ L }}[/tex] = 0.035 M
The ICE Table for the above reaction can be constructed as follows:
[tex]\mathtt{CH_3COOH \ \ \ + \ \ \ \ OH^- \ \ \to \ \ CH_3COO^- \ \ \ + \ \ \ H_2O}[/tex]
Initial 0.10 0.035 0.10 -
Change -0.035 -0.035 + 0.035 -
Equilibrium 0.065 0 0.135 -
By using Henderson-Hasselbalch equation:
The pH of this solution = pKa + log [tex]\mathtt{\dfrac{CH_3COO^-}{CH_3COOH}}[/tex]
The pH of this solution = 4.74 + log [tex]\mathtt{\dfrac{0.135}{0.065}}[/tex]
The pH of this solution = 4.74 + log (2.076923077 )
The pH of this solution = 4.74 + 0.3174
The pH of this solution = 5.0574
The pH of this solution = 5.06 to two decimal places
Which of the following elements can't have an expanded octet? answers A. oxygen B. phosphorous C. chlorine d. sulfer
answer is oxygen .
oxygen is an exception in octet rule
Among the following given elements,oxygen is an element which cannot have an expanded octet.
What is an expanded octet?Expanded octet is a condition where an octet has more than 8 electrons and which is called as hyper-valency state. This concept is related to hybrid orbital theory and Lewis theory. Hyper-valent compounds are not less common and are of equal stability as the compounds which obey octet rule.
Expansion of octet is possible for elements from third period on wards only as they have low-lying empty d - orbitals which can accommodate more than eight electrons.
Expanded octet is not applicable to oxygen as it is second period of periodic table and has less than ten electrons and even does not have the 2d -orbitals due to which it does not fulfill the criteria of an element to have an expanded octet.
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Suppose a small amount of a solid is added to water and, after a short time, all the solid has dissolved. Which of the following statements is most likely to be true?
A) The solution is supersaturated with solute.
B) The solution is saturated with solute.
C) The solution is unsaturated with solute.
D) The solution is either unsaturated or supersaturated with solute.
E) The solution is either saturated or supersaturated with solute.
Answer:
the option e is correct I think
The solution is either saturated or supersaturated with solute
Answer:
the option e is correct I think
The solution is either saturated or supersaturated with solute
Explanation:
A qualitative researcher may select a ______ case that is unusually rich in information pertaining to the research question.
A) critical
B) typical
C) deviant
D) rare
Answer:
D. Rare
Explanation:
Qualitative research has to do with non-numerical data and is used to understand the beliefs of a group of people which can be gotten from surveys, questionnaires, interviews, etc.
A qualitative researcher may select a RARE case that is unusually rich in information pertaining to the research question.
This is because he wants to get an insight into the why, how, where and when of that particular ase as it's not a usual occurrence.
The rate constant for the decay of a radioactive element is 3.68 × 10⁻³ day⁻¹. What is the half-life of this element?
Answer:
half life=0.693/rate constant =188.3
The half-life of this element is 188.32 years
The formula for calculating the half-life of an element is expressed according to the equation:
[tex]t_{1/2}=\frac{ln 2}{\lambda}[/tex]
[tex]\lambda[/tex] is the decay constantt1/2 is the half-lifeGiven the following parameters:
The rate constant for the decay = 3.68 × 10⁻³ day⁻¹.
Substitute into the formula for calculating the half-life as shown:
[tex]3.68\times 10^{-3}=\frac{0.693}{\lambda}\\ 0.00368=\frac{0.693}{\lambda}\\\lambda=\frac{0.693}{0.00368}\\\lambda = 188.32 years[/tex]
Hence the half-life of this element is 188.32 years
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Write an equation to show how the base NaOH(s) behaves in water. Include states of matter in your answer. Click in the answer box to open the symbol palette.
Answer:
The reaction is given as:
[tex]NaOH(s)\rightarrow Na^+(aq)+OH^-(aq)[/tex]
Explanation:
Bases are defined as those chemical substances which give hydroxide ions in their aqueous solutions.
[tex]BOH(s)\rightarrow B^+(aq)+OH^-(aq)[/tex]
When sodium hydroxide is added to water it gets dissociated into two ions that are sodium ions and hydroxide ions. Along with this heat energy also releases during this reaction.
The reaction is given as:
[tex]NaOH(s)\rightarrow Na^+(aq)+OH^-(aq)[/tex]
The equation to show how NaOH behaves in water is NaOH → Na⁺ + (OH)⁻
The compound that produce negative hydroxide (OH−) ions when dissolved
in water are called bases .
This compounds NaOH (sodium hydroxide) is an example of a base.
When it dissolves in water it dissociate to form negative hydroxide (OH−)
ions and positive sodium (Na+) ions.
It can be represented by the following equation:
NaOH → Na⁺ + (OH)⁻
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Which describes the molecule below
Answer:
Option D. A lipid with three unsaturated fatty acid.
Explanation:
The molecule in the diagram above contains three fatty acid.
A careful observation of the molecule reveals that each of the three fatty acids contains a double bond.
The presence of a double bond in a compound shows that the compound is unsaturated.
Thus, we can say that the molecule is a lipid with three unsaturated fatty acid.
What is the change in entropy (AS) when a solid substance decomposes and
produces gaseous products?
Explanation:
Entropy refers to the degree of disorderliness in a system. Generally solids have a greater degree of order when compared to gas in which the molecules/particles move randomly in all directions.
The change in entropy from solid to gaseous phase is a positive change. Because there is increase disorderliness of the system.
26.0 g of a liquid that has a density of 1.44 g/mL needs to be measured out in a graduated cylinder . What volume should be used
Answer: The volume of liquid used will be 18.055 mL.
Given the following equivalents, make the following conversion: 1.00 knop = ? knips
4 clips = 5 blips
1 knop = 6 bippy
3 blip = 18 pringle
1 clip = 10 knip
10 bippy = 8 pringle
Answer:
[tex]6.4knips[/tex]
Explanation:
Hello,
In this case, given the stated equivalences, we can use the following proportional factor in order to compute the required knips:
[tex]knips=1.00knop*\frac{6bippy}{1knop} *\frac{8pringle}{10bippy}* \frac{3blip}{18pringle} *\frac{4clips}{5blips} *\frac{10knip}{1clip} \\\\=6.4knips[/tex]
Regards.
1 knop=6.4 knips
First convert knop to bippy:-
[tex]1\ knop\times\frac{6\ bippy}{1\ knop} =6\ bippy[/tex]
Now, Convert 6 bippy to pringle:-
[tex]6\ bippy\times\frac{8\ pringle}{10\ bippy} =4.8\ pringle[/tex]
Now, convert 4.8 pringle to blip:-
[tex]4.8\ pringle\times\frac{3\ blip}{18\ priangle} =0.8\ blip[/tex]
Now, convert 0.8 blip to clips as follows:-
[tex]0.8\ blip\times\frac{4\ clips}{5\ blip} =0.64\ clip[/tex]
Now, convert 0.64 clip to knips:-
[tex]0.64\ clip\times\frac{10\ knip}{1\ clip} =6.4\ knip[/tex]
Hence, the following conversion is as follows:-
1.00 knop=6.4 knips
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Calculate the amount of heat needed to melt of ice () and bring it to a temperature of . Round your answer to significant digits. Also, be sure your answer contains a unit symbol.
Answer:
your answer is given below
Explanation:
Take it stepwise, and you are going to have to look up the various heat values.
You have 36.5grams of ice, presumably at 0C. You will need to add heat to take the ice at 0 C to water at 0 C. (Latent heat of melting) Then you add in a different heat value to take the water at 0 C to water at 82.3 C. (Specific heat of water)
Add the two heat amounts together.
According to specific heat capacity, to calculate the amount of heat needed to melt ice given mass of ices is multiplied by specific heat of ice and the temperature change which it undergoes.
What is specific heat capacity?Specific heat capacity is defined as the amount of energy required to raise the temperature of one gram of substance by one degree Celsius. It has units of calories or joules per gram per degree Celsius.
It varies with temperature and is different for each state of matter. Water in the liquid form has the highest specific heat capacity among all common substances .Specific heat capacity of a substance is infinite as it undergoes phase transition ,it is highest for gases and can rise if the gas is allowed to expand.
It is given by the formula ,
Q=mcΔT
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Will a precipitate (ppt) form when 20.0 mL of 1.1 × 10 –3 M Ba(NO 3) 2 are added to 80.0 mL of 8.4 × 10 –4 M Na 2CO 3?
Answer:
A precipitate will form, BaCO₃
Explanation:
When Ba²⁺ and CO₃²⁻ ions are in an aqueous media, BaCO₃(s), a precipitate, is produced following its Ksp expression:
Ksp = 5.1x10⁻⁹ = [Ba²⁺] [CO₃²⁻]
Where the concentrations of the ions are the concentrations in equilibrium
For actual concentrations of a solution, you can define Q, reaction quotient, as:
Q = [Ba²⁺] [CO₃²⁻]
If Q > Ksp, the ions will react producing BaCO₃, if not, no precipitate will form.
Actual concentrations of Ba²⁺ and CO₃²⁻ are:
[Ba²⁺] = [Ba(NO₃)₂] = 1.1x10⁻³ × (20.0mL / 100.0mL) = 2.2x10⁻⁴M
[CO₃²⁻] = [Na₂CO₃] = 8.4x10⁻⁴ × (80.0mL / 100.0mL) = 6.72x10⁻⁴M
100.0mL is the volume of the mixture of the solutions
Replacing in Q expression:
Q = [Ba²⁺] [CO₃²⁻]
Q = [2.2x10⁻⁴M] [6.72x10⁻⁴M]
Q = 1.5x10⁻⁷
As Q > Ksp
A precipitate will form, BaCO₃
A complexometric titration can also be used to determine the amount of calcium in milk. The calcium concentration in milk is typically 1,200 mg/L. How would you alter the procedure used in this experiment to determine milk calcium content
Answer:
d
Explanation:answer is d on edg 2020
what are the monomers of bakelite
Answer:
Bakelite is a polymer made up of the monomers phenol and formaldehyde. This phenol-formaldehyde resin is a thermosetting polymer.
Answer: The monomers of bakelite are formaldehyde and phenol
Explanation:
If a reaction is first-order with respect to a particular reactant, when the concentration of that reactant is increased by a factor of 2, the reaction rate will _____.
Answer:
It would increase by a factor of 2
Explanation:
The rate law for a fist order reaction is given as;
A --> B
rate = k [A]
upon doubling the concentration, we have;
rate = k [A]
rate 2 = k 2 [A]
Dividing both equations;
rate 2 / rate 1 = k 2 [A] / k [A]
rate 2 / rate 1 = 2 / 1
The ratio between rate 2 and rate 1 is 2 : 1. This means that the reaction rate would also increase by a factor of 2.
Calculate the molar hydrogen ion concentration of each of the following biological solutions given the pH:
(a) gastric juice, pH= 1.80
(b) urine, pH 4.75 56.
Answer: A
Explanation: Calculate the molar hydrogen ion concentration of each of the following biological solutions given the pH, Urine pH= 4.90
Calculate the solubility of Mg(OH)2 in water at 25 C. You'll find Ksp data in the ALEKS Data tab. Round your answer to significant digits.
Answer:
1.12 × 10⁻⁴ M
Explanation:
Step 1: Write the reaction for the solution of Mg(OH)₂
Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)
Step 2: Make an ICE chart
We can relate the solubility product constant (Ksp) with the solubility (S) through an ICE chart.
Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2 OH⁻(aq)
I 0 0
C +S +2S
E S 2S
The solubility product constant is:
Ksp = 5.61 × 10⁻¹² = [Mg²⁺] × [OH⁻]² = S × (2S)² = 4S³
S = 1.12 × 10⁻⁴ M
Ozone (O 3) in the atmosphere can react with nitric oxide (NO): O 3(g) + NO(g) → NO 2(g) + O 2(g). Calculate the ΔG° for this reaction at 25°C. (ΔH° = –199 kJ/mol, ΔS° = –4.1 J/K·mol)
Answer:
ΔG° = 1022. 8 kJ
Explanation:
ΔH° = –199 kJ/mol
ΔS° = –4.1 J/K·mol
T = 25°C = 25 + 273 = 298K (Converting to kelvin temperature)
ΔG° = ?
The relationship between these varriables are;
ΔG° = ΔH° - TΔS°
ΔG° = –199 - 298 (–4.1)
ΔG° = -199 + 1221.8
ΔG° = 1022. 8 kJ
To calculate ΔG, the following equation should be used:
ΔG° = ΔH° - TΔS°
Given:
ΔH° = –199 kJ/mol,
ΔS° = –4.1 J/K·mol
T=25+273K=298K
Substitute the respective values:
ΔG° = ΔH° - TΔS°
=–199 kJ/mol-(298K×(-4.1 J/K·mol*(1KJ/1000J)))
=-197.78kJ/mol
Thus, we can conclude the value of ΔG°=-197.78kJ/mol
Learn more about ΔG° here:
https://brainly.com/question/2491198
Name four types of salts
Answer:
Any ionic molecule formed of a base and an acid, which dissolves in water to produce ions is known as a salt. The four common types of salts are:
1. NaCl or sodium chloride is the most common kind of salt known. It is also known as table salt.
2. K2Cr2O7 or potassium dichromate refers to an orange-colored salt formed of chromium, potassium, and oxygen. It is toxic to humans and is also an oxidizer, which is a fire hazard.
3. CaCl2 or calcium chloride looks like table salt due to its white color. It is broadly used to withdraw ice from roads. It is hygroscopic.
4. NaHSO4 or sodium bisulfate produces from hydrogen, sodium, oxygen, and sulfur. It is also known as dry acid. It has commercial applications like reducing the pH of swimming pools and spas and others.
03/08/2020
Question
1. (a) State the definition of a chemical formula.
(b) What does it tell about a compound?
(c) What information is conveyed by the formulation H2SO4?
Explanation:
According to your question,
no. a. ans would be like; chemical formula is defined as the an expression which determines no. and type of molecule of a compound.
b. no. ans; it tells that what type of compound is formed with the type and no. of atoms present in the atom.
c. no ans; the formulation of h2so4 states that it is acid named as hydrochloric acid which is formed by reacting of hydrogen (2 atoms ) ,sulpher (*1atom) and oxygen(4atoms).
Hope it helps...
A rock has a mass of 15.8 g and causes the water level in a graduated cylinder to raise from 22.3 mL to 32.5 mL. What is the density of the rock in Kg/mL?
Answer:
Explanation:
mass - 15.8 g = 0.0158 kg
volume = 32.5 - 22.5 = 10.2 ml
density = mass / volume
= 0.0158 / 10.2
= 0.00154 kg/ml
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