Answer: 1.2 millimoles will be left after 2250 years
Explanation:
Expression for rate law for first order kinetics is given by:
[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]
where,
k = rate constant
t = age of sample
a = let initial amount of the reactant
a - x = amount left after decay process
a) for completion of half life:
Half life is the amount of time taken by a radioactive material to decay to half of its original value.
[tex]t_{\frac{1}{2}}=\frac{0.693}{k}[/tex]
[tex]k=\frac{0.693}{5730}=0.00012years^{-1}[/tex]
b) Amount left after 2250 years
[tex]2250=\frac{2.303}{k}\log\frac{1.6}{a-x}[/tex]
[tex]2250=\frac{2.303}{0.00012}\log\frac{1.6}{a-x}[/tex]
[tex]\log\frac{1.6}{a-x}=0.117[/tex]
[tex]\frac{1.6}{a-x}=1.31[/tex]
[tex]{a-x}=\frac{1.6}{1.31}=1.2[/tex]
Thus 1.2 millimoles will be left after 2250 years
We discussed the different types of intermolecular forces in this lesson, which can affect the boiling point of a substance.
1. Which of these has the highest boiling point?
A) Ar
B) Kr
C) Xe
D) Ne
2. Which substance has the highest boiling point?
A) CH4
B) He
C) HF
D) Cl2
Answer:
1, C, Xe 2, B,He
Explanation:
1, cause as u go down a group the boiling point increases.
2, boiling point of single element is greater than a compound
According to periodic trends in periodic table boiling point increases down the group and hence Xe has highest boiling point and more amount of heat is required to boil an element hence He has highest boiling point.
What is periodic table?
Periodic table is a tabular arrangement of elements in the form of a table. In the periodic table, elements are arranged according to the modern periodic law which states that the properties of elements are a periodic function of their atomic numbers.
It is called as periodic because properties repeat after regular intervals of atomic numbers . It is a tabular arrangement consisting of seven horizontal rows called periods and eighteen vertical columns called groups.
Elements present in the same group have same number of valence electrons and hence have similar properties while elements present in the same period show gradual variation in properties due to addition of one electron for each successive element in a period.
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I add a 50. g piece of Al (c = 0.88 J/g-deg) that is at 225°C to 100. mL of water at 20°C. What is the final temperature of the water in °C? The density of water is approximately 1g/mL.
Answer:
THE FINAL TEMPERATURE OF WATER IS -4.117 °C
Explanation:
Mass of the aluminium = 50 g
c = 0.88 J/g C
Initial temperature of aluminium = 225 °C
Volume of water = 100 ml
Density of water = 1 g/ml
Mass of water = density * volume of water
Mass of water = 1 * 100 = 100 g of water
Initial temperature of water = 20 C
It is worthy to note that the heat of a system is constant and conserved as no heat is lost or gained by a closed system,
So therefore,
heat lost by aluminium = heat gained by water
H = mass * specific heat capacity * temeprature change
So:
m c ( T2- T1) = m c (T2-T1)
50 * 0.88 * ( T2 - 225) = 100 * 4.18 *( T2 - 20)
44 ( T2 - 225 ) = 418 ( T2 - 20)
44 T2 - 9900 = 418 T2 - 8360
-9900 + 8360 = 418 T2 - 44 T2
-1540 = 374 T2
T2 = - 4.117
So therefore the final temperature of water is -4.117 °C
Which of the following goes through the largest volumetric change? Question 4 options: A) Water when it's heated from 1oC to 99oC B) Water when it freezes into ice C) Ice when it melts into water D) Water when it boils into steam
Answer:
Water when it freezes into ice
Explanation:
Most liquids expand when heated and contract when cooled, water behaves in an anomalous fashion. Water rather expands when cooled and contracts when heated.
Water usually contracts on cooling from any temperature until 4°C, after 4°C, the water begins to expand rapidly. Hence water has its least volume at 4°C and increases rapidly afterwards.
Thus the largest volume change for water occurs during freezing since it expands when cooled.
The solubility of lead(II) iodide is 0.064 g/100 mL at 20ºC. What is the solubility product for lead(II) iodide?
Answer:
[tex]Ksp=1.07x10^{-8}[/tex]
Explanation:
Hello,
In this case, the dissociation reaction is:
[tex]PbI_2(s)\rightleftharpoons Pb^{2+}(aq)+2I^-(aq)[/tex]
For which the equilibrium expression is:
[tex]Ksp=[Pb^{2+}][I^-]^2[/tex]
Thus, since the saturated solution is 0.064g/100 mL at 20 °C we need to compute the molar solubility by using its molar mass (461.2 g/mol)
[tex]Molar solubility=\frac{0.064g}{100mL}*\frac{1000mL}{1L}*\frac{1mol}{461.2g}=1.39x10^{-3}M[/tex]
In such a way, since the mole ratio between lead (II) iodide to lead (II) and iodide ions is 1:1 and 1:2 respectively, the concentration of each ion turns out:
[tex][Pb^{2+}]=1.39x10^{-3}M[/tex]
[tex][I^-]=1.39x10^{-3}M*2=2.78x10^{-3}M[/tex]
Thereby, the solubility product results:
[tex]Ksp=(1.39x10^{-3}M)(2.78x10^{-3}M)^2\\\\Ksp=1.07x10^{-8}[/tex]
Regards.
Solubility product constant for the product of lead(II) iodide is [tex]\bold { 1.07x 10^-^8}[/tex].
The dissociation reaction for lead (II) iodide
[tex]\bold {Pb I^2 (s) \leftrightharpoons Pb^2^+ + 2I^- }[/tex]
Solubility product constant at equilibrium.
[tex]\bold {Ksp = [Pb^2^++[I^-]^2}[/tex]
The molar solubility of the substance can be calculated by using the molar mass,
[tex]\bold {s = \dfrac {0.064}{100 mL} \times 461.2 g/mol = 1.39x10^-^3}[/tex]
Molar ratio between between PbI to lead and iodide ions is 1:1 and 1:2 respectively.
Thus Ksp will be,
[tex]\bold {Ksp =(1.39x10^-^3)(2.78x10^-^3 )^2}\\\\\bold {Ksp = 1.07x 10^-^8}[/tex]
Therefore, solubility product constant for the product of lead(II) iodide is [tex]\bold { 1.07x 10^-^8}[/tex].
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11mg of cyanide per kilogram of body weight is lethal for 50% of domestic chickens. How many grams per kilogram of body weight is a lethal dose for 50% of domestic chickens?
Answer:
[tex]0.033g[/tex]
Explanation:
Hello,
In this case, since 11 mg per kilogram of body weight has the given lethality, the mg that turn out lethal for a chicken weighting 3 kg is computed by using a rule of three:
[tex]11mg\longrightarrow 1kg\\\\x\ \ \ \ \ \ \longrightarrow 3kg[/tex]
Thus, we obtain:
[tex]x=\frac{3kg*11mg}{1kg}\\ \\x=33mg[/tex]
That in grams is:
[tex]=33mg*\frac{1g}{1000mg} \\\\=0.033g[/tex]
Regards.
A sample of ammonia gas was allowed to come to equilibrium at 400 K. 2NH3(g) <----> N2(g) 3H2(g) At equilibrium, it was found that the concentration of H2 was 0.0484 M, the concentration of N2 was 0.0161 M, and the concentration of NH3 was 0.295 M. What was the initial concentration of ammonia
Answer:
0.327 M
Explanation:
Step 1: Write the balanced equation
2 NH₃(g) ⇄ N₂(g) + 3H₂(g)
Step 2: Make an ICE chart
2 NH₃(g) ⇄ N₂(g) + 3 H₂(g)
I x 0 0
C -2y +y +3y
E x-2y y 3y
Step 3: Find the value of y
The concentration of N₂ at equilibrium is 0.0161 M. Then,
y = 0.0161
Step 4: Find the value of x
The concentration of NH₃ at equilibrium is 0.295 M. Then,
x-2y = 0.295
x-2(0.0161) = 0.295
x = 0.327
The direction of the functional group is called?
Explanation:
they are called hydrocarbyls
pls mark me brainliest
Answer:
The first carbon atom that attaches to the functional group is referred to as the alpha carbon.
If 1 mol of a pure triglyceride is hydrolyzed to give 2 mol of RCOOH, 1 mol of R'COOH, and 1 mol of glycerol, which of the following compounds might be the triglyceride?
CHOC(O)R
A. CHOC(O)R
CHOC(O)R
CH,OC(O)R
B. CHOC(O)R
CH2OC(O)R
CHOC(O)R
C. CHOC(O)R
CHOC(O)R
CHOC(O)R
D. CHOC(O)R
CHOC(O)R
Answer:
The correct option is C.
Note the full question and structure of the moleculesis found in the attachment below.
Explanation:
Triglycerides or triacylglycerols are non-polar, hydrophobic lipid molecules composed of three fatty acids linked by ester bonds to a molecule of glycerol.
The fatty acids linked to the glycerol molecule are denoted by R and may be of the same kind or different. when the R group is the same, the R is attached in all the three positions for ester bonding in the glycerol molecule but when they are different are denoted by R, R' and R'' respectively.
During the hydrolysis of triglycerides, the three fatty acids molecules are obtained as well as a glycerol molecule.
From the question, when 1 mole of the triglyceride is hydrolysed, 2 moles of RCOOH, 1 mole of R'COOH and 1 mole of glycerol is obtained. The triglyceride must then be composed of two fatty acids which are the same denoted by R, and a different fatty acid molecule denoted by R'.
The correct option therefore, is C
105/22 • (1.251 - 0.620)=
Answer:
105/22*(1.251-0.620)
105/22*0.631
4.772*0.631
3.011132
Hope it helps
Answer:
3.0
Explanation:
First, complete the operations inside the parenthesis according to the normal rules for significant figures. Because there are subsequent calculations, keep at least one extra significant figure when possible: (4.7727) × (0.631).
The final product will be rounded to two significant figures because it can’t be more precise than the least precise number in the problem, 22. The final product is 3.0.
From the graph of Density vs. Concentration, created in Graph 1, what was the relationship between the concentration of the sugar solution and the density of the sugar solution?
The graph is not given in the question, so, the required graph is attached below:
Answer:
According to the graph, the relationship between the density of the sugar solution and the concentration of the sugar solution is directly proportional to each other as they both are increasing exponentially.
The graph shows that, the density of sugar solution will increase with the increase in concentration of sugar in the solution.
Calculate the equilibrium constant K c for the following overall reaction: AgCl(s) + 2CN –(aq) Ag(CN) 2 –(aq) + Cl –(aq) For AgCl, K sp = 1.6 × 10 –10; for Ag(CN) 2 –, K f = 1.0 × 10 21.
Answer:
1.6x10¹¹ = Kc
Explanation:
For the reaction:
AgCl(s) + 2CN⁻(aq) ⇄ Ag(CN)₂⁻(aq) + Cl⁻(aq)
Kc is defined as:
Kc = [Ag(CN)₂⁻] [Cl⁻] / [CN⁻]²
Ksp of AgCl is:
AgCl(s) ⇄ Ag⁺(aq) + Cl⁻(aq)
Where Ksp is:
Ksp = [Ag⁺] [Cl⁻] = 1.6x10⁻¹⁰
In the same way, Kf of Ag(CN)₂⁻ is:
Ag⁺(aq) + 2CN⁻ ⇄ Ag(CN)₂⁻
Kf = [Ag(CN)₂⁻] / [CN⁻]² [Ag⁺] = 1.0x10²¹
The multiplication of Kf with Ksp gives:
[Ag⁺] [Cl⁻] * [Ag(CN)₂⁻] / [CN⁻]² [Ag⁺] = Ksp*Kf
[Ag(CN)₂⁻] [Cl⁻] / [CN⁻]² = Ksp*Kf
Obtaining the same expression of the first reaction
That means Ksp*Kf = Kc
1.6x10⁻¹⁰*1.0x10²¹ = Kc
1.6x10¹¹ = KcCandle wax melts low temperature, it is not conductive to electricity, it is insoluble in water and partially soluble in solvents nonpolar, like gasoline. Than type of links are present in the candle wax?
A. Electrostatics.
B. Apolar.
C. lónicos.
D. Hydrogen bridges.
electrostatic and ionic are definitely not the answer because they have high melting point
hydrogen bonds are too weak and not permanent.
so the answer is apolar as it is soluble in polar solvents (water)
Answer:
B. Nonpolar
Explanation:
The low melting point tells you the compound is not ionic, metallic, or a network solid.
It is almost certainly a molecular solid.
It does not conduct electricity, so it is not metallic (which we have already ruled out).
It is insoluble in polar solvents (water) and soluble in nonpolar solvents (gasoline).
Since like dissolves like, the molecule is nonpolar.
The type of links must be nonpolar.
i) Briefly discuss the strengths and weaknesses of the four spectroscopy techniques listed below. Include in your answer the specific structural information you get from each method.
IR
UV-VIS
NMR
Mass Spec
delete please .....................................
Calculate the molar solubility of CaF2 at 25°C in a solution that is 0.010 M in Ca(NO3)2. The Ksp for CaF2 is 3.9 x 10-11.
Answer:
[tex]Molar \ solubility=3.12x10^{-5}M[/tex]
Explanation:
Hello,
In this case, for the dissociation of calcium fluoride:
[tex]CaF_2(s)\rightleftharpoons Ca^{2+}+2F^-[/tex]
The equilibrium expression is:
[tex]Ksp=[Ca^{2+}][F^-]^2[/tex]
In such a way, via the ICE procedure, including an initial concentration of calcium of 0.01 M (due to the calcium nitrate solution), the reaction extent [tex]x[/tex] is computed as follows:
[tex]3.9x10^{-11}=(0.01+x)(2*x)^2\\\\x=0.0000312M[/tex]
Thus, the molar solubility equals the reaction extent [tex]x[/tex], therefore:
[tex]Molar \ solubility=3.12x10^{-5}M[/tex]
Regards.
The molar solubility of Calcium fluoride has been calculated as [tex]3.12\;\times\;10^-^5\;\rm M[/tex].
The dissociation of calcium fluoride has been given by:
[tex]\rm CaF_2\;\rightarrow\;Ca^2^+\;+\;2\;F^-[/tex]
The solubility constant, ksp has been given as:
[tex]ksp=\rm[Mg^2^+]\;[F^-]^2[/tex]
From the dissociation of Calcium nitrate, the concentration of Ca ion in the solution has been 0.01 M.
The dissociation of Calcium fluoride x M has been resulted in x M Ca and 2x M F ions.
The concentration of Ca in the solution has been resulted as x + 0.01 M.
The solubility product can be given as:
[tex]3.9\;\times\;10^-^1^1=[x+0.01]\;[2x]^2\\3.9\;\times\;10^-^1^1=[x+0.01]\;4x^2\\x=3.12\;\times\;10^-^5[/tex]
The molar solubility of Calcium fluoride has been calculated as [tex]3.12\;\times\;10^-^5\;\rm M[/tex].
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How are Math, Physics, Chemistry, and Biology all related?
Answer:
- you have to do maths in all 3
- atoms make up everything even parts of a cel and theyre studied in chem and physics
- chemistry is used in biology by finding out what different substances are eg cytoplasm in a cell
Given that π = n M R T, rearrange the equation to solve for V
Answer:
V= n/M
Explanation:
From;
π = nRT/V = MRT
Where;
n= number of moles
R= gas constant
T= absolute temperature
M= molar mass
V= volume of the solution
π= osmotic pressure
Thus;
nRT/V = MRT
nRT = VMRT
V= nRT/MRT
V= n/M
Increasing which factor will cause the gravitational force between two objects to decrease?
weights of the objects
distance between the objects
acceleration of the objects
masses of the objects
Answer:
B
Explanation:
Increasing distance between the objects factor will cause the gravitational force between two objects to decrease. Therefore, option B is correct.
What causes gravitational force to decrease?The gravitational force grows in proportion to the size of the masses . The gravitational force weakens rapidly as the distance between masses grows. Unless at least one of the objects has a lot of mass, detecting gravitational force is extremely difficult.
Gravity is affected by object size and distance between objects. Mass is a unit of measurement for the amount of matter in an object.
The force of gravity is proportional to the masses of the two objects and inversely proportional to the square of the distance between them. This means that the force of gravity increases with mass but decreases as the distance between objects increases.
Thus, option B is correct.
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Consider the following reaction at 298.15 K: Co(s)+Fe2+(aq,1.47 M)⟶Co2+(aq,0.33 M)+Fe(s) If the standard reduction potential for cobalt(II) is −0.28 V and the standard reduction potential for iron(II) is −0.447 V, what is the cell potential in volts for this cell? Report your answer with two significant figures.
Answer:
The correct answer is 0.186 V
Explanation:
The two hemirreactions are:
Reduction: Fe²⁺ + 2 e- → Fe(s)
Oxidation : Co(s) → Co²⁺ + 2 e-
Thus, we calculate the standard cell potential (Eº) from the difference between the reduction potentials of cobalt and iron, respectively, as follows:
Eº = Eº(Fe²⁺/Fe(s)) - Eº(Co²⁺/Co(s)) = -0.28 V - (-0.447 V) = 0.167 V
Then, we use the Nernst equation to calculate the cell potential (E) at 298.15 K:
E= Eº - (0.0592 V/n) x log Q
Where:
n: number of electrons that are transferred in the reaction. In this case, n= 2.
Q: ratio between the concentrations of products over reactants, calculated as follows:
[tex]Q = \frac{ [Co^{2+} ]}{[Fe^{2+} ]} = \frac{0.33 M}{1.47 M} = 0.2244[/tex]
Finally, we introduce Eº= 0.167 V, n= 2, Q=0.2244, to obtain E:
E= 0.167 V - (0.0592 V/2) x log (0.2244) = 0.186 V
Which of the following types of electromagnetic radiation have higher frequencies than visible light and which have shorter frequencies than visible light? Sort them accordingly. ltems (6 items) (Drag and drop into the appropriate area below)
a. Gamma rays
b. Infrared radiation
c. Ultraviolet liht
d. X-rays
e. Microwaves
f. Radio waves
Answer:
Higher frequency than visible light - Ultraviolet light, X-rays, and Gamma rays
Lower frequency than visible light - Infrared radiation, microwaves, and Radio waves
Explanation:
The frequencies of electromagnetic radiations vary according to their wavelengths. The relationship between the frequency and wavelength of the waves is expressed such that:
λ = c/f, where λ = wavelength, c = speed of light, and f = frequency.
Thus, there is an inverse relationship between the wavelength and the frequency of electromagnetic waves.
The order of the electromagnetic waves based on their frequency from the lowest to the highest is radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma-rays
Hence, electromagnetic waves with higher frequencies than visible light include ultraviolet light, X-rays, and Gamma rays while those with lower frequencies include Infrared radiation, microwaves, and Radio waves.
Answer:
need points
Explanation:
Which of the following happens to a molecule of an object when the object is heated? (1 point)
Answer:
They get more energy, so they vibrate!
Explanation:
The complex ion Fe(CN)63- is paramagnetic with one unpaired electron. The complex ion Fe(SCN)63- has five unpaired electrons. Where does SCN- lie in the spectrochemical series with respect to CN-?
Answer:
SCN- is a weak field ligand while CN- is a strong field ligand
Explanation:
The spectrochemical series is an arrangement of ligands according to their magnitude of crystal field splitting. Ligands that cause only a small degree of crystal field splitting are called weak field ligands while ligands that cause large crystal field splitting are called strong field ligands.
Strong field ligands often lead to the formation of low spin complexes with the least number of unpaired electrons while high spin complexes are formed by weak field ligands.
CN- is a strong field ligand as it lies towards the right hand side of the spectrochemical series.
SCN- is a weak field ligand hence it forms a high spin complex having the maximum number of unpaired electrons for Fe^3+, hence the answer.
SCN⁻ lies in the lower (weak field) region of the spectrochemical series while CN⁻ lies in the higher (stronger field) region.
CN⁻ is a strong field ligand with a large splitting constant, and it is high up in the spectrochemical series.
Conversely, SCN⁻ is a weak field ligand with a low splitting constant, and it is lower in the spectrochemical series.
Hence, SCN⁻ lies in the lower (weak field) region of the spectrochemical series while CN⁻ lies in the higher (stronger field) region.
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A soft drink contains 63 g of sugar in 378 g of H2O. What is the concentration of sugar in the soft drink in mass percent
Answer:
[tex]\% m/m= 14.3\%[/tex]
Explanation:
Hello,
In this case, the by mass percent is computed as shown below:
[tex]\% m/m=\frac{m_{solute}}{m_{solute}+m_{solvent}} *100\%[/tex]
Whereas the solute is the sugar and the solvent the water, therefore, the concentration results:
[tex]\% m/m=\frac{63g}{63g+378g} *100\%\\\\\% m/m= 14.3\%[/tex]
Best regards.
what is ammonium nitrate
Answer:
Ammonium nitrate is a chemical compound with the chemical formula NH₄NO₃. It is a white crystalline solid consisting of ions of ammonium and nitrate.
How many molecules are there in 3.5 moles of carbon dioxide? A. 63.21 x 10^23 B. 21.07 x 10^23 C. 42.14 x 10^23 D. 6.02 x 10^23
Answer:
B. 21.07 x 10²³ molecules
Explanation:
Avogadro's Number: 6.022 x 10²³
Step 1: Set up equation
[tex]3.5 mols CO_2(\frac{6.022(10^{23}) moleculesCO_2}{1 mol CO_2})[/tex]
Step 2: Multiply and cancel out units
3.5(6.022 x 10²³) = 21.07 x 10²³ molecules CO₂
Step 3: Convert to proper scientific notation
≈ 2.11 x 10²³ molecules CO₂
What is the pOH of a solution at 25.0∘C with [H3O+]=4.8×10−6 M?
Answer:
8.68
Explanation:
pOH = 8.68
all you need is contained in the sheet
Answer:
Approximately [tex]8.68[/tex].
Explanation:
The [tex]\rm pOH[/tex] of a solution can be found from the hydroxide ion concentration [tex]\rm \left[OH^{-}\right][/tex] with the following equation:
[tex]\displaystyle \rm pOH = -\log_{10} \rm \left[OH^{-}\right][/tex].
On the other hand, the ion-product constant of water, [tex]K_{\text{w}}[/tex], relates the hydroxide ion concentration [tex]\rm \left[OH^{-}\right][/tex] of a solution to its hydronium ion concentration [tex]\rm \left[{H_3O}^{+}\right][/tex]:
[tex]K_\text{w} = \rm \left[{H_3O}^{+}\right] \cdot \rm \left[OH^{-}\right][/tex].
At [tex]25 \; ^\circ \rm C[/tex], [tex]K_{\text{w}} \approx 1.0 \times 10^{-14}[/tex]. For this particular [tex]25 \; ^\circ \rm C[/tex] solution, [tex]\rm \left[{H_3O}^{+}\right] = 4.8 \times 10^{-6}\; \rm mol \cdot L^{-1}[/tex].Hence the [tex]\rm \left[OH^{-}\right][/tex] of this solution:
[tex]\begin{aligned}\left[\mathrm{OH}^{-}\right] &= \frac{K_\text{w}}{\rm \left[{H_3O}^{+}\right]} \\ &= \frac{1.0 \times 10^{-14}}{4.8 \times 10^{-6}}\; \rm mol\cdot L^{-1} \approx 2.08333 \times 10^{-9}\; \rm mol\cdot L^{-1}\end{aligned}[/tex].
Therefore, the [tex]\rm pOH[/tex] of this solution would be:
[tex]\begin{aligned}\rm pOH &= -\log_{10} \rm \left[OH^{-}\right] \\ &\approx -\log_{10} \left(4.8 \times 10^{-6}\right) \approx 8.68\end{aligned}[/tex].
Note that by convention, the number of decimal places in [tex]\rm pOH[/tex] should be the same as the number of significant figures in [tex]\rm \left[OH^{-}\right][/tex].
For example, because the [tex]\rm \left[{H_3O}^{+}\right][/tex] from the question has two significant figures, the [tex]\rm \left[OH^{-}\right][/tex] here also has two significant figures. As a result, the [tex]\rm pOH[/tex] in the result should have two decimal places.
Identify a homogeneous catalyst. Identify a homogeneous catalyst. Pd in H2 gas N2 and H2 catalyzed by Fe SO2 over vanadium (V) oxide Pt with methane H2SO4 with concentrated HCl
Answer:
H2SO4 with concentrated HCl
Chlorine monoxide and dichlorine dioxide are involved in the catalytic destruction of stratospheric ozone. They are related by the equation:
2ClO(g) ⇌ Cl2O2(g) for which Kc is 4.96×10^11 at 273 K.
For an equilibrium mixture in which [Cl2O2] is 6.00 x 10^-6M, what is [ClO]?
Answer:
[ClO] = 3.48×10¯⁹ M.
Explanation:
The following data were obtained from the question:
Equilibrium constant (Kc) = 4.96×10¹¹
Concentration of Cl2O2, [Cl2O2] = 6x10¯⁶ M.
Concentration of ClO, [ClO] =.?
The equation for the reaction is given below:
2ClO(g) ⇌ Cl2O2(g)
The equilibrium constant for a reaction is simply defined as the ratio of the concentration of product raised to their coefficient to the concentration of the reactant raised to their coefficient.
The equilibrium constant, Kc for the reaction is given by:
Kc = [Cl2O2] / [ClO]²
Thus, we can calculate the concentration of ClO, [ClO] as follow:
Kc = [Cl2O2] / [ClO]²
4.96×10¹¹ = 6x10¯⁶ / [ClO]²
Cross multiply
4.96×10¹¹ × [ClO]² = 6x10¯⁶
Divide both side by 4.96×10¹¹
[ClO]² = 6x10¯⁶ / 4.96×10¹¹
[ClO]² = 1.21×10¯¹⁷
Take the square root of both side
[ClO] = √ (1.21×10¯¹⁷)
[ClO] = 3.48×10¯⁹ M
Therefore, the concentration of ClO, [ClO] is 3.48×10¯⁹ M.
Why will the conjugate base of a weak acid affect pH? Select the correct answer below: it will react with hydroxide
Answer:
It will react with water
Explanation:
I know this is an older question, but I just wanted to provide the correct answer.
Since we are dealing with a weak base, and the acid is somewhat stronger, it will react with the water molecules to produce hydronium, which will affect the pH of the solution.
The conjugate base of a weak acid affect pH because it will react with hydronium ion.
A weak acid is an acid that does not dissociate completely in water. On the other hand, a strong acid achieves almost 100% dissociation in water.
Acids dissociate in water to yield the hydronium ion and a conjugate base. For instance, the weak acid, acetic acid is dissociated as follows;
CH3COOH(aq) + H2O(l) ⇄ CH3COO-(aq) + H3O+(aq)
We can see that the conjugate base( CH3COO-) could react with the hydronium ions thereby moving the equilibrium position to the left hand side and affecting the pH by decreasing the hydronium ion concentration.
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Missing parts;
Why will the conjugate base of a weak acid affect pH? Select the correct answer below: O it will react with hydroxide O it will react with water O it will react with hydronium O none of the above
If the heat of combustion for a specific compound is −1320.0 kJ/mol and its molar mass is 30.55 g/mol, how many grams of this compound must you burn to release 617.30 kJ of heat?
Answer:
14.297 g
Explanation:
From the question;
1 mo of the compound requires 1320.0 kJ
From the molar mass;
1 ml of the compound weighs 30.55g
How many grams requires 617.30kJ?
1 ml = 1320
x mol = 617.30
x = 617.30 / 1320
x = 0.468 mol
But 1 mol = 30.55
0.468 mol = x
x = 14.297 g
What is advertising used for? Check all that apply. influencing consumer tastes tracking product popularity increasing product awareness promoting company branding gathering data about potential consumers
Answer:
influencing consumer tastes
increasing product awareness
promoting company branding
Explanation:
Advertising is basically a form of communication using creative ideas and communicating benefits of the products. Advertising plays a very crucial role in product business and some of the important uses of advertising are as follows:
Creative advertisements, influence customers or consumers to buy the product.Advertisings involve information regarding the product and so increases product awareness.Advertising on social media platforms, TVs, radio and newspapers, promotes company branding.Hence, the correct options are:
influencing consumer tastesincreasing product awarenesspromoting company branding
Answer:
1,3,4
Explanation:
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