Solution :
A cell that is concentrated is constructed by the same half reaction for the anode as well as he cathode.
We know,
In a standard cell,
the reduction half cell reaction is :
[tex]$Ag^+(aq)+e^- \rightarrow Ag(s) E^0 = -0.80 \ V$[/tex]
The oxidation half ell reaction :
[tex]$Ag(s) \rightarrow Ag^+(aq) + e^- \ E^0= +0.80 \ V$[/tex]
Thus the complete reaction of the cell is :
[tex]$Ag^+(aq)+ Ag(s) \rightarrow Ag^+(aq)+Ag(s)$[/tex]
[tex]$E^0 $[/tex] cell = [tex]$E_R - E_L = 0.00 \ \text{volts}$[/tex]
An elementary step is defined as a chemical collision in a reaction mechanism. A collection of different types of collisions makes up the reaction mechanism, so elementary steps provide a molecular view of the overall reaction.
a. True
b. False
Two planets have similar masses but differ in their distances from the Sun.
Which planet experiences a greater gravitational force with the Sun?
A. The planet that is smaller in volume
B. The planet that is larger in volume
C. The planet that is farther from the Sun
D. The planet that is closer to the Sun
Answer:
D. The planet that is closer to the Sun
2KClO3 (s)⇄2KCl (s)+ 3O2 (aq) equilibrium constant
Answer: The equilibrium constant for the given chemical reaction is [tex][O_2]^3[/tex]
Explanation:
The equilibrium constant is defined as the ratio of the concentration of the products to the concentration of reactants each raised to the power of their stoichiometric coefficients.
The concentration of all the solids and liquids are considered to be 1 in the expression of equilibrium constant
For the given chemical equation:
[tex]2KClO_3(s)\rightleftharpoons 2KCl(s)+3O_2(aq)[/tex]
The expression of equilibrium constant follows:
[tex]K_{eq}=[O_2]^3[/tex]
Hence, the equilibrium constant for the given chemical reaction is [tex][O_2]^3[/tex]
help me plsssssssss.... im timed
Answer:
a
Explanation:
a
Answer:
B. Isotopes of the element.
Explanation:
Isotopes are basically atoms of the same element that contain the same number of protons, but a different number of neutrons.
The information code that an organism inherits can best be referred to as its -
O A genotype
B. territory
C. species
D. kingdom
Answer:
it will be no.A genotype
If 0.5 L of O2(g) reacts with H, to produce 1 L of H2O(g), what is the volume of
H2O(g) obtained from 1 L of O2(g)?
-
0.5 L
2.5 L
2 L
1.5 L
Answer:
2 L
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
2H₂ + O₂ —> 2H₂O
From the balanced equation above, we can see clearly that 1 L of O₂ reacted to 2 L of H₂O.
This implies that 2 L of H₂O can be obtained by the reaction of 1 L of O₂.
Thus, option 3 gives the correct answer to the question.
As a result of the particles in a gas being in constant motion, gas has a _______.
variable volume
variable Pressure
variable Shape
variable mass
Answer:
i think it's variable pressure
if not soo advance sorry :)
The limiting reactant in a chemical reaction is the reactant __________ Select one: A. for which you have the lowest mass in grams. B. which has the lowest coefficient in the balanced equation. C. which has the lowest molar mass. D. which is left over after the reaction has gone to completion. E. None of the above.
Answer:
i think its A
An aqueous sucrose (C12H22O11) solution must be created for an experiment. If 100.00 mL of 0.200 M solution is needed, what amount of sucrose (in grams) must be weighed out
Answer:
6.85 g
Explanation:
Step 1: Given data
Molar concentration of the solution: 0.200 M (0.200 mol/L)Volume of the solution: 100.00 mL (0.10000 L)Step 2: Calculate the moles of sucrose (solute) required
Molarity is equal to the moles of solute divided by the liters of solution.
M = moles of solute / liters of solution
moles of solute = M × liters of solution
moles of solute = 0.200 mol/L × 0.10000 L = 0.0200 mol
Step 3: Calculate the mass corresponding to 0.0200 moles of sucrose
The molar mass of sucrose is 342.3 g/mol.
0.0200 mol × 342.3 g/mol = 6.85 g
The amount of sucrose (in grams) must be weighed out for the given reaction is 6.846g.
How we calculate mass from moles?Mass of any substance will be calculated by using the moles as:
n = W/M, where
W = required mass
M = molar mass
Given that, molarity of sucrose = 0.2 M
Volume of solution = 100mL = 0.1 L
Relation between moles and molarity is represented as:
M = n/V
On putting values on the above equation we get,
n = (0.2)(0.1) = 0.02 moles
We know that molar mass of sucrose (C₁₂H₂₂O₁₁) = 342.3 g/mole
Now we calculate the required mass by putting values on the first equation as:
W = (0.02)(342.3) = 6.846g
Hence, the required mass of sucrose is 6.846g.
To know more about moles, visit the below link:
https://brainly.com/question/24639749
A buffer solution contains 0.475 M nitrous acid and 0.302 M sodium nitrite . If 0.0224 moles of potassium hydroxide are added to 150 mL of this buffer, what is the pH of the resulting solution
Answer: The pH of the resulting solution will be 3.001
Explanation:
Molarity is calculated by using the equation:
[tex]\text{Molarity}=\frac{\text{Moles}}{\text{Volume}}[/tex] ......(1)
We are given:
Moles of NaOH = 0.0224 moles
Molarity of nitrous acid = 0.475 M
Molarity of sodium nitrite = 0.302 M
Volume of solution = 150 mL = 0.150 L (Conversion factor: 1 L = 1000 mL)
Putting values in equation 1, we get:
[tex]\text{Moles of nitrous acid}=(0.475mol/L\times 0.150L)=0.07125mol[/tex]
[tex]\text{Moles of sodium nitrite}=(0.302mol/L\times 0.150L)=0.0453mol[/tex]
The chemical equation for the reaction of nitrous acid and NaOH follows:
[tex]HNO_2+NaOH\rightleftharpoons NaNO_2+H_2O[/tex]
I: 0.07125 0.0224 0.0453
C: -0.0224 -0.0224 +0.0224
E: 0.04885 - 0.0677
The power of the acid dissociation constant is the negative logarithm of the acid dissociation constant. The equation used is:
[tex]pK_a=-\log K_a[/tex] ......(2)
We know:
[tex]K_a[/tex] for nitrous acid = [tex]7.2\times 10^{-4}[/tex]
Using equation 2:
[tex]pK_a=-\log (7.2\times 10^{-4})=3.143[/tex]
To calculate the pH of the acidic buffer, the equation for Henderson-Hasselbalch is used:
[tex]pH=pK_a+ \log \frac{\text{[conjugate base]}}{\text{[acid]}}[/tex] .......(3)
Given values:
[tex][NaNO_2]=\frac{0.0677}{0.150}[/tex]
[tex][HNO_2]=\frac{0.04885}{0.150}[/tex]
[tex]pK_a=3.143[/tex]
Putting values in equation 3. we get:
[tex]pH=3.143-\log \frac{(0.0677/0.150)}{(0.04885/0.150)}\\\\pH=3.143-0.142\\\\pH=3.001[/tex]
Hence, the pH of the resulting solution will be 3.001
Determine the electron geometry, molecular geometry, and idealized bond angles for each of the following molecules. CF4CF4 NF3NF3 OF2OF2 H2SH2S In which cases do you expect deviations from the idealized bond angle
Answer:
CF4
Molecular geometry- tetrahedral
Electron geometry- tetrahedral
NF3
-molecular geometry - trigonal pyramidal
Electron geometry - tetrahedral
OF2
Molecular geometry - bent
Molecular geometry - tetrahedral
H2S
Molecular geometry- bent
Electron geometry - tetrahedral
Explanation:
According to Valence Shell Electron Pair Repulsion Theory, the shape of a molecule depends on the number of electron pairs on the valence shell of the central atom in the molecule.
For all the compounds listed, the central atom has four points of electron density. This correspond to a tetrahedra electron pair geometry. The presence of lone pairs on the central atom of OF2,NF3 and H2S accounts for the departure of the observed molecular geometry from the geometry and idealized bond angle predicted on the basis of the VSEPR theory.
Please select the word from the list that best fits the definition
Include skin, the respiratory system, the circulatory system, and inflammation
A. antibody
B. antigen
C.Natural Defences
D.active immunity
Answer:
I choose D option because may be it's correct
I believe it is c
Explanation:
because if u think about it and also do some research you would see that the circulatory system is a strong part of your body which can help u through natural defences if this sounds weird it's all in research but if it ain't c dont blame me for ruining your life- lol but yeah I think its c
Given the balanced reaction: Zn + 2HCl → H2 + ZnCl2
If 5 grams of each reactant are available for the reaction and HCl is known
to be the limiting reactant, which of the following is correct?
O Both reactants will be completely used up.
O There will be excess of both reactants remaining.
O HCl will be completely used up while Zn will remain in excess.
O Zn will be completely used up while HCl will remain in excess.
Answer:
O HCl will be completely used up while Zn will remain in excess.
Explanation:
Zn + 2HCl → H₂ + ZnCl₂In reactions involving two reactants, if one of them is the limiting reactant then the other one has to be the reactant in excess.
Meaning that in this case, the reaction will proceed until HCl is completely used up, and a certain amount of Zn will remain (thus being the reactant in excess).
Francine makes several measurements of the mass of a metal block. The data set is shown in the table below.
Measurement
Mass of metal block (g)
1
20.73
2
20.76
3
20.68
4
20.75
After analyzing this data set, Francine calculates a value of 20.73 g.
Which of these characteristics has been calculated?
mean
median
mode
range
Answer:
Mean
Explanation:
The mean of a series of measurements is calculated when all the measurements are added up and then divided by the number of measurements taken, as follows:
Sum of Measurements = 20.73 + 20.76 + 20.68 + 20.75 = 82.92As there are 4 measurements, the mean is:
Mean = 82.92 / 4 = 20.73Answer:
A
Explanation:
what climate zone is asia?
Answer:
The Tundra Climate
Explanation:
:) hope this helps
is it a physical or chemical change when a candle is lit
What happens when you increase the pressure on the following
reaction
2X(g) + 3Y(g) --> 5Z(g) + heat
Answer:
When pressure increases, there is no effect on the reaction because volume is constant.
A rod “X”, has a positive charge of 5. An otherwise identical rod, “Y”, has a negative charge of 12.
The rods are touched together and then separated.
a) When they touched, what particles moved between them?
b) Did the particles move from “X” to “Y” or from “Y” to “X”.
Answer:
a) electrons
b) from Y to X
Explanation:
positive protons are the cores of atoms in relative to electrons very, very heavy.
the outer electrons of atoms can move, under certain conditions, away from the atom, leaving it electrically unbalanced -> positively charged
there can also be a surplus of electrons on many surfaces, leading to a static negative charge. you know this when you are charged and you discharge with an object or another person, electrically balancing the two bodies charge.
electrons are also much smaller. they are the "things" to move, let it be trough the air or trough a wire, while the heavy protons will stay in place (unless the materials is melted of course, extreme heat brakes the bounds between atoms relatively well)
Question 13 of 32
The pH of a sample of seawater is 8.0. What is the hydrogen ion concentration of seawater?
A. 8.0 M
B. 1 x 108 M
C. 6.0 M
D. 1 x 10-8 M
Answer:
D
Explanation:
Phosphorus -32 is a commonly used radioactive nuclide in biochemical research, particularly in studies of nucleic acids. The half-life of phosphorus-32 is 14.3 days. What mass of phosphorus-32 is left from an original sample of 175 mg of Na332PO4 after 35.0 days
Answer: The mass of P-32 left from the original sample is 32.07 mg
Explanation:
All radioactive decay processes follow first-order reactions.
Calculating rate constant for first order reaction using half life:
[tex]t_{1/2}=\frac{0.693}{k}[/tex] .....(1)
[tex]t_{1/2}[/tex] = half life period = 14.3 days
k = rate constant = ?
Putting values in equation 1:
[tex]k=\frac{0.693}{14.3days}\\\\k=0.0485days^{-1}[/tex]
The integrated rate law equation for first-order kinetics:
[tex]k=\frac{2.303}{t}\log \frac{a}{a-x}[/tex] ......(2)
Given values:
a = initial concentration of reactant = 175 mg
a - x = concentration of reactant left after time 't' = ? mg
t = time period = 35.0 days
Putting values in equation 2:
[tex]0.0485days^{-1}=\frac{2.303}{35.0 days}\log (\frac{175}{a-x})\\\\\log (\frac{175}{a-x})=\frac{0.0485\times 35.0}{2.303}\\\\\log (\frac{175}{a-x})=0.737\\\\\frac{175}{a-x}=10^{0.737}\\\\a-x=\frac{175}{5.457}=32.07mg[/tex]
Hence, the mass of P-32 left from the original sample is 32.07 mg
Iodine-131 is administered orally in the form of NaI(aq) as a treatment for thyroid cancer. The half-life of iodine-131 is 8.04 days. If you begin with 34.7 mg of this isotope, what mass remains after 8.52 days have passed
Answer:
16.6 mg
Explanation:
Step 1: Calculate the rate constant (k) for Iodine-131 decay
We know the half-life is t1/2 = 8.04 day. We can calculate the rate constant using the following expression.
k = ln2 / t1/2 = ln2 / 8.04 day = 0.0862 day⁻¹
Step 2: Calculate the mass of iodine after 8.52 days
Iodine-131 decays following first-order kinetics. Given the initial mass (I₀ = 34.7 mg) and the time elapsed (t = 8.52 day), we can calculate the mass of iodine-131 using the following expression.
ln I = ln I₀ - k × t
ln I = ln 34.7 - 0.0862 day⁻¹ × 8.52 day
I = 16.6 mg
a solution that contains a large amount of solute would be described as what
Answer:
A concentrated solution is one that has a relatively large amount of dissolved solute. A dilute solution is one that has a relatively small amount of dissolved solute.
the density of gold is 19.3 g/ml. if you had a cube made of pure gold that weighs 58.1 pounds what would be the lengh of a side of this cube
Answer:
11.1 cm
Explanation:
First we convert 58.1 pounds into grams:
58.1 lb * [tex]\frac{453.592g}{1lb}[/tex] = 26353.69 gThen we calculate the volume of the gold cube, using the given density:
26353.69 g ÷ 19.3 g/mL = 1365.48 mL
1365.48 mL is equal to 1365.48 cm³.With the volume of a cube we can calculate the side length:
Length = ∛(1365.48 cm³) = 11.1 cmHydrogen bonds within liquid water are attractions between protons and hydroxide ions. are dipole-dipole attractions. are ion-induced dipole attractions. are attractions between protons and oxygen nuclei. are attractions between two hydrogen atoms.
Answer:
true because the bonds cannot be broken down
A molecule of composition is replicated in a solution containing unlabeled (not radioactive) GTP, CTP, and TTP plus adenine nucleoside triphosphate with all its phosphorus atoms in the form of the radioactive isotope 32P. Will both daughter molecules be radioactive
Answer:
Please find the complete question in the attached file.
Explanation:
It would only be radioactive if the DNA molecule that employed the poly-T rand as templates. Its other molecule of the daughter would not have been radioactive as it did not need dATP for its replication. While each strand of the second molecule includes t, simultaneous reproduction dATP from both daughter molecules is needed so that each of those is radioactive.
Compound A has the formula C8H8. It reacts rapidly with acidic KMnO4 but reacts with only 1 equivalent of H2 over a palladium catalyst. On hydrogenation under conditions that reduce aromatic rings, A reacts with 4 equivalents of H2, and hydrocarbon B, C8H16, is produced. The reaction of A with KMnO4 gives CO2 and a carboxylic acid C, C7H6O2.
Required:
Draw the structure of compound B below.
Answer:
C8H16 (Ethylcyclohexane).
Explanation:
From the given information:
Compound A is an alkene because it interacts with 1 unit of hydrogen across a palladium catalyst.
Also, we are given another hint that:
Compound A needs 4 equivalence of H2 to hydrogenate under circumstances that decrease aromatic rings, indicating that it is a phenyl substituted alkene.
Compound A with formula C8H8 reacts instantly with KMnO4 to produce CO2, as well as carboxylic acid, points out that Compound acts as a terminal alkene.
Therefore, we can opine that compound A is a terminal phenyl substituted alkene whose formula = C8H8 (Styrene)
The diagrammatic expression of the compound can be seen below.
However, in the presence of the palladium catalyst, the reduction of Compound A with 4 units of hydrogen produces Compound B: C8H16 (Ethylcyclohexane).
gress.
Newton's third law of motion
Is the following statement true or false?
When a large truck pushes a small car with a given force, the small car is applying an equal and opposite
force on the truck.
Choose 1 answer:
True
False
Answer:
true
Explanation:
the small car also has gravity making it heavy
Which commercial technology commonly uses plasmas?
a radio
a race car
a television
a microwave oven
Answer:
A television is commercial technology commonly uses plasmas.The element tin has the following number of electrons per shell: 2, 8, 18, 18, 4. Notice that the number of electrons in the outer shell of a tin atom is the same as that for a carbon atom. Therefore, what must be true of tin
The question is incomplete, the complete question is:
The element tin has the following number of electrons per shell: 2.8. 18, 18, 4. Notice that the number of electrons in the outer shell of a tin atom is the same as that for a carbon atom. Therefore, what must be true of tin? Tin is a polar atom and can bind to other polar atoms. Tin has a high molecular weight to give tin-containing molecules greater stabilty. All of the above Tin conform single covalent bonds with other elements, but not double or triple covalent bonds Tincan bind to up to four elements at a time
Answer:
Tin can bind to up to four elements at a time
Explanation:
Certain important points were made in the question about tin and one of them is that tin is an element in the same group as carbon hence it has the same number of valence electrons as carbon.
Carbon is always tetra valent. The tetra valency of carbon is the idea that carbon forms four bonds.
If tin has the same number of valence electrons as carbon, then, tin can bind to up to four elements at a time
A 250ml sample of sir at 373.15k is warmed at 473.15 at constant pressure. What is the volume of the air sample at the new temperature
Answer:
317 mL.
Explanation:
From the question given above, the following data were obtained:
Initial volume (V₁) = 250 mL
Initial temperature (T₁) = 373.15 K
Final temperature (T₂) = 473.15 K
Pressure = Constant
Final volume (V₂) =?
The final volume of the air sample can be obtained by using the Charles' law equation as illustrated below:
V₁/T₁ = V₂/T₂
250 / 373.15 = V₂ / 473.15
Cross multiply
373.15 × V₂ = 250 × 473.15
373.15 × V₂ = 118287.5
Divide both side by 373.15
V₂ = 118287.5 / 373.15
V₂ = 317 mL
Therefore, the final volume of the air sample is 317 mL
