Answer: The empirical formula of the compound is [tex]KMnO_4[/tex]
Explanation:
The empirical formula is the chemical formula of the simplest ratio of the number of atoms of each element present in a compound.
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
Let the mass of the compound be 100 g
Given values:
% of K = 24.7%
% of Mn = 34.76%
% of O = 40.50%
Mass of K = 24.7 g
Mass of Mn = 34.76 g
Mass of O = 40.50 g
To calculate the empirical formula of a compound, few steps need to be followed:
Step 1: Calculating the number of moles of each elementWe know:
Molar mass of K = 39.10 g/mol
Molar mass of Mn = 54.94 g/mol
Molar mass of O = 16 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of K}=\frac{24.7g}{39.10g/mol}=0.632 mol[/tex]
[tex]\text{Moles of Mn}=\frac{34.76g}{54.94g/mol}=0.633 mol[/tex]
[tex]\text{Moles of O}=\frac{40.50g}{16g/mol}=2.53 mol[/tex]
Step 2: Calculating the mole fraction of each element by dividing the calculated moles by the least calculated number of moles that is 0.632 moles[tex]\text{Mole fraction of K}=\frac{0.632}{0.632}=1[/tex]
[tex]\text{Mole fraction of Mn}=\frac{0.633}{0.632}=1[/tex]
[tex]\text{Mole fraction of O}=\frac{2.53}{0.632}=4[/tex]
Step 3: Writing the mole fraction as the subscripts of each of the elementThe empirical formula of the compound becomes [tex]K_1Mn_1O_4=KMnO_4[/tex]
Hence, the empirical formula of the compound is [tex]KMnO_4[/tex]
The empirical formula of the compound is KMnO₄
Data obtained from the question Potassium (K) = 24.74%Manganese (Mn) = 34.76%Oxygen (O) = 40.50%Empirical formula =? How to determine the empirical formulaDivide by their molar mass
K = 24.74 / 39 = 0.634
Mn = 34.76 / 55 = 0.632
O = 40.50 / 16 = 2.531
Divide by the smallest
K = 0.634 / 0.632 = 1
Mn = 0.632 / 0.632 = 1
O = 2.531 / 0.632 = 4
Therefore, the empirical formula of the compound is KMnO₄
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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 cmgress.
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
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
Boric acid, B(OH)3, acts as an acid in water, but does not do so via ionization of a proton from boric acid. Use a balanced chemical equation to show why boric acid may be described as an acid in water.
Answer:
B(OH)₃ + H₂O = B(OH)₄⁻ + H⁺
Explanation:
Let's consider Arrhenius acid-base theory:
An acid is a substance that in aqueous media releases H⁺.A base is a substance that in aqueous media releases OH⁻.Boric acid, B(OH)₃ reacts with water according to the following equation.
B(OH)₃ + H₂O = B(OH)₄⁻ + H⁺
As we can see, boric acid releases H⁺ in aqueous media. Thus, it is an acid.
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).
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
what climate zone is asia?
Answer:
The Tundra Climate
Explanation:
:) hope this helps
Hydrogen 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
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
1,500 grams is equal to:
1.5 kg
1.5L
1.5 cm
all of the above
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).
Upon decomposition, a sample of sodium bicarbonate produces 0.0118 g of sodium carbonate, Na2CO3. How many grams of water does it produce
Answer:
2.00 × 10⁻³ g
Explanation:
Step 1: Write the balanced decomposition reaction
2 NaHCO₃ ⇒ Na₂CO₃ + CO₂ + H₂O
Step 2: Calculate the moles corresponding to 0.0118 g of Na₂CO₃
The molar mass of Na₂CO₃ is 105.99 g/mol.
0.0118 g × 1 mol/105.99 g = 1.11 × 10⁻⁴ mol
Step 3: Calculate the moles of H₂O produced with 1.11 × 10⁻⁴ moles of Na₂CO₃
The molar ratio of Na₂CO₃ to H₂O is 1:1. The moles of H₂O produced are 1/1 × 1.11 × 10⁻⁴ mol = 1.11 × 10⁻⁴ mol.
Step 4: Calculate the mass corresponding to 1.11 × 10⁻⁴ moles of H₂O
The molar mass of H₂O is 18.02 g/mol.
1.11 × 10⁻⁴ mol × 18.02 g/mol = 2.00 × 10⁻³ g
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 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
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.
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)
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
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
how many moles of H2o are equivelant to 97.3g h2o
Answer:
Number of mole in H2O = 5.4 moles
Explanation:
Given:
Mass of H2O = 97.3 gram
Find:
Number of mole in H2O
Computation:
We know that molar mass of H2O = 18 g/mole
So,
Number of mole = Given Mass / Molar mass
Number of mole in H2O = Mass of H2O / Molar mass of H2O
Number of mole in H2O = 97.3 / 18
Number of mole in H2O = 5.4055
Number of mole in H2O = 5.4 moles
if the radius of a potassium atom is 2.27 x 10-7 mm what is the radius in μm
Explanation:
the answer is in the Image above
The radius of the potassium atom is [tex]2.27\times 10^{-4} \mu m[/tex].
Explanation:
Given:
The radius of potassium atom =[tex]2.27\times 10^{-7} mm[/tex]
To find:
The radius of potassium atom in micrometers.
Solution:
The radius of potassium atom =[tex]2.27\times 10^{-7} mm[/tex]
In 1 millimeter there are 1000 micrometers.
[tex]1 mm = 1000 \mu m[/tex]
Then in [tex]2.27\times 10^{-7} mm[/tex]:
[tex]=2.27\times 10^{-7} \times 1000\\=2.27\times 10^{-4} \mu m[/tex]
The radius of the potassium atom is [tex]2.27\times 10^{-4} \mu m[/tex].
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If the initial amount of indium-117 is 5.2 g, how much indium-117 is left in the body after 86 min
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:
In which of the following molecules is hydrogen bonding likely to be the most significant component of the total intermolecular forces??
a. C6H13NH2
b. CH3OH
c. CH4
d. C5H11OH
e. CO2
Answer:
The given molecules are:
a. C6H13NH2
b. CH3OH
c. CH4
d. C5H11OH
e. CO2
Explanation:
The hydrogen bond is a weak electrostatic force of attarction that exists between covalently bonded hydrogen (of -OH or -NH2 or HF) with a highly electronegative atom like N,O and F.
Hydrogen bonding is of two types:
Intermolecular hydrogen bond (exists between two molecules)
Intramolecular hydrogen bond(exists within a molecule).
For example intermolecular hydrogen bond in water is shown below:
Among the given options,
a. C6H13NH2 has -NH2 linkage which leads to hydrogen bond formation.
b. CH3OH has -OH bond and it leads to hydrogen bond fomation.
d. C5H11OH has also -OH bond and it leads to hydrogen bond fomation.
Reamining molecules, CH4 and CO2 do not form hydrogen bond.
Hence, answer is:
options a,b,d.
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.
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:
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
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
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.
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.
