Answer:
Vibrational energies of atoms
Explanation:
Planck applied quantization to Vibrational energies of atoms because in Black body spectrum prediction a blackbody at equilibrium is expected to radiate energies at various Frequencies ( i.e. increase in radiated energy ∝ increase in frequency ) but towards the ultraviolet region of the spectrum the energy radiated begins to drop as frequency increases. The phenomenon of drop in energy with increase in frequency is termed Ultraviolet catastrophe. hence to solve this phenomenon Planck applied quantization to Vibrational energies of atoms
A change of state is a(n)
process.
A. irreversible
B. reversible
Answer:
Changes of states are reversible, you can go from a solid to liquid and liquid to solid.Answer:
Reversible
Explanation:
Changes of state are physical changes in matter. Common changes of the state include melting, freezing, sublimation, deposition, condensation, and vaporization.
0
Which is not one of Earth's layers?
A А
crust
B)
inner core
mantle
D
ocean
The ocean is not a part of Earth's layers.
Answer:
Ocean
Explanation:
An atom has 81 electrons, 84 neutrons, and 82 protons. What element is this atom?
Answer:
Lead
Explanation:
The subatomic particles within an atom can be used to know the atom or element given.
Of particular interest is the number of protons within the atom.
The periodic table is based on the atomic number of atoms. This atomic number is the number of protons within an atomic space.
So; If we know the number of protons within an atom, we can know the element.
The number of protons given is 82, the element is therefore lead.
Answer:
The atomic number of polonium is 84. The atomic number lead is 82.
Explanation:
A particular term in an atom in which LS coupling is a good approximation splits into three levels, each having the same L and same S but different J. If the relative spacings between the levels are in the proportion 5:3, find L and S.
Answer:
Explanation:
From the information given;
Consider using Lande's Interval rule which can be expressed as:
[tex]\Delta E = E_{j+1} - E_jj \ = \alpha (j+1)[/tex]
here;
[tex]j+1[/tex] = highest level of j
and
[tex]\dfrac{\Delta E_1}{\Delta E_2} = \dfrac{(j+2)}{(j+1)}[/tex]
[tex]\dfrac{5}{3} = \dfrac{(j+2)}{(j+1)}[/tex]
[tex]5(j+1) = 3(j+2)[/tex]
[tex]5j+5 = 3j+6[/tex]
[tex]2j = 1\\ \\ j = \dfrac{1}{2}[/tex]
recall that:
[tex]j = |S-L| \ \to \ |S+L |[/tex]
So;
[tex]S-L = \dfrac{1}{2} --- (1)[/tex]; &
[tex]S+L = \dfrac{5}{2} --- (1)[/tex]
Using the elimination method, we have:
[tex]2S = \dfrac{6}{2}[/tex]
[tex]S = \dfrac{3}{2}[/tex]
Since [tex]S = \dfrac{3}{2}[/tex]; then from (1)
[tex]\dfrac{3}{2} -L = \dfrac{1}{2}[/tex]
[tex]L = \dfrac{2}{2}[/tex]
[tex]L = 1[/tex]
A sample of PCl5 weighting 2.69 gram was placed in 1.00 Litter container and completely vaporized at 250C. The pressure observed at that temperature was 1.00 atm. The possibility exists that some of the PCl5 dissociated according to PCl5 (g) ! PCl3 (g) Cl2 (g) . What must be the partial pressures of PCl5 PCl3 and Cl2 under these experimental conditions
Answer:
Partial pressures:
PCl₅ = 0.558 atm
PCl₃ = 0.22 atm
Cl₂ = 0.22 atm
Explanation:
From the given information:
The number of moles of PCl₅ associated with the evaporation is:
[tex]n_{PCl_5}= \dfrac {weight \ of \ PCl_5} {M.Wt. \ of \ PCl_5}[/tex]
[tex]n_{PCl_5}= \dfrac {2.69 \ g} {208.5 \ g/mol}[/tex]
[tex]n_{PCl_5}= 0.013 \ mol[/tex]
Temperature of the gas = 250° C = (250 + 273.15) K
= 523.15 K
Using the Ideal gas equation to determine the pressure exerted by the completely vaporized PCl₅
PV = nRT
[tex]P = \dfrac{nRT}{V}[/tex]
[tex]P = \dfrac{0.0013 \ mol \times 0.082 \ Latm^0 K^{-1} . mol ^{-1} \times 523.15 \ K}{1.0 \ L}[/tex]
P = 0.558 atm
Thus, at 250° C, decomposition of PCl₅ occurs.
In the container, PCl₅ decomposes to PCl₃ and Cl₂.
i.e.
[tex]PCl_{5(g)} \to PCl_{3(g)}+ Cl_{2(g)}[/tex]
Using Dalton's Law:
[tex]P_{total } =P_1 + P_2+P_3 +...[/tex]
[tex]P_1 = P_{Total} \times X_1[/tex]
where;
X = mole fraction
Then, the total no. of moles in the container is:
[tex]n = \dfrac{PV} {RT}[/tex]
[tex]n = \dfrac{1\ atm \times 1.0\ L}{0.0821 \ L \ atm \ K^{-1}.mol \times 523.15\ K}[/tex]
n = 0.023 mol
Now, the container contains a total amount of 0.023 mol where initially 0.013 mol are that of PCl₅ and remaining 0.005 mol of PCl₃ and 0.005 mol of Cl₂.
Thus, the partial pressure of PCl₃ is:
[tex]P__{PCL_3} }= P_{total} \times \dfrac{no. \ of \ moles \ of PCl_5}{total \ no. \ of \ moles}[/tex]
[tex]P__{PCL_3}} = 1 \ atm \times \dfrac{0.005}{0.023}[/tex]
[tex]P__{PCL_3}} = 0.22 \ atm[/tex]
Thus, since the no of moles of PCl₃ and Cl₂ are the same, then the partial pressure for Cl₂ is = 0.22 atm
Which of the following choices is not evidence supporting the theory of plate tectonics?
Answer:
B
Explanation:
Molecule A undergoes isomerization to molecule B in acetone. Using curved arrows, showing key intermediates and any formal charges, propose a detailed mechanism for this isomerization. Provide a brief explanation why this isomerization occurs.
Answer:
hello your question is incomplete attached below is the complete question
answer :
we use Isomerization because conjugated allylic carbocation is more stable when compared to a Non-conjugated Allylic carbocation
Explanation:
Reason for the mechanism
we use Isomerization because conjugated allylic carbocation is more stable when compared to a Non-conjugated Allylic carbocation
attached below is the detailed mechanism
You want to clean a 500-ml flask that has been used to store a 0.9M solution. Each time the flask is emptied, 1.00 ml of solution adheres to the walls, and thus remains in the flask. For each rinse cycle, you pour 9.00 ml of solvent into the flask (to make 10.00 ml total), swirl to mix uniformly, and then empty it. What is the minimum number of such rinses necessary to reduce the residual concentration of 0.00001 M or below
Answer:
In the 5th cycle rinse, the residual concentration of the solution is < 0.00001M
Explanation:
In each rinse cycle, the dilution that you are doing of the solution is from 1.00mL to 10.00mL, that is a dilution of 10
In the first rinse the concentration must be of 0.9M 10 = 0.09M
2nd = 0.009M
3rd = 0.0009M
4th = 0.00009M
5th = 0.000009M →
In the 5th cycle rinse, the residual concentration of the solution is < 0.00001MHow many orbitals in an atom can have each of the following designations:
(a) 1s;
(b) 4d;
(c) 3p;
(d) n=3?
Answer:
(a) 1s; has one orbital
(b) 4d; has five orbitals
(c) 3p; has three orbitals
(d) n=3 has nine orbitals
Explanation:
Electrons in an atom are always in constant motion, making it hard to predict there exact position. However, the most probable locations electrons can be be found are described with the terms shells, subshells and orbitals. A shell contains subshells and orbitals are found within subshells. The shells are given names such as K, L, M, N, which correspond to the principal quantum numbers, n = 1, 2, 3, and 4 respectively. There are 4 major types of subshells that can be found in a shell. They are named as s, p, d, f. Each subshell is composed of several orbitals.
a. 1s; the s subshell has only one orbital. Therefore, the 1s subshell has one orbital
b. 4d; the d subshell has five orbitals. Therefore, the 4d subshell has five orbitals
c. 3p; the p subshell has three orbitals. Therefore, the 3d subshell has three orbitals
d. n = 3; the shell with n = 3 has the following subshells, 3s, 3p, 3d.the number of orbitals will be 1 + 3 + 5 = 9 orbitals. Therefore, the number of orbitals in n = 3 is nine orbitals
To determine the concentration of citric acid, you will need to titrate this solution with 0.100 M NaOH. You are given a 1.00 M NaOH stock solution and will need to make enough 0.100 M NaOH to perform 3 titrations. For each titration, you will use 20.0 mL of 0.100 M NaOH solution.
Calculate the total volume (in mL) of the diluted solution you will need to prepare for the 3 titrations.
Determine the minimum volume (in mL) of 1.00 M NaOH stock solution needed to prepare the 0.100 M NaOH solution.
Answer:
60.0mL of the diluted solution are needed
6.00mL of the 1.00M NaOH stock solution is the minimum volume needed to prepare the diluted solution.
Explanation:
As in each titration we need to use 20.0mL of the diluted 0.100M solution. As there are 3 titration, the volume must be:
3 * 20.0mL = 60.0mL of the diluted solution are needed
Now, to prepare a 0.100M NaOH solution from a 1.00M NaOH stock solution the dilution must be of:
1.00M / 0.100M = 10 times must be diluted the solution.
As we need at least 60.0mL, the minimum volume of the stock solution must be:
60.0mL / 10 times =
6.00mL of the 1.00M NaOH stock solution is the minimum volume needed to prepare the diluted solution.Classify each of the following compounds as a strong acid, weak acid, strong base, or weak base, and write the Ka expression for any weak acid or weak base:
1. [ Select ] ["strong base", "weak base", "strong acid", "weak acid"] LiOH
2. [ Select ] ["weak acid", "strong acid", "strong base", "weak base"] HF
3. [ Select ] ["strong acid", "weak acid", "strong base", "weak base"] HCl
4. [ Select ] ["weak base", "strong base", "weak acid", "strong acid"] NH3
Ka expression: [ Select ] ["[H+][F-] / [HF]", "[Li+][OH-]/ [LiOH]", "[H+][Cl-} / [HCl]", "[NH4+] / [NH3]", "[HF] / [H+][F-}", "[LiOH] / [Li+][OH-]", "[HCl] / [H+][Cl-}", "none"]
Calculate the concentration of OHLaTeX: -? in a solution that has a concentration of H+ = 7 x 10LaTeX: -?6 M at 25°C. Multiply the answer you get by 1010 and enter that into the field to 2 decimal places.
Answer:
See explanation below
Explanation:
There are several ways to know if an acid or base is strong. One method is calculating the pH. If the pH is really low, is a strong acid, and if it's really high is a strong base.
However we do not have a pH value here.
The other method is using bronsted - lowry theory. If an acid is strong, then his conjugate base is weak. Same thing with the bases.
Now, Looking at the 4 compounds, we can say that only two of them is weak and the other two are strong compounds. Let's see:
LiOH ---> Strong. If you try to dissociate :
LiOH ------> Li⁺ + OH⁻ The Li⁺ is a weak conjugate acid.
HF -----> Weak
HF --------> H⁺ + F⁻ The Fluorine is a relatively strong conjugate base.
HCl -----> Strong
This is actually one of the strongest acid.
NH₃ ------> Weak
Now writting the Ka and Kb expressions:
Ka = [H⁺] [F⁻] / [HF]
Kb = [NH₄⁺] [OH⁻] / [NH₃]
Finally, to calculate the [OH⁻] we need to use the following expression:
Kw = [H⁻] [OH⁻]
Solving for [OH⁻] we have:
[OH⁻] = Kw / [H⁺]
Remember that the value of Kw is 1x10⁻¹⁴. So replacing:
[OH⁻] = 1x10⁻¹⁴ / 7x10⁻⁶
[OH⁻] = 1.43x10⁻⁹ M
And now, multiplying by 10¹⁰ we have:
[OH⁻] = 1.429x10⁻⁹ * 1x10¹⁰
[OH⁻] = 14.29Hope this helps
Strong acids and bases are those which completely ionized in body fluid, and weak acids and bases are those who does not completely ionized in body fluid.
Ka expression is used to differentiate between strong and weak acids.
Which are strong acids and base and weak acids and bases?LiOH - strong baseHF - weak acidHCl - strong acidNH3 - weak baseWhat are the Ka expression of the following?Weak acid – HF[tex]\bold{\dfrac{[H+][F-]}{[HF]}}[/tex]
Weak base – NH3[tex]\bold{\dfrac{[NH_4^+] [OH^-]}{[NH_3]} }[/tex]
Calculate the concentration of OH?Given, [tex]\bold{ [H^+]=1\times10^-^6\; at \;25^oC}[/tex]
We know, [tex]\bold{ [H^+]\times[OH^-]=1\times10^-^6\; at \;25^oC}[/tex]
[tex]\bold{[OH^-]=\dfrac{1\times10^-^1^4}{6.2\times10^-^6} = 1.43\times10^-^9}[/tex]
Now, multiplying the value by [tex]10^1^0[/tex]
[tex]\bold{( 1.429\times10^-^9) \times 1\times10^1^0= 14.29}[/tex]
Thus, the value is 14.29.
Learn more about acid and base, here:
https://brainly.com/question/10468518
A chemist prepares a solution of aluminum sulfate by weighing out of aluminum sulfate into a volumetric flask and filling the flask to the mark with water. Calculate the concentration in of the chemist's aluminum sulfate solution. Be sure your answer has the correct number of significant digits.
Answer:
25.8 g/dL
Explanation:
A chemist prepares a solution of aluminum sulfate by weighing out 116.0 g of aluminum sulfate into a 450. mL volumetric flask and filling the flask to the mark with water. Calculate the concentration in g/dL of the chemist's aluminum sulfate solution. Be sure your answer has the correct number of significant digits.
Step 1: Given data
Mass of aluminum sulfate (m): 116.0 gVolume of the solution (V): 450. mLStep 2: Convert "V" to dL
We will use the following conversion factors.
1 L = 1000 mL1 L = 10 dL450. mL × 1 L/1000 mL × 10 dL/1 L = 4.50 dL
Step 3: Calculate the concentration (C) of aluminum sulfate if g/dL
We will use the following expression.
C = m/V = 116.0 g/4.50 dL = 25.8 g/dL
A species of desert plant produces flowers that only bloom at night. How does this enhance the survival of the species?
A. This allows the plants to conserve water and not bloom during the heat of the day.
B. This species relies on nocturnal animals like moths for pollination and reproduction
C. This species relies on moonlight for photosynthesis.
D.This allows flowers to stay closed durng the day when herbivores are more likely to eat them.
Od
Answer:
A. This allows the plants to conserve water and not bloom during the heat of the day
Explanation:
Most desert plants only bloom at night because they take advantage of animals like moths and insects that fly at night for pollination and reproduction.
Because these plants are in the desert and do not get enough water except from short occasional rainfalls, they conserve water and not bloom during the heat of the day. They bloom at night when the temperature is low and this enhances their water conservation and survival.
water is a unique material in that the density of the solid is lower than the density of the liquid (which is why ice forms at the top of a pond and why ice floats in our drinks). if the density for ice at 0C is .917g/mL and the density for water at 0C is .999 g/mL, what is the calculated free space (as %) for each of these materials. you will need to estimate the volume of water as the sum of 2 H atoms and 1 O atom with radii 37 and 66 pm respectively. note that you will also have to assume a quantity of water to perform this exercise
Answer:
% Free space in water = [tex]\frac{9.945* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.45%
% Free space in ice = [tex]\frac{9.98* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.8%
Explanation:
As given ,
Density for ice at 0⁰C = 0.917 g/ml
Density for water at 0⁰C = 0.999 g/ml
Radii of H atoms = 37 pm
Radii of O atoms = 66 pm
Now,
Consider 1 ml of water = 1 cm²
As , we know that mass of water in 1 cm² = 0.999 g
Moles of water = [tex]\frac{0.999}{18} = 0.056[/tex]
Volume of H₂O = 1.624×[tex]10^{-31}[/tex] m²
Now,
Volume occupied by water = 0.056×6.022×[tex]10^{23}[/tex]× 1.624×[tex]10^{-31}[/tex] m²
= 5.48×[tex]10^{-9}[/tex] m²
⇒Volume occupied by water = 5.48×[tex]10^{-9}[/tex] m²
Now,
Free space = 1×[tex]10^{-6}[/tex] - 5.48×[tex]10^{-9}[/tex] = 9.95×[tex]10^{-7}[/tex] m²
% Free space = [tex]\frac{9.945* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.45%
Now,
Consider 1 ml of ice = 1 cm²
S.I unit of ice = 1×[tex]10^{-6}[/tex] m²
As , we know that mass of water in 1×[tex]10^{-6}[/tex] m² = 0.917 g
Moles of ice = [tex]\frac{0.917}{18} = 0.012[/tex]
Volume of H₂O = 6.022×[tex]10^{23}[/tex] ×0.012
Volume of ice unit = [tex]\frac{4}{3} \pi (37*10^{-12})^{3} *2 + \frac{4}{3} \pi (66*10^{-12})^{3} = 1.624*10^{-31}m^{3}[/tex]
Now,
Volume occupied by water = 0.012×6.022×[tex]10^{23}[/tex]× 1.624×[tex]10^{-31}[/tex] m²
= 1.17×[tex]10^{-9}[/tex] m²
⇒Volume occupied by water = 1.17×[tex]10^{-9}[/tex] m²
Now,
Free space = 1×[tex]10^{-6}[/tex] - 1.17×[tex]10^{-9}[/tex] = 9.98×[tex]10^{-7}[/tex] m²
% Free space = [tex]\frac{9.98* 10^{-7} }{1*10^{-6} }[/tex]×100 = 99.8%
A reaction between liquid reactants takes place at in a sealed, evacuated vessel with a measured volume of . Measurements show that the reaction produced of sulfur hexafluoride gas. Calculate the pressure of sulfur hexafluoride gas in the reaction vessel after the reaction. You may ignore the volume of the liquid reactants. Round your answer to significant digits.
Answer:
0.41 atm
Explanation:
A reaction between liquid reactants takes place at 10.0 °C in a sealed, evacuated vessel with a measured volume of 5.0 L. Measurements show that the reaction produced 13. g of sulfur hexafluoride gas. Calculate the pressure of sulfur hexafluoride gas in the reaction vessel after the reaction. You may ignore the volume of the liquid reactants. Round your answer to 2 significant digits.
Step 1: Given data
Temperature (T): 10.0 °CVolume of the vessel (V): 5.0 LMass of sulfur hexafluoride gas (m): 13. gStep 2: Convert "T" to Kelvin
We will use the following expression.
K = °C + 273.15
K = 10.0 °C + 273.15 = 283.2 K
Step 3: Calculate the moles (n) of SF₆
The molar mass of SF₆ is 146.06 g/mol.
13. g × 1 mol/146.06 g = 0.089 mol
Step 4: Calculate the pressure (P) of SF₆
We will use the ideal gas equation.
P × V = n × R × T
P = n × R × T/V
P = 0.089 mol × (0.0821 atm.L/mol.K) × 283.2 K/5.0 L = 0.41 atm
How do the valence electrons of an element determine how they will combine with other elements to produce a compound? Please help this is urgent :)
Answer:
See explanation
Explanation:
The valence electrons are electrons found on the valence (outermost) shell of an atom.
When an atoms form compounds, there is an exchange of valence electrons between the atoms of one element and the atoms of another element.
Let us consider a typical example, sodium has one valence electron and chlorine has seven valence electrons. This means that chlorine needs one electron to complete its octet while sodium needs to release one electron in order to attain the octet structure.
So, sodium gives out its one electron and becomes a stable sodium ion and chlorine accepts that electron and becomes a stable chloride ion. This is how the compound sodium chloride is formed.
balance the following equation by oxidation reduction method FeSO4
+
KMnO4+ H2SO4 → Fe2 (SO4)3+ k2SO4+MnSO4+H2O
Answer:
[tex]10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + 8\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + K_2SO_4 + 2\, {Mn}SO_4 + 8\, H_2O[/tex].
Explanation:
Identify the elements with oxidation state changes:
Oxidation states of iron, [tex]\rm Fe[/tex]:
[tex]+2[/tex] in [tex]\rm FeSO_4[/tex] among the reactants.[tex]+3[/tex] in [tex]\rm Fe_2(SO_4)_3[/tex] among the products.Change to the oxidation state: [tex]+1[/tex] (oxidation) for each [tex]\rm Fe[/tex] atom.Oxidation state of manganese, [tex]\rm Mn[/tex]:
[tex]+7[/tex] in [tex]\rm KMnO_4[/tex] among the reactants.[tex]+2[/tex] in [tex]\rm MnSO_4[/tex] among the products.Change to the oxidation state: [tex](-5)[/tex] (reduction) for each [tex]\rm Mn[/tex] atom.The change in the oxidation state of [tex]\rm Mn[/tex] is five times the opposite of the change to the oxidation state of [tex]\rm Fe[/tex]. If there are one mole of [tex]\rm Mn\![/tex] atoms in each mole of this reaction, there would be five times as many [tex]\rm Fe\![/tex] atoms per mole reaction. In other words:
[tex]\displaystyle 5\, \overset{+2}{\rm Fe}\rm SO_4 + 1\, \rm K \overset{+7}{Mn} O_4 + ?\, H_2SO_4\\ \to \frac{5}{2}\, \overset{+3}{Fe} (SO_4)_3 + ?\, K_2SO_4 + 1\, \overset{+2}{Mn}SO_4 + ?\, H_2O[/tex].
(Notice that each mole of this reaction would include five times as many [tex]\rm Fe[/tex] atoms as [tex]\rm Mn[/tex] atoms.)
Multiply the coefficients by [tex]2[/tex] to eliminate the fraction:
[tex]\displaystyle 10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + ?\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + ?\, K_2SO_4 + 2\, {Mn}SO_4 + ?\, H_2O[/tex].
Find the unknown coefficients using the conservation of atoms.
Reactants:
[tex]2[/tex] potassium [tex]\rm K[/tex] atoms in two [tex]\rm K_2SO_4[/tex] formula units.Therefore, among the products:
[tex]2[/tex] potassium [tex]\rm K[/tex] atoms in one [tex]\rm K_2SO_4[/tex] formula unit.[tex]\displaystyle 10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + ?\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + {1}\, K_2SO_4 + 2\, {Mn}SO_4 + ?\, H_2O[/tex].
Products:
[tex]5 \times 3 + 2 + 1 = 18[/tex] sulfur [tex]\rm S[/tex] atoms in five [tex]\rm Fe_2(SO_4)_3[/tex] formula units, two [tex]\rm K_2 SO_4[/tex] formula units, and one [tex]\rm MnSO_4[/tex] formula unit.Reactants:
There are already ten [tex]\rm S[/tex] atoms in that ten [tex]\rm Fe(SO_4)_2[/tex] formula units. The other [tex]18 - 10 = 8[/tex] formula units would correspond to eight [tex]\rm H_2SO_4[/tex] molecules among the reactants of this reaction.[tex]\displaystyle 10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + 8\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + {1}\, K_2SO_4 + 2\, {Mn}SO_4 + ?\, H_2O[/tex].
Products:
There are [tex]8 \times 2 = 16[/tex] hydrogen [tex]\rm H[/tex] atoms in that eight [tex]\rm H_2SO_4[/tex] molecules.Therefore, among the products:
There would be [tex]16 / 2 = 8[/tex] molecules of [tex]\rm H_2O[/tex], with two [tex]\rm H[/tex] atoms in each [tex]\rm H_2O\![/tex] molecule.[tex]\displaystyle 10\, {\rm Fe}\rm SO_4 + 2\, \rm K {Mn} O_4 + 8\, H_2SO_4\\ \to 5\, {Fe} (SO_4)_3 + {1}\, K_2SO_4 + 2\, {Mn}SO_4 + 8\, H_2O[/tex].
Calculate the number of oxygen atoms in a 50.0g sample of scheelite CaWO4
Answer:
0.696 atoms of oxygen
Explanation:
We'll begin by calculating the number of mole in 50 g of scheelite CaWO₄. This can be obtained as follow:
Mass of CaWO₄ = 50 g
Molar mass of CaWO₄ = 40 + 184 + (4×16)
= 40 + 184 + 64
= 288 g/mol
Mole of CaWO₄ =?
Mole = mass / Molar mass
Mole of CaWO₄ = 50 / 288
Mole of CaWO₄ = 0.174 mole
Finally, we shall determine the number of oxygen atom in 50 g (i.e 0.174 mole) of CaWO₄. This can be obtained as follow:
1 mole of CaWO₄ contains 4 atoms of oxygen.
Therefore, 0.174 mole of CaWO₄ will contain = 0.696 atoms of oxygen.
Thus, 50 g (i.e 0.174 mole) of CaWO₄ contains 0.696 atoms of oxygen.
chemistry
Definition in your own words. I will check if you got it from online.
Word:
Malleable
(malleability)
Which statement defines the enthalpy of solution?
the total number of particles in a solution
the phase change from a solid to a liquid
o the difference in characteristics between reactants and products
the change in energy when one substance dissolves in another
Answer:
the change in energy when one substance dissolves in another
Explanation:
The enthalpy changes are the heat changes accompanying physical and chemical changes. Actually, an enthalpy change is the difference between the sum of the heat contents of products (final state) and sum of the heat contents of reactants (initial state).
There are basically two types of heat changes that accompanies are reaction which are:
Exothermic changes Endothermic changesTherefore, the change in energy when one substance dissolves in another defines the enthalpy of solution.
Enthalpy of solution is the heat liberated or absorbed when one mole of a substance (solute) is dissolved in a specified volume of solvent (water).
Calculate the percent composition (percent by mass of each element) of NH4Cl.
Round to the nearest ONES place ((example: 12.34% = 12%))
Answer:
[tex]\%N=26.2\%\\\\\%H=7.5\%\\\\\%Cl=66.3\%[/tex]
Explanation:
Hello!
In this case, since the calculation of the percent composition of an element in a chemical compound is computing considering its atomic mass, subscript in the formula and molecular mass of the compound it is; for nitrogen, hydrogen and chlorine we have that ammonium chloride has a molar mass of 53.49 g/mol so the percent compositions are:
[tex]\%N=\frac{14.01*1}{53.49}*100\% =26.2\%\\\\\%H=\frac{1.01*4}{53.49}*100\% =7.5\%\\\\\%Cl=\frac{35.45*1}{53.49}*100\% =66.3\%[/tex]
Best regards!
Vinegar is insoluble in vegatable oil. Does this mean that vinegar is a totally insoluble substance?
Answer:
No
Explanation:
This does not mean that vinegar is insoluble totally. In fact, vinegar is soluble in water because water is a polar solvent.
For a substance to be soluble in another, it must obey the rule of solubility.
The rule states that "like dissolves like"
It implies that polar solvent will only dissolve polar solute.
Also, non-polar solvent will only dissolve non-polar solute.
Vegetable oil is a non-polar solventIt cannot dissolve a polar solute such as vinegarTherefore, the answer is no, vinegar will dissolve in water.
Consider the following chemical equilibrium:
NH4SH(s) ⇌ H2SgNH3(g) Now write an equation below that shows how to calculate Kp from Kc for this reaction at an absolute temperature T. You can assume T is comfortably above room temperature. If you include any common physical constants in your equation be sure you use their standard symbols, found in the ALEKS Calculator.
Answer:
See explanation below
Explanation:
First, let's write the reaction:
NH₄SH(s) <------> H₂S(g) + NH₃(g)
The reaction is already balanced so we don't need to do anything else.
Second, let's take into account the following. The Kc expression for this reaction, only compounds in gaseous state are the only ones that contribute to the equilibrium. Solid and liquid do not contribute to the Kc expression. This is because solid and liquid have a constant concentration near to 1, so, it won't do any difference.
Knowing this, the Kc expression for this reaction is:
Kc = [H₂S] [NH₃]
Now, to calculate Kp from Kc, there's an expression that helps a lot to do this. The expression is the following:
Kp = Kc (RT)ᵃⁿ (1)
Where:
R: universal constant of gases
T: Temperature in K
ᵃⁿ = difference of the coefficients of the reaction.
This expression comes from the fact that Kp is an expression that instead of working with concentrations, it works with pressure.
If we use the ideal gas equation we have:
PV = nRT
Solving for P:
P = nRT/V and C = n/V so
P = CRT
If we now replace this, in the Kp expression of equilibrium we have:
Kp = pH₂S * pNH₃
Kp = ([H₂S]RT)¹ ([NH₃]RT)¹
Kp = (RT)¹⁺¹ ([H₂S] [NH₃])
Kp = (RT)²Kc
So finally the expression for Kp would be:
Kp = Kc (RT)²Hope this helps
Which pair of elements would you expect to exhibit the greatest similarity in their physical
and chemical properties?
Select one:
O a. No
O b. Mg, Al
O c. Br, Kr
O d. As, Br
O e. I, AT
Answer:
e. I, At
Explanation:
Hello!
In this case, since the periodic trends of a series of elements belonging to the same group towards physical and chemical properties tend to be the same when closer in period, we notice that Mg and Al, Br and Kr and As and Br are close but in period, not in the same group; therefore e. I, At, iodine and astatine, are going to tend to exhibit the greatest similarity in their physical and chemical properties.
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PLZ HELP ASAP WILL GIVE BRAINLISTS TO RIGHT ANSWER
How many molecules of carbon dioxide are in 12.2 L of the gas at STP?
A) 3.28 x 10^23 molecules
B) 5.01 X 10^23 molecules
C)2.24 x 10^23 molecules
D)8.12 x 10^22 molecules
Answer:
c
Explanation:
ok than not c than b maybe
LaKeisha is measuring the density of a solid piece of metal using the graduated cylinder method. She initially measures a volume of water in the cylinder to be 3.28 mL. After placing the metal into the graduated cylinder, the new volume was 8.72 mL. The mass of the metal was 42.26 g on a top loading balance.
Required:
What is the density of the metal calculated to the correct number of significant figures?
Answer: 7.77 g/ml
Explanation:
Volume of cylinder with only water = 3.28 mL
Volume of cylinder with water and metal = 8.72 mL
Volume of metal = (Volume of cylinder with water and metal ) -(Volume of cylinder with only water)
=8.72-3.28
=5.44 ml
Mass of metal = 42.26 g
Formula of Density = [tex]\dfrac{\text{Mass}}{\text{Volume}}[/tex]
i.e. the density of the metal = [tex]\dfrac{42.26}{5.44}\approx7.77\text{ g/ml}[/tex]
Hence, the density of metal = 7.77 g/ml
Gravity pulls rain and snow down to Earth from the atmosphere through a paire
process called precipitation Water is pulled from elevated areas such as
mountains and hills into lakes, oceans, and water reserviors. What is this
describing?*
role of gravity in the water cycle
role of gravity in condensation
O
role of gravity in evaporation
role of gravity in precipitation
Explain the differences between an ideal gas and a real gas.
Answer:
Ideal Gas
The ideal gas is extremely small and the mass is almost zero and no volume Ideal gas is also considered as a point mass.
Real Gas
The molecules of real gas occupy space though they are small particles and also have volume.
anation:
The differences between an ideal gas and a real gas are that the ideal gas follows the gas laws perfectly under all conditions. Whereas a real gas deviates from ideal gas behaviors.
The ideal gas law, also known as the general gas equation, is a fundamental principle in thermodynamics and relates the pressure, volume, temperature, and number of moles of an ideal gas.
An ideal gas is a theoretical gas that follows the gas laws perfectly under all conditions of temperature and pressure. It is assumed to have no volume, no intermolecular forces, and elastic collisions between its particles. An ideal gas also obeys the ideal gas law.
On the other hand, a real gas is a gas that does not follow the gas laws perfectly under all conditions of temperature and pressure. Real gases have volume and intermolecular forces that affect their behavior. These forces cause deviations from ideal gas behavior, especially at high pressures and low temperatures.
In summary, while an ideal gas is a theoretical gas that follows the gas laws perfectly under all conditions, a real gas is a gas that deviates from ideal gas behavior due to its volume, intermolecular forces, and non-elastic collisions between its particles.
Learn more about ideal gas law here:
https://brainly.com/question/33342075
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In a space shuttle, the Carbon dioxide, CO2 that the crew exhales is removed from the air by a reaction within canisters of Lithium Hydroxide, LiOH. The LiOH is only 85% efficient. On average, each astronaut exhales around 20.0 mol of CO2 every day. What volume of water is produced when the CO2 reacts with the excess LiOH
Answer:
What volume of water is produced when the CO2 reacts with the excess LiOH
X = 360 mL H2O
Explanation:
CO2 (g) + 2 LiOH(s) ⇒ Li2CO3 (aq) + H2O(l)
20.0 mol excess x g
X = 360 mL H2O
x mL H20 = 20.0 mol CO2 (1 mol H2O /1 mol CO2)(18 g H2O/1 mol H2O)
(1 mL H2O /1 g H2O)
X = 360 mL H2O
The number of atoms in 10. grams of calcium is equal to 6.0 x 1023 multiplied by which number?
Answer:
1.51×10²³ atoms.
Explanation:
From the question given above, the following data were obtained:
Mass of Calcium (Ca = 10 g
Number of atom of Calcium (Ca) =?
The number of atoms present in 10 g of Ca be obtained as follow:
From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02×10²³ atoms. This implies that 1 mole Ca contains 6.02×10²³ atoms.
1 mole of Ca = 40 g.
Now, if 40 g of Ca contains 6.02×10²³ atoms.
Therefore, 10 g of Ca will contain = (10 × 6.02×10²³) / 40 = 1.51×10²³ atoms.
Thus, 10 g of calcium contains 1.51×10²³ atoms.
