The percent composition of the compound.
A O-18.18%, N-21.21%, H-60.60%
Further explanationGiven
6.00 grams of oxygen,
7.00 grams of nitrogen,
20.00 grams of hydrogen.
Required
The percent composition
Solution
Total mass :
= mass of O + mass of N + mass of H
= 6 + 7 + 20
= 33 g
% O = 6/33 x 100%= 18.18%
% N = 7/33 x 100%=21.21%
% H = 20/33 x 100% = 60.6 %
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.
chemistry
Definition in your own words. I will check if you got it from online.
Word:
Malleable
(malleability)
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.
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Suppose a 500.mL flask is filled with 0.40mol of N2 and 1.0mol of NO. The following reaction becomes possible:
N2g+O2g ->2NOg
The equilibrium constant K for this reaction is 5.93 at the temperature of the flask. Calculate the equilibrium molarity of N2. Round your answer to two decimal places.
Answer:
[N₂] = 1.1M
Explanation:
Based on the chemical reaction:
N₂(g) + O₂(g) ⇄ 2 NO(g)
Equilibrium constant, K, is defined as:
K = 5.93 = [NO]² / [N₂] [O₂]
Where [] are equilibrium concentrations of each specie
As initial concentrations are:
N₂ = 0.40mol / 0.500L = 0.8M
NO = 1mol / 0.500L = 2M
The equilbrium concentrations are:
[NO] = 2M - 2X
[N₂] = 0.8M +X
[O₂] = X
Replacing:
5.93 = [2 - 2X]² / [0.8+X] [X]
5.93 = 4 - 8X + 4X² / 0.8X + X²
4.744X + 5.93X² = 4 - 8X + 4X²
1.93X² + 12.744X - 4 = 0
Solving for X:
X = -6.9M → False solution. There are no negative concentrations
X = 0.3M. Real solution.
[N₂] in equilibrium is:
[N₂] = 0.8M +0.3M
[N₂] = 1.1M
1. Each substance written to the right of the arrow in a chemical equation is a
(1 point)
O catalyst
O reactant
O precipitate
O product
Answer: product
Explanation:
Each substance written to the right of the arrow in a chemical equation is referred to as a product.
When writing a chemical equation, the substance that's written to the left of arrow in the equation is the reactants.
On the other hand which is the right side is the product.
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:
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.16. Using the average atomic masses given in the inside front cover of this book, calculate the indicated quantities.
d. the number of moles of cobalt represented by 5.99 x 1021 cobalt atoms e. the mass of 4.23 mol of cobalt
f. the number of cobalt atoms in 4.23 mol of cobalt
g. the number of cobalt atoms in 4.23 g of cobalt
Answer:
d. 9.95 × 10⁻³ mol
e. 249 g
f. 2.55 × 10²⁴ atoms
g. 4.32 × 10²² atoms
Explanation:
d. the number of moles of cobalt represented by 5.99 x 10²¹ cobalt atoms
We will use Avogadro's number: there are 6.02 × 10²³ atoms of cobalt in 1 mole of atoms of cobalt.
5.99 x 10²¹ atoms × 1 mol/6.02 × 10²³ atoms = 9.95 × 10⁻³ mol
e. the mass of 4.23 mol of cobalt
The molar mass of cobalt is 58.93 g/mol.
4.23 mol × 58.93 g/mol = 249 g
f. the number of cobalt atoms in 4.23 mol of cobalt
We will use Avogadro's number: there are 6.02 × 10²³ atoms of cobalt in 1 mole of atoms of cobalt.
4.23 mol × 6.02 × 10²³ atoms/1 mol = 2.55 × 10²⁴ atoms
g. the number of cobalt atoms in 4.23 g of cobalt
First, we will calculate the moles of cobalt using the molar mass of cobalt.
4.23 g × 1 mol/58.93 g = 0.0718 mol
Then, we will calculate the number of cobalt atoms using Avogadro's number.
0.0718 mol × 6.02 × 10²³ atoms/1 mol = 4.32 × 10²² atoms
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
How can you model the cycling of matter in the Earth system?
Answer:
The cycling of matter is important to many Earth processes and to the survival of organisms the existing matter must cycle continuously for this planet to support life Water, carbon, nitrogen, phosphorus, and even rocks move through cycles If these materials did not cycle, Earth could not support life.
Explanation:
Earth activities depend on matter cycling, and for organisms to survive, this planet's surface must cycle with the flow of matter.
What is Earth system?Rocks, as well as water, carbon, nitrogen, and phosphorus, go through cycles. The planet Earth could not support life if these materials did not cycle.
Subsystems exist within the Earth system. These subsystems include the exosphere, atmosphere, hydrosphere, lithosphere and geosphere, also referred to as the lithosphere, and the living environment (biosphere).
These systems are powered by energy that comes from both the Sun and the interior of the Earth. Through processes known as biogeochemical cycles, nutrients and elements also move through these systems along with energy.
Therefore, Earth activities depend on matter cycling, and for organisms to survive, this planet's surface must cycle with the flow of matter.
To learn more about Earth, refer to the link:
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What produces the magnetic force of an electromagnet?
O magnetic fields passing through the device
O static charged particles on the wire
O movement of charged particles through the wire
O positive and negative charges repelling each other
Answer:
movement of charged particles through the wire .
Explanation:
When electricity is passed through the wire of electromagnet , moving electrons of the wire produces magnetic field . This magnetic field in increased due to high permeability of soft iron of the electromagnet . It is this magnetic field which creates magnetic force .
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]
How many moles of water can be formed from 0.57 moles of hydrogen gas?
Answer:
0.57 water
Explanation:
To solve this problem, we need to write the reaction expression first.
The reactants are oxygen gas and hydrogen gas.
They react to give a product of water
2H₂ + O₂ → 2 H₂O
Given that;
Number of moles of hydrogen gas = 0.57moles
From the balanced reaction expression;
2 moles of hydrogen gas produces 2 moles of water
So;
0.57mole of hydrogen gas will also produce 0.57 water
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
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].
definition of solubility
(science)
Answer:
th relative ability of a solute to devolve into a solvent
How many moles of hydrogen gas are present in 65.0 liters at STP?
1456 moles
1.45 moles
3.00 moles
2.90 moles
Answer:
2.9moles of hydrogen gas
Explanation:
convert liters to dm³
since 1liter= 1dm³
thus, 65.0liters = 65.0dm³
number of moles = volume given/22.4dm³
= 65.0/22.4
=2.9moles
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
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.
Best regards!
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 series of dilute NaCl solutions are prepared starting with an initial stock solution of 0.100 M NaCl. Solution A is prepared by pipeting 10 mL of the stock solution into a 250-mL volumetric flask and diluting to volume. Solution B is prepared by pipeting 25 mL of solution A into a 100-mL volumetric flask and diluting to volume. Solution C is prepared by pipeting 20 mL of solution B into a 500-mL volumetric flask and diluting to volume. What is the molar concentration of NaCl in solutions A, B and C
Answer:
Solution A: 0.00400M
Solution B: 0.00400M
Solution C: 4.00x10⁻⁵M
Explanation:
Solution A is diluting the 0.100M NaCl from 10mL to 250mL. That is:
250mL / 10mL = 25 times.
That means molar concentration of sln A is:
0.100M / 25 = 0.00400M
Solution B is obtained diluting 25mL to 100mL:
100mL / 25mL = 4 times
0.00400M / 4 times = 0.00100M
And solution C is obtained diluting the solution C from 20mL to 500mL:
500mL / 20mL = 25 times
Solution C:
0.00100M / 25 times = 4.00x10⁻⁵M
The formula for serial dilution can be used to obtain the molarity of solution A, B , C.
For solution AM1V1 = M2V2
M2 = 0.100 M × 10 mL/250-mL
M2 = 0.004 M
For solution BM1V1 = M2V2
M2 = 0.004 M × 25 mL/100-mL
M2 = 0.001 M
For solution CM1V1 = M2V2
M2 = 0.001 M × 20 mL/500-mL
M2 = 0.00004 M
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How 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
is C5H10 ionic or covalent?
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 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.
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.
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
what is the formula for H-H
Answer:
H-H equation is written as follows:
pH=pK + log
{HCO3-}(base)
{H2CO3}(acid)
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:
