Answer:
a. minus 10
Explanation:
An element in group 13 = Boron ,valence electrons = 3 , therefore, valence electrons in group 13 = group no. -10
An element in group 18 = Neon, valence electrons = 8 , therefore, valence electrons in group 18 = group no. - 10
For groups 13 through 18, the number of valence electrons in an atom is equal to the group number minus 10. Therefore, option (A) is correct.
What is a valence electron?Valence electrons in an atom can be described as the electrons occupying the outer most electron shell of an atom while the electrons in the inner shell are core electrons. Lewis structures can be helpful to calculate the number of valence electrons.
Valence electrons can be filled in several electron shells as they are caused interaction between atoms and responsible for the formation of chemical bonds. Only valence electrons can contribute to the formation of a chemical bond and decide the reactivity of the element.
The general electronic configuration of group 13 is ns²np¹ has three valence electrons. It can be determined as group number - 10 = 13 - 10 = 3.
The general electronic configuration of group 18 is ns²np⁶ has eight valence electrons. It can be determined as group number - 10 = 18 - 10 = 8.
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A gas within a piston–cylinder assembly undergoes an isothermal process at 400 K during which the change in entropy is −0.3 kJ/K. Assuming the ideal gas model for the gas and negligible kinetic and potential energy effects, evaluate the work, in kJ.
Answer:
W = -120 KJ
Explanation:
Since the piston–cylinder assembly undergoes an isothermal process, then the temperature is constant.
Thus; T1 = T2 = 400K
change in entropy; ΔS = −0.3 kJ/K
Formula for change in entropy is written as;
ΔS = Q/T
Where Q is amount of heat transferred.
Thus;
Q = ΔS × T
Q = -0.3 × 400
Q = -120 KJ
From the first law of thermodynamics, we can find the workdone from;
Q = ΔU + W
Where;
ΔU is Change in the internal energy
W = Work done
Now, since it's an ideal gas model, the change in internal energy is expressed as;
ΔU = m•C_v•ΔT
Where;
m is mass
C_v is heat capacity at constant volume
ΔT is change in temperature
Now, since it's an isothermal process where temperature is constant, then;
ΔT = T2 - T1 = 0
Thus;
ΔU = m•C_v•ΔT = 0
ΔU = 0
From earlier;
Q = ΔU + W
Thus;
-120 = 0+ W
W = -120 KJ
How is magma formed?
Answer:
“Magma” is exclusively found and formed beneath the earth’s surface. Once magma is on or above the surface of the earth it is referred to as “lava.” Magma is typically formed by extreme temperature melting solid rock within the earth. Pressure and rock composition can also affect magma formation. High pressure can help magma be “squeezed” from partially molten rock. Likewise, as rocks are usually composed of different minerals with different melting points, magma formation from rocks is usually only partial and uneven.
Explanation:
Based on the standard EMF series and your knowledge of half-reactions, determine the cell potential and spontanei ty of a cell that consists of a pure cobalt electrode in a solution of Co^2+ ions; the other half is a lead electrode immersed in a Pb^2+ solution.
Pb +2e- Pb Sn +2e Sn Ni 2e Ni Co 2e -0.126 -0.136 -0.250 -0.277 Co
a. +0.403, spontaneous
b. -0.403, nonspontaneous
c. +0.151, spontaneous
d. -0.151, nonspontaneous
Answer:
+0.151, spontaneous
Explanation:
Given that;
Co^2+(aq) + 2e ---->Co(s) -0.28 V
Pb^2+(aq) + 2e ---->Pb(s). -0.13 V
Hence Co is the anode and Pb is the cathode
E°cell = E°cathode - E°anode
So;
E°cell = -0.13 V - (-0.28 V)
E°cell = 0.15 V
The cell reaction is spontaneous since E°cell is positive.
Please help fast
All four referenced Greek thinkers: Democritus, Aristotle, Archimedes, and Anaxagoras, observed Nature and argued for his theory of
the composition of matter and natural laws. Only one of them tested his hypothesis and proposed a natural laws based on reproducible
observations, controlled experiments, and mathematical reasoning. All others used logic and thought experiments, as philosophers do,
to support their theories. Who is the experimental scientist in this group?
O Democritus
O Aristotle
O Archimedes
O Anaxagoras
Answer:
Anaxagoras was perhaps the first literate person to attempt to explain physical phenomena rationally, basing his ideas upon careful observations and simple experiments. This is fundamental to modern science and is the sine qua non of environmental study.
100 mL of 0.2 mol/L sodium carbonate solution and 200 mL of 0.1 mol/L calcium nitrate solution are mixed together. Calculate the mass of calcium carbonate that would precipitate and the concentration of the sodium nitrate solution that will be produced.
Answer:
Explanation:
Na2CO3+Ca(NO3)2=CaCO3+2NaNO3
nNa2CO3=0.02
nCa(NO3)2=0.02
mCaCO3=0.02*100=2 gram
nNaNo3=0.04
Cm=2/15
From the calculation, the mass of the product is 2 g.
What is a reaction?A chemical reaction occurs when two more substances are mixed together. In this case, the reaction is shown by; Ca(NO3)2 + Na2CO3 ----> CaCO3(s) + 2NaNO3.
Number of moles of Na2CO3 = 100/1000 L * 0.2 mol/L = 0.02 moles
Number of moles of Ca(NO3)2 = 200/1000 L * 0.1 mol/L = 0.02 moles
Since the reaction is equimolar, amount of the product = 0.02 moles * 100 g/mol = 2 g
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Arrange the following compounds in order of increasing reactivity (least reactive first.) to electrophilic aromatic substitution:.
Bromobenzene Nitrobenzene Benzene Phenol
a. Bromobenzene < Nitrobenzene < Benzene < Phenol
b. Nitrobenzene < Bromobenzene < Benzene < Phenol
c. Phenol < Benzene < Bromobenzene < Nitrobenzene
d. Nitrobenzene < Benzene < Bromobenzene < Phenol
Answer:
Nitrobenzene < Bromobenzene < Benzene < Phenol
Explanation:
Aromatic compounds undergo electrophilic aromatic substitution reaction in the presence of relevant electrophiles. Certain substituents tend to increase or decrease the tendency of an aromatic compound towards electrophilic aromatic substitution reaction.
Substituents that increase the electron density around the ring such as in phenol tends to make the ring more reactive towards electrophilic substitution. Halogens such as bromine has a -I inductive effect as well as a +M mesomeric effect.
However the -I(electron withdrawing effect) of the halogens supersedes the +M electron donation due to mesomeric effect.
Putting all these together, the order of increasing reactivity of the compounds towards electrophilic aromatic substitution is;
Nitrobenzene < Bromobenzene < Benzene < Phenol
Is a 4p S 4s transition allowed in sodium? If so, what is its wavelength? If not, why not? b. Is a 3d S 4s transition allowed in sodium? If so, what is its wavelength? If not, why not? g
Answer:
a) 4p ⇒ 4s transition is Allowed
b) 3d ⇒ 4s transition not allowed
Explanation:
a) 4p ⇒ 4s transition
This transition is allowed because for a 4p state; l = 1 and for a 4s state I = 0
hence Δl = 1 - 0 = 1
Energy of 4p ( Ei ) = 3.75eV
Energy of 4s ( E2 ) = 3.19 eV
where : λ = 1240 eV nm / ( E₂ - E₁ )
= 2214 nm ≈ 2.214 μm
b) 3d ⇒ 4s transition
This transition is not allowed
a 3d state , l = 2 while for 4s state l = 0
hence Δl = 2 - 0 = 2
therefore the transition is not allowed
Different vinegars can be 5-20% acetic acid solutions and have been used for medicinal purposes for thousands of years. If a person takes 2.0 tablespoons of vinegar a day and the Molarity of the vinegar is .84 M, then how many grams of acetic acid (HC2H3O2) will be consumed? 1 Tablespoon is 15 mL.
.013 g
.026 g
.76 g
1.5 g
Answer:
1.5g
Explanation:
Remember that Molarity = (#moles of solute)/(#liters of solution)
This problem informs us that the Molarity of the vinegar is 0.84 and that the solution is 15mL.
First let's get your SI units to the correct ones.
15mL (1L/1000mL) = 0.015L
Molarity = (#moles of solute)/(#liters of solution) ~
(Molarity)(#liters of solution) = #moles of solute
(0.84M)(.015L) = 0.0126moles of acetic acid per tablespoon
2 tablespoons a day = 0.0126moles*2 = 0.0252 moles of acetic acid.
Now that we have the # of moles of acetic acid we need to get our answer into grams. The molecular weight of HC2H3O2 is 60g/mole.
0.0252mole HC2H3O2 (60g HC2H3O2/1mole HC2H3O2) = 1.512g ~ 1.5g HC2H3O2.
An ice cube, measured at 260 Kelvin, is dropped into a cup of tea that is 350 Kelvin. The temperature of the tea is recorded every 30 seconds and shows the temperature dropping for 4 minutes. After 4 minutes the temperature stays steady at 300 Kelvin. What is this called?
A. Thermal equilibrium
B. Specific heat capacity
C. Latent heat
D. Temperature transfer
Answer:
Specific Heat Capacity
The half life of radium-226 is 1600 years. If you have 200 grams of radium today how many grams would be present in 8000 years?
Answer:
Half life is the time taken by a radio active isotope to reduce by half of its original amount. Radium-226 has a half life of 1602 years meaning that it would take 1602 years for a mass of radium to reduce by half.
Number of half lives in 9612 years = 9612/1602 = 6 half lives
New mass = Original mass x (1/2)n where n is the number of half lives.
Therefore, New mass= 500 x (1/2)∧6
= 500 x 0.015625
= 7.8125 g
Hence the mass of radium after 9612 years will be 7.8125 grams.
Explanation:
Answer:
[tex]\boxed {\boxed {\sf 6.25 \ grams}}[/tex]
Explanation:
We are asked to find the mass of a sample of radium-226 after half-life decay. We will use the following formula:
[tex]A= A_o *\frac{1}{2}^{\frac{t}{h}}[/tex]
In this formula, [tex]A_o[/tex] is the initial amount, t is the time, and h is the half-life.
For this problem, the initial amount is 200 grams of radium-226, the time is 8,000 years, and the half-life is 1,600 years.
[tex]\bullet \ A_o= 200 \ g \\\\bullet \ t= 8,000 \ \\\bullet \ h= 1,600[/tex]
Substitute the values into the formula.
[tex]A= 200 \ g * \frac{1}{2} ^{\frac{8.000}{1,600}[/tex]
Solve the fraction in the exponent.
[tex]A= 200 \ g * \frac{1}{2}^{5}[/tex]
Solve the exponent.
[tex]A= 200 \ g *0.03125[/tex]
[tex]A= 6.25 \ g[/tex]
In addition, we can solve this another way. First, we find the number of half-lives by dividing the total time by the half-life.
8,000/1,600= 5 half-livesEvery half-life, 1/2 of the mass decays. Divide the initial mass in half, then that result in half, and so on 5 times.
1. 200 g/2= 100 g2. 100 g / 2 = 50 g3. 50 g / 2 = 25 g 4. 25 g / 2 = 12.5 g5. 12.5 g / 6.25 gAfter 8,000 years, 6.25 grams of radium-226 remains.
15.27
The following equilibria were attained at 823 K:
COO(s) + H2() Co(s) + H2O(g) K = 67
COO(s) + CO(8) = Co(s) + CO2(8) K = 490
Based on these equilibria, calculate the equilibrium con-
stant for
H2(g) + CO2(g) = CO(g) + H2O(g) at 823 K.
The equilibrium constant for the reaction is K = 0.137
We obtain the equilibrium constant considering the following equilibria and their constants:
COO(s) + H₂(g) → Co(s) + H₂O(g) K₁ = 67
COO(s) + CO(g) → Co(s) + CO₂(g) K₂ = 490
We write the first reaction in the forward direction because we need H₂(g) in the reactants side:
(1) COO(s) + H₂(g) → Co(s) + H₂O(g) K₁ = 67
Then, we write the second reaction in the reverse direction because we need CO₂(g) in the reactants side. Thus, the equilibrium constant for the reaction in the reverse direction is the reciprocal of the constant for the reaction in the forward direction (K₂):
(2) Co(s) + CO₂(g) → COO(s) + CO(g) K₂ = 1/490
From the addition of (1) and (2), we obtain:
COO(s) + H₂(g) → Co(s) + H₂O(g) K₁ = 67
+
Co(s) + CO₂(g) → COO(s) + CO(g) K₂ = 1/490
-------------------------------------------------
H₂(g) + CO₂(g) → CO(g) + H₂O(g)
Notice that Co(s) and COO(s) are removed that appear in the same amount at both sides of the chemical equation.
Now, the equilibrium constant K for the reaction that is the sum of other two reactions is calculated as the product of the equilibrium constants, as follows:
K = K₁ x K₂ = 67 x 1/490 = 67/490 = 0.137
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9. Consider a magnesium atom with charge +2. How many overall electrons are on this particle?
Hint: Magnesium's atomic number is 12.
10
12
14
3. Oxalic acid, C H20, is a toxic substance found in rhubarb leaves. When mixed with sufficient
quantities of a strong base, this weak diprotic acid loses two protons to form a polyatomic ion
called oxalate, C2022. Write a balanced equation that describes the reaction between oxalic acid
and sodium hydroxide.
Answer:
COOHCOOH + 2OH⁻ ⇄ C₂O₄²⁻ + 2H₂O
Explanation:
The reaction of oxalic acid with a strong base like sodium hydroxide is the following:
COOHCOOH + OH⁻ ⇄ COOHCOO⁻ + H₂O (1)
In this first reaction, the oxalic acid loses one proton. In a second reaction with NaOH, the ion COOHCOO⁻ loses its second proton to form ion oxalate as follows:
COOHCOO⁻ + OH⁻ ⇄ C₂O₄²⁻ + H₂O (2)
The general reaction between oxalic acid and NaOH is (eq 1 + eq 2):
COOHCOOH + 2OH⁻ ⇄ C₂O₄²⁻ + 2H₂O
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importance of hematology
Answer:
Haematology is the specialty important for the diagnosis and management of a wide range of benign and malignant disorders of the red and white blood cells, platelets and the coagulation system in adults and children.
Select the correct relationship among the concentrations of species present in a 1.0 M aqueous solution of the weak acid represented by HA. A. [H2O] > [HA] > [A-] > [H3O ] > [OH-] B. [H2O] > [A-] ~ [H3O ] > [HA] > [OH-] C. [HA] > [H2O] > [A-] > [H3O ] > [OH-] D. [H2O] > [HA] > [A-] ~ [H3O ] > [OH-] E. [HA] > [H2O] > [A-] ~ [H3O ] > [OH-]
Answer:
D
Explanation:
We have to bear in mind that the acid is a weak acid. A weak acid does not dissociate completely in solution. We will have more concentration of undissociated acid than A^- and H3O^+ and OH^- in the system at equilibrium.
Being a weak acid, there is maximum concentration of water molecules followed by that of undissiociated acid.
Hence, for this solution, the concentration of ions in solution follows the order;
[H2O] > [HA] > [A-] ~ [H3O ] > [OH-]
Calculate the numerical value of the equilibrium constant, Kc, for the reaction below if the equilibrium concentrations for CO, H2 , CH4 and H2O are 0.989 M, 0.993 M, 1.078 M and 0.878 M, respectively. (calculate your answer to three sig figs)
CO(g) + 3 H2(g) ⇌ CH4(g) + H2O(g)
Kc = [CH4]×[H2O] / [CO]×[H2]^3
Kc = 1.078×0.878 / (0.989×0.933^3)
Kc = 0.977
The numerical value of the equilibrium constant, Kc, for the given reaction is found to be 0.977.
What is Equilibrium constant?The Equilibrium constant may be defined as the numerical value that significantly indicated the correlation between the amounts of products and reactants present at equilibrium in a reversible chemical reaction at a definite temperature.
According to the question, the reaction is as follows:
[tex]CO +3H_2[/tex] ↔ [tex]CH_4+ H_2O[/tex].
The equilibrium concentrations are 0.989 M, 0.993 M, 1.078 M and 0.878 M, respectively.
Now, the equilibrium constant is calculated by the following formula:
Kc = [CH4]×[H2O] / [CO]×[tex][H_2]^3[/tex]= 1.078×0.878 / (0.989×0.93[tex]3^3[/tex]).
= 0.9464/(0.989 × 0.8121)
= 0.977.
Therefore, the numerical value of the equilibrium constant, Kc, for the given reaction is found to be 0.977.
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The radioactivity due to carbon-14 measured in a piece of a wood from an ancient site was found to produce 20 counts per minute from a given sample, whereas the same amount of carbon from a piece of living wood produced 160 counts per minute. The half-life of carbon-14, a beta emitter, is 5730 y. The age of the artifact is closest to
Answer:
The answer is "17200 years".
Explanation:
Given:
[tex]A = 20 \ \frac{counts}{minute}\\\\A_{o} = 160\ \frac{counts}{minute}[/tex]
Let the half-life of carbon-14, is beta emitter, is [tex]T = 5730\ years[/tex]
Constant decay [tex]\ w = \frac{0.693}{ T}[/tex]
[tex]= 1.209 \times 10^{-4} \ \frac{1}{year}\\[/tex]
The artifact age [tex]t= ?[/tex]
[tex]A = A_{o} e^{-wt} \\\\e^{-wt} = \frac{A}{A_{o}}\\\\-wt = \ln \frac{A}{A_{o}}\\\\= -2.079\\\\t = 1.7199 \times 10^{4} \ years\\\\\sim \ 17200\ years\\[/tex]
help me in my hw,wt is physical change and chemical change Answer it asap plz don't spam
Answer:
Sorry but i don't undertsnad the question.
Explanation:
Answer:
A physical change is a change to the physical—as opposed to chemical—properties of a substance. They are usually reversible. The physical properties of a substance include such characteristics as shape, color, texture, flexibility, density, and mass.
A chemical change happens when one chemical substance is transformed into one or more different substances, such as when iron becomes rust.
Do u want examples ?
Write a balanced chemical equation for the reaction that occurs
when:
(a) titanium metal reacts with O21g2;
(b) silver(I) oxide decomposes into silver metal and oxygen gas when heated;
(c) propanol, C3H7OH1l2 burns in air;
(d) methyl tert-butyl ether, C5H12O1l2, burns in air.
Answer:
Explanation:
A balanced chemical equation refers to the reaction taking place whereby the number of atoms associated in the reactants side is equivalent to the number of atoms on the products side.
From the given information, the balanced equations are as follows:
[tex]\mathbf{(a) \ \ \ Ti(s) + O_{2(g)} \to TiO_{2(s)}}[/tex]
[tex]\mathbf{(b) \ \ \ 2Ag_{2}O \to 4Ag_{(s)} + O_{2(g)}}[/tex]
[tex]\mathbf{(c) \ \ \ 2C_3H_7OH + 9O_2 \to 6CO_2+8H_2O}[/tex]
[tex]\mathbf{(d) \ \ \ 2C_5 H_{12}O \to 10 CO_2 + 12 H_2O}[/tex]
convert 100kcals to kilojoules
Answer:
Explanation:
418.4kj is the correct answer
Dung dịch nào sau đây chỉ chứa các ion (bỏ qua sự điện li của nước, các chất điện li mạnh phân li hoàn toàn)?
A. HBr, Na2S, Mg(OH)2, Na2CO3.
B. H2SO4, NaOH, NaCl, HF.
C. HNO3, H2SO4, KOH, K2SiO3.
D. Ca(OH)2, KOH, CH3COOH, NaCl.
Answer:
Dung dịch nào sau đây chỉ chứa các ion (bỏ qua sự điện li của nước, các chất điện li mạnh phân li hoàn toàn)?
A. HBr, Na2S, Mg(OH)2, Na2CO3.
B. H2SO4, NaOH, NaCl, HF.
C. HNO3, H2SO4, KOH, K2SiO3.
D. Ca(OH)2, KOH, CH3COOH, NaCl.
A solution is made by dissolving 5.84 grams of NaCl in enough distilled water to give a final volume of 1.00 L. What is the molarity of the solution
Group of answer choices
0.0250 M
0.400 M
0.100 M
1.00 M
Answer:
Explanation:
1. A solution is made by dissolving 5.84g of NaCl is enough distilled water to a give a final volume of 1.00L. What is the molarity of the solution? a. 0.100 M b. 1.00 M c. 0.0250 M d. 0.400 M 2. A 0.9% NaCl (w/w) solution in water is a. is made by mixing 0.9 moles of NaCl in a 100 moles of water b. made and has the same final volume as 0.9% solution in ethyl alcohol c. a solution that boils at or above 100°C d. All the above (don't choose this one) 3. In an exergonic process, the system a. gains energy b. loses energy c. either gains or loses energy d. no energy change at all
Answer:
[tex]\boxed {\boxed {\sf 0.100 \ M }}[/tex]
Explanation:
Molarity is a measure of concentration in moles per liter.
[tex]molarity = \frac{moles \ of \ solute}{liters \ of \ solution}}[/tex]
The solution has 5.84 grams of sodium chloride or NaCl and a volume of 1.00 liters.
1. Moles of SoluteWe are given the mass of solute in grams, so we must convert to moles. This requires the molar mass, or the mass of 1 mole of a substance. These values are found on the Periodic Table as the atomic masses, but the units are grams per mole, not atomic mass units.
We have the compound sodium chloride, so look up the molar masses of the individual elements: sodium and chlorine.
Na: 22.9897693 g/mol Cl: 35.45 g/molThe chemical formula (NaCl) contains no subscripts, so there is 1 mole of each element in 1 mole of the compound. Add the 2 molar masses to find the compound's molar mass.
NaCl: 22.9897693 + 35.45 = 58.4397693 g/molThere are 58.4397693 grams of sodium chloride in 1 mole. We will use dimensional analysis and create a ratio using this information.
[tex]\frac {58.4397693 \ g\ \ NaCl} {1 \ mol \ NaCl}[/tex]
We are converting 5.84 grams to moles, so we multiply by that value.
[tex]5.84 \ g \ NaCl *\frac {58.4397693 \ g\ NaCl} {1 \ mol \ NaCl}[/tex]
Flip the ratio. It remains equivalent and the units of grams of sodium chloride cancel.
[tex]5.84 \ g \ NaCl *\frac {1 \ mol \ NaCl}{58.4397693 \ g\ NaCl}[/tex]
[tex]5.84 *\frac {1 \ mol \ NaCl}{58.4397693 }[/tex]
[tex]0.09993194823 \ mol \ NaCl[/tex]
2. MolarityWe can use the number of moles we just calculated to find the molarity. Remember there is 1 liter of solution.
[tex]molarity= \frac{moles \ of \ solute}{liters \ of \ solution}[/tex]
[tex]molarity= \frac{ 0.09993194823 \ mol \ NaCl}{1 \ L}[/tex]
[tex]molarity= 0.09993194823 \ mol \ NaCl/L[/tex]
3. Units and Significant FiguresThe original measurements of mass and volume have 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandths place. The 9 in the ten-thousandths place tells us to round the 9 to a 0, but then we must also the next 9 to a 0, and the 0 to a 1.
[tex]molarity \approx 0.100 \ mol \ NaCl/L[/tex]
1 mole per liter is 1 molar or M. We can convert the units.
[tex]molarity \approx 0.100 \ M \ NaCl[/tex]
The molarity of the solution is 0.100 M.
What Volume of silver metal will weigh exactly 2500.0g. The density of silver
Answer:
cm3 = 2500.0 g / 10.5 g/cm3 = 238 cm3
How are all planets in the solar system similar?
A. They have a gas atmosphere.
B. They have a water atmosphere.
C. They have a gas-surface composition.
D. They have a rock surface composition.
THIS IS FOR SCIENCE!!!!!!
PLEASE HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Answer:
They have a rock surface composition.
Explanation:
I hope this will help you
385 x 42.13 x 0.079 is (consider significant figures):
385 x 42.13 x 0.079 = 1281.38395
Calculate the number of iron atoms contained in 434.52 g of iron: Question 7 options: 2.620 x 10 26 Fe atoms 5.769 x 10 24 Fe atoms 4.685 x 10 24 Fe atoms 3.223 x 10 25 Fe atoms 1.169 x 10 23 Fe atoms
Answer:
[tex]\boxed {\boxed {\sf 4.685 \times 10^{24} \ Fe \ atoms}}[/tex]
Explanation:
We are asked to convert grams of iron to atoms of iron.
1. Convert Grams to MolesFirst, we convert grams to moles. We use the molar mass or the mass in 1 mole of a substance. This is found on the Periodic Table because the molar mass is equal to the atomic mass, but the units are grams per mole instead of atomic mass units. Look up iron's molar mass.
Fe: 55.84 g/molWe convert using dimensional analysis, so we must set up a ratio using the molar mass.
[tex]\frac {55.84 \ g \ Fe}{ 1 \ mol \ Fe}[/tex]
We are converting 434.52 grams to moles, so we multiply by this value.
[tex]434.52 \ g \ Fe *\frac {55.84 \ g \ Fe}{ 1 \ mol \ Fe}[/tex]
Flip the ratio so the units of grams of iron cancel.
[tex]434.52 \ g \ Fe *\frac { 1 \ mol \ Fe}{55.84 \ g \ Fe}[/tex]
[tex]434.52 *\frac { 1 \ mol \ Fe}{55.84}[/tex]
[tex]\frac { 434.52}{55.84} \ mol \ Fe[/tex]
[tex]7.781518625 \ mol \ Fe[/tex]
2. Convert Moles to AtomsNext, we convert moles to atoms. We use Avogadro's Number or 6.02 ×10²³. It is the number of particles (atoms, molecules, formula units, etc) in 1 mole of a substance. For this problem, the particles are atoms of iron. We set up another ratio using this number.
[tex]\frac {6.02 \times 10^{23} \ atoms \ Fe }{ 1 \ mol \ Fe}[/tex]
Multiply by the number of moles we calculated.
[tex]7.781518625 \ mol \ Fe *\frac {6.02 \times 10^{23} \ atoms \ Fe }{ 1 \ mol \ Fe}[/tex]
The units of moles of iron cancel.
[tex]7.781518625 *\frac {6.02 \times 10^{23} \ atoms \ Fe }{ 1 }[/tex]
[tex]4.68447421 \times 10^{24} \ atoms \ Fe[/tex]
The correct answer choice is Choice 3: 4.685 × 10²⁴ atoms of iron.
What is the balanced form of the following equation?
Br2 + S2O32- + H2O → Br1- + SO42- + H+
Answer:
5 Br₂ + S₂O₃²⁻ + 5 H₂O ⇒ 10 Br⁻ + 2 SO₄²⁻ + 10 H⁺
Explanation:
We will balance the redox reaction through the ion-electron method.
Step 1: Identify both half-reactions
Reduction: Br₂ ⇒ Br⁻
Oxidation: S₂O₃²⁻ ⇒ SO₄²⁻
Step 2: Perform the mass balance, adding H⁺ and H₂O where appropriate
Br₂ ⇒ 2 Br⁻
5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺
Step 3: Perform the charge balance, adding electrons where appropriate
2 e⁻ + Br₂ ⇒ 2 Br⁻
5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺ + 10 e⁻
Step 4: Make the number of electrons gained and lost equal
5 × (2 e⁻ + Br₂ ⇒ 2 Br⁻)
1 × (5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺ + 10 e⁻)
Step 5: Add both half-reactions
5 Br₂ + S₂O₃²⁻ + 5 H₂O ⇒ 10 Br⁻ + 2 SO₄²⁻ + 10 H⁺
How many moles of (CH3)3NH+ are in 6.0 g of (CH3)3NH+?
Answer:
0.1 mol
Explanation:
6/(15*3+15)
0.1 mol moles of (CH3)3NH+ are in 6.0 g of (CH3)3NH+
What is mole?
The mole, symbol mol, exists as the SI base unit of the amount of substance. The quantity amount of substance exists as a measure of how many elementary entities of a provided substance exist in an object or sample.A mole corresponds to the mass of a substance that includes 6.023 x 1023 particles of the substance. The mole exists the SI unit for the amount of a substance. Its symbol stands mol.
The compound trimethylamine, (CH3 )3N, exists as a weak base when dissolved in water.
A mole exist expressed as 6.02214076 × 1023 of some chemical unit, be it atoms, molecules, ions, or others. The mole exists as a convenient unit to utilize because of the great number of atoms, molecules, or others in any substance.
To find the amount of the substance (CH3)3NH+ to calculate its molar mass:
M((CH3)3NH+) = (12+3)*3 + 14+1 = 60 g/mol
n((CH3)3NH+) = m/M
m((CH3)3NH+) = 6g
Thus,
n((CH3)3NH+) = 6g/60 g/mol = 0.1 mol
Hence,
n((CH3)3NH+) = 0.1 mol
To learn more about mole refer to:
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Suppose that you add 24.3 g of an unknown molecular compound to 0.250 kg of benzene, which has a K f of 5.12 oC/m. With the added solute, you find that there is a freezing point depression of 3.14 oC compared to pure benzene. What is the molar mass (in g/mol) of the unknown compound
Solution :
We know that :
[tex]$\Delta T_f = k_f.m$[/tex] and [tex]$m=\frac{w_2}{m_2 \times w_1}$[/tex]
Then, [tex]$\Delta T_f = k_f.\frac{w_2}{m_2.w_1}$[/tex] ..................(1)
Where,
[tex]w_1[/tex] = amount of solvent (in kg)
[tex]w_2[/tex] = amount of solute (in kg)
[tex]m_2[/tex] = molar mass of solute (g/mole)
[tex]m[/tex] = molality of solution (mole/kg)
Given :
[tex]\Delta T_f[/tex] = [tex]3.14\ ^\circ C[/tex], [tex]k_f= 5.12\ ^\circ C/m[/tex]
[tex]=5.12 \ ^\circ C/mole/kg[/tex]
[tex]=5.12 \ ^\circ C \ kg/mole[/tex]
[tex]w_1[/tex] = 0.250 kg, [tex]w_2[/tex] = 24.3 g
Then putting this values in the equation is (1),
[tex]$3.14 = \frac{5.12 \times 24.3}{m_2 \times 0.250}$[/tex]
[tex]$m_2 = \frac{5.12 \times 24.3}{3.14 \times 0.250}$[/tex]
[tex]m_2= 158.49[/tex] g/mole
So, the molar mass of the unknown compound is 158.49 g/mole.
A solution is made by dissolving 0.565 g of potassium nitrate in enough water to make up 250. mL of solution. What is the molarity of this solution?
Please explain and show work.
[tex]\\ \large\sf\longmapsto KNO_3[/tex]
[tex]\\ \large\sf\longmapsto 39u+14u+3(16u)[/tex]
[tex]\\ \large\sf\longmapsto 53u+48u[/tex]
[tex]\\ \large\sf\longmapsto 101u[/tex]
[tex]\\ \large\sf\longmapsto 101g/mol[/tex]
Now
[tex]\boxed{\sf No\:of\:moles=\dfrac{Given\:mass}{Molar\:mass}}[/tex]
[tex]\\ \large\sf\longmapsto No\:of\:moles=\dfrac{0.565}{101}[/tex]
[tex]\\ \large\sf\longmapsto No\:of\:moles=0.005mol[/tex]
We know
[tex]\boxed{\sf Molarity=\dfrac{Moles\:of\:solute}{Vol\:of\:Solution\:in\:L}}[/tex]
[tex]\\ \large\sf\longmapsto Molarity=\dfrac{0.005}{\dfrac{250}{1000}L}[/tex]
[tex]\\ \large\sf\longmapsto Molarity=\dfrac{0.005}{0.250}[/tex]
[tex]\\ \large\sf\longmapsto Molarity=0.02M[/tex]
[tex] \: \: \: \: \: \: \: \: \: [/tex]