In the first 15.0 s of the reaction, 1.7×10−2 mol of O2 is produced in a reaction vessel with a volume of 0.440 L . What is the average rate of the reaction over this time interval?

Answer:

1s²,2s²,2p⁶,3s²,3p⁶,4s²,3d¹⁰,4p⁶,5s⁰,4d¹⁰.

Explanation:

Palladium is a chemical element with the symbol Pd and atomic number 46.

The electronic configuration is;

[Kr] 4d¹⁰

The full electronic configuration observed for palladium is given as;

1s²,2s²,2p⁶,3s²,3p⁶,4s²,3d¹⁰,4p⁶,5s⁰,4d¹⁰.

The reason for for the anomlaous electron configuration is beacuse;

1. Full d orbitals are more stable than partially filled ones.

2. At higher energy levels, the levels are said to be degenerated which means that they have very close energies and then electrons can jump from one orbital to another easily.

Answer:

Explanation:

1. A student mixes baking soda and vinegar in a glass. The results are shown at left. ... Yes I do belive that new substances are being formed because there is a chemical reaction between the baking soda and vinegar turning it into a bubbly substances instead of a powder and liquid.

Yes, there are new substances created from this mixture.

Answer:

Rate of diffusion is same .

Explanation:

As we know that Rate of the diffusion is directly proportional to the [tex]\frac{1}{\sqrt{M} }[/tex] .They have same mass if there is same rate and similar condition therefore the mass of carbon (iv) oxide,propane and nitrogen (i) oxide will be similar.

Answer and Explanation:

Sodium hydroxide (NaOH) is a strong base and nitric acid (HNO₃) is a strong acid. That means that they dissociates in water by giving the ions:

NaOH ⇒ Na⁺(ac) + OH⁻(ac)

HNO₃ ⇒ H⁺(ac) + NO₃⁻(ac)

The reaction between an acid and a base is called neutralization. In this case, HNO₃ loses its proton and it is converted in NO₃⁻ (nitrate anion). NaOH loses its hydroxyl anion (OH⁻) by giving Na⁺ cations.

Na⁺ cations with NO₃⁻ anions form the salt NaNO₃ (sodium nitrate); whereas H⁺ and OH⁻ form water molecules. The complete equation is the following:

HNO₃(ac) + NaOH(ac) ⇒ NaNO₃(ac) + H₂O(l)

The ionic equation is:

H⁺(ac) + NO₃⁻(ac) + Na⁺(ac) + OH⁻(ac) ⇄ Na⁺(ac) + NO₃⁻(ac) + H₂O(ac)

If we cancel the repeated ions at both sides of the equation, it gives the following ionic reaction:

H⁺(ac) + OH⁻(ac) ⇄ H₂O(ac)

Answer:

32.062

Explanation:

The following data were obtained from the question:

Mass of isotope A (32S) = 31.97207 u

Abundance of isotope A (A%) = 95.0%

Mass of isotope B (33S) = 32.97146 u Abundance of isotope B (B%) = 0.76%

Mass of isotope C (34S) = 33.96786 u

Abundance of isotope C (C%) = 4.22%

Mass of isotope D (36S) = 35.96709 u Abundance of isotope D (D%) = 0.014%

Average atomic mass of S =..?

The average atomic mass of sulphur, S can be obtained as follow:

Average atomic mass = [(Mass of A x A%)/100] + [(Mass of B x B%)/100] + [(Mass of C x C%)/100] + [(Mass of D x D%)/100]

Average atomic mass of sulphur =

[(31.97207 x 95)/100] + [(32.97146 x 0.76)/100] + [(33.96786 x 4.22)/100] + [(35.96709 x 0.014)/100]

= 30.373 + 0.251 + 1.433 + 0.005

= 32.062

Therefore, the average atomic mass of sulphur is 32.062

Answer:

HCO3- (aq) + H2O (I) <--> H2CO3 (aq) + OH- (aq)

Explanation:

The equation to distinguish between cation and anion hydrolysis is given below :  

HCO3- (aq) + H2O (I) <--> H2CO3 (aq) + OH- (aq)

The important thing to remember is their origin. The anions can react with water and can produce hydroxide ions while hydroxide ions make a solution basic.

Answer: Mass of hydrogen produced is 0.0376 g.

Explanation:

The reaction equation will be as follows.

[tex]NaH(aq) + H_{2}O(l) \rightarrow H_{2}(g) + NaOH(aq)[/tex]

Now, formula for total pressure will be as follows.

  [tex]P_{total} = P_{H_{2}} + P_{H_{2}O}[/tex]

Hence,    [tex]P_{H_{2}} = P_{total} - P_{H_{2}O}[/tex]

                            = 755 mm Hg - 42.23 mm Hg

                            = 712.77 mm Hg

[tex]P_{H_{2}} = \frac{712.77 \times 1 atm}{760 mm Hg}[/tex]

             = 0.937 atm

Now, we will calculate the moles of [tex]H_{2}[/tex] as follows.

   [tex]P_{H_{2}}V = nRT[/tex]

   [tex]0.937 atm \times 0.505 L = n \times 0.0821 \times 308.15 K[/tex]

      n = [tex]\frac{0.473}{25.29}[/tex] mol

         = 0.0187 mol

Therefore, mass of [tex]H_{2}[/tex] will be calculated as follows.

        [tex]m_{H_{2}} = \frac{0.0187 mol \times 2.0158 g}{1 mol}[/tex]

                   = 0.0376 g

Thus, we can conclude that mass of hydrogen produced is 0.0376 g.

Answer:

K₂SO₄ + Pb(C₂H₃O₂)₂ →PbSO₄ + 2KC₂H₃O₂

A chemical equation is a symbolic representation of a chemical reaction. The chemical equation for the reaction between potassium sulfate ([tex]K_2SO_4[/tex]) and lead(II) acetate ([tex]Pb(CH_3COO)_2[/tex]) can be written as follows:

[tex]K_2SO_4 + Pb(CH_3COO)_2 = PbSO_4 + 2CH_3COOK[/tex]

A basic chemical equation consists of two main parts: the reactant side (left side) and the product side (right side), separated by an arrow indicating the direction of the reaction. Reactants are substances that undergo a chemical change, while products are substances formed as a result of the reaction.

In this reaction, potassium sulfate reacts with lead(II) acetate to form lead(II) sulfate and potassium acetate. It is important to note that the equation is balanced with stoichiometric coefficients, ensuring that the number of atoms of each element is the same on both sides of the equation.

Therefore, the chemical equation for the reaction between potassium sulfate ([tex]K_2SO_4[/tex]) and lead(II) acetate ([tex]Pb(CH_3COO)_2[/tex]) can be written as follows:

[tex]K_2SO_4 + Pb(CH_3COO)_2 = PbSO_4 + 2CH_3COOK[/tex]

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Answer:

a) HC₂H₃O₂, C₂H₃O₂⁻, H₃O⁺, H₂O, OH⁻

b) HC₂H₃O₂ + LiOH ⇄ H₂O + LiC₂H₃O₂

c) C₂H₃O₂⁻ + HBr ⇄ HC₂H₃O₂ + Br⁻

Explanation:

a) In a HC₂H₃O₂/C₂H₃O₂⁻ buffer system, the following reactions take place:

HC₂H₃O₂ + H₂O ⇄ C₂H₃O₂⁻ + H₃O⁺

C₂H₃O₂⁻ + H₂O ⇄ HC₂H₃O₂ + OH⁻

Thus, the species present are: HC₂H₃O₂, C₂H₃O₂⁻, H₃O⁺, H₂O, OH⁻.

b) When LiOH is added to the buffer system, it is partially neutralized according to the following equation.

HC₂H₃O₂ + LiOH ⇄ H₂O + LiC₂H₃O₂

c) When HBr is added to the buffer system, it is partially neutralized according to the following equation.

C₂H₃O₂⁻ + HBr ⇄ HC₂H₃O₂ + Br⁻

Strong acids are completely ionised and weak acids are partly ionised

Answer:

Como forman los iones en soluciión

Explanation:

Los ácidos fuertes y las bases fuertes se refieren a especies que se disocian completamente para formar los iones en solución. Por el contrario, los ácidos y bases débiles se ionizan solo parcialmente y la reacción de ionización es reversible.

Answer:

31.2g

Explanation:

The following data were obtained from the question:

Volume of bromine = 10mL

Density of bromine = 3.12 g/mL

Mass of bromine =...?

The Density of the substance is related to it's mass and volume by the following equation:

Density = Mass /volume

With the above equation, we can calculate the mass of bromine as follow:

Density = Mass /volume

Volume of bromine = 10mL

Density of bromine = 3.12 g/mL

Mass of bromine =...?

Density = Mass /volume

3.12 = Mass /10

Cross multiply

Mass of bromine = 3.12 x 10

Mass of bromine = 31.2g

Therefore, the mass of bromine is 31.2g

Answer:potassium is more reactive than Mg because both lie in the same group and the element potassium has more electropositivity than magnesium

Explanation:

I hope it will help you

Answer: B. Potassium(K)

Explanation:

Answer:

density = 8.824g/mL

Explanation:

given

mass = 75g

volume = 8.5mL

density = mass/volume

density = 75g/8.5mL

density = 8.824g/mL

Answer:0.088g/ml

Explanation:

Density=mass/volume

d=0.75g/8.5ml

d=0.088g/ml

Answer:

both are the types of mixture and both are impure substances that donot have fixed composition and the composition of constituents  is not uniform

Answer:

Their components van be separated by physical processes

Explanation:

Out of the answers im given, it makes the most sense. I would double check before submitting though

Answer:

Alkene form hexan-1-ol with oxidation in presence of NaOH with highest yield  

Explanation:

Answer:

The concentration of  fructose-6-phosphate F6P ≅ 1.35 mM

Explanation:

Given that:

ΔG°′ is the  conversion of glucose-6-phosphate to fructose-6-phosphate (F6P)   = +1.67 kJ/mol = 1670 J/mol

concentration of glucose-6-phosphate at equilibrium = 2.65 mM

Assuming temperature = 25.0°C

=( 25 + 273)K

= 298 K

We are to find the concentration of fructose-6-phosphate

Using the relation;

ΔG' = -RT In K_c

where;

R = 8.314 J/K/mol

1670 = - (8.314 × 298 ) In K_c

1670 = -2477.572   × In K_c

1670/ 2477.572 =  In K_c

0.67 = In K_c

[tex]K_c = e^{-0.67}[/tex]

[tex]K_c =[/tex] 0.511

Now using the equilibrium constant [tex]K_c[/tex]

[tex]K_c = \dfrac{[F6P]}{[G6P]}[/tex]

[tex]0.511 = \dfrac{[F6P]}{[2.65]}[/tex]

F6P = 0.511 × 2.65

F6P = 1.35415

F6P ≅ 1.35 mM

Answer:

Take a look at the attachment below

Explanation:

Hope that helps!

Answer:

35.578g or 36g if you round

Explanation:

Q=mc ∆∅ where ∅ is temperature difference

1160= m x 1.716 x (42-23)

m = 1160/ 1.716 x19

m=35.578g

m = 36g to nearest whole number

Answer: C. 36 g

Explanation: I got this right on Edmentum.

Answer:

B. PROTONS EXHIBIT STRONGER PULL ON OUTER f ORBITALS

Explanation:

Lanthanide contraction is the greater than normal decrease in the ionic radius of the lanthanide series from atomic number 57 to atomic number 71. This decrease is rather not expected of the ionic radii of these elements and they result in the greater decrease in the subsequent series of the lanthanides from the atomic number 72. The cause of which is as a result of the poor shielding effects of the nuclear charge around the electrons of the f orbitals. So therefore, protons are strongly pulled out of the 4f orbital and as a result of the poor shielding effect which causes the electrons of the 6s orbitals to be drawn more closer to the nucleus and hence resulting in a smaller atomic radii. It is worthy to note that the shielding effects of the inner electrons decreasing from s orbital to the f orbital; that is s > p > d > f. So from the decrease in the shielding effects from s to the f orbitals, lanthanide contraction results from the inability of the orbitals far away from s like the 4f orbiatls to shield the outermost shells of the lanthanide elements. So the cause of lanthanide contraction is the action of the protons which strongly pull the electrons of the f orbitals because of the poor shielding effects due to the distance of this orbital from the nucleus.

Answer:

B) Protons exhibit a stronger pull on outer f orbitals than on d orbitals.

Explanation:

Answer:

Group 1 or akali metals have the greatest metallic property.

Group 17 has the lowest metallic character.

C. As you move from right to lefton the periodic table, metallic character increases which is the ability to lose electrons. Ionization energy decrease as we move from right to left on the periodic table.

Explanation:

Akali metals in group 1 have the greatest metallic property and they are the most reactive metals. Francium metal on the group has the most metallic characteristics. It is rare and very radioactive. Group 17 has the lowest metallic character. This is because while moving across the period, the number of electrons in the outermost shell increases. This make it difficult for atoms to leave see electrons and become electropositive . Group 17 has the highest tendency of accepting electrons.

Ionization energy is the energy use to remove electron from an atom in gaseous stage. Ionization energy decrease as we move from right to left on the periodic table and metallic character increases as we move from right to left on the periodic table.

Answer:

4–octene.

Explanation:

To name the compound given in the question, we must determine the following:

1. Determine the functional group of the compound and locate its position by giving it the lowest possible count.

2. Locate the longest continuous carbon. This gives the parent name of the compound.

3. Combine the above to obtain the name of the compound.

Now, let us determine the name of the compound bearing in mind the information given above. This is illustrated below:

1. The functional group of the compound is double bond i.e alkene and it located at carbon 4.

2. The longest continuous carbon chain is 8. Since the compound is an alkene, the name becomes octene.

3. Therefore, the name of the compound is:

4–octene.

Answer:

Concentration of the H₂SO₄ solution is 0.0737 M

Explanation:

Equation of the neutralization reaction between the acid, H₂SO₄, and the base, NaOH, is given below:

H₂SO₄ + 2NaOH -----> Na₂SO₄ + 2H₂O

From the above equation, one mole of acid requires 2 moles of base for complete neutralization which occurs at phenolphthalein endpoint.

mole ratio of acid to base, nA/nB = 1:2

Concentration of the base, Cb = 0.1311 M

Volume of base, Vb, = 28.11 mL

Concentration of acid, Ca = ?

Volume of acid, Va + 25.0 mL

Using the formula, CaVa/CbVb = nA/nB

making Ca subject of the formula, Ca = Cb*Vb*nA/Va*nB

substituting the values into the equation

Ca = (0.1311 * 28.11 * 1) / 25.0 * 2 = 0.0737 M

Therefore, concentration of the H₂SO₄ solution is 0.0737 M

Answer:

-30.7 kj/mol

Explanation:

The standard free energy for the given reaction that is the hydrolysis of ATP is calculated using the formula:  ∆Go ’= -RTln K’eq

where,  

R = -8.315 J / mo

T = 298 K

For reaction,

1. K′eq1=270,

∆Go ’= -RTln K’eq

= - 8.315 x 298 x ln 270

=  - 8.315 x 298 x 5.59

= - 13,851.293 J / mo

= - 13.85 kj/mol

2. K′eq2=890

∆Go ’= -RTln K’eq

= - 8.315 x 298 x ln 890

=  - 8.315 x 298 x 6.79

=  - 16.82 kj/mol

therefore, total standard free energy

= - 13.85 + (-16.82)

=  -30.7 kj/mol

Thus, -30.7 kj/mol is the correct answer.

Answer:

1.) zinc and aluminum

2.)

a.) The old coating was made of tin.

b.) Aluminum would be a good choice for repairing the fracture.

Answer:

For every given mass of Vanadium, the relative number of oxygen atoms present or the mole ratio of Oxygen to Vanadium is:

A. 1:1

B. 3:2

C. 2:1

D. 5:2

Note: The question is stated more clearly below:

Vanadium (V) and oxygen (O) form a series of compounds with the following compositions: Mass % V 76.10 67.98 61.42 56.02 Mass % O 23.90 32.02 38.58 43.98 Compound Mass % N 1 33.28 2 39.94.

What are the relative numbers of atoms of oxygen in the compounds for a given mass of vanadium?

Explanation:

Number of moles in 100 g mass = % mass / molar mass

Molar mass of Vanadium, V = 51 g/mol

Molar mass of oxygen atom, O = 16 g/mol

1. Percentage mass of V and O is 76.10% and 23.90% respectively.

Number of moles of each atom;

V = 76.10/51.0 = 1.5 moles

O = 23.9/16 = 1.5 moles

Mole ratio of oxygen to vanadium = 1.5/1.5 = 1 : 1

2. Percentage mass of V and O is 67.98% and 32.02% respectively

Number of moles of each atom:

V = 67.98/51 = 1.33

O = 32.02/16 = 2

Mole ratio of oxygen to vanadium = 2/1.33 = 1.5 : 1 = 3 : 2

3. Percentage mass of V and O is 61.42% and 38.58% respectively

Number of moles of each atom:

V = 61.42/51 = 1.2

O = 38.58/16 = 2.4

Mole ratio of oxygen to vanadium = 2.4/1.2 = 2 : 1

4. Percentage mass of V and O is 56.02% and 43.98% respectively

Number of moles of each atom:

V = 56.02/51 = 1.10

O = 43.98/16 = 2.75

Mole ratio of oxygen to vanadium = 2.75/1.10 = 2.5 : 1 = 5 : 2

Mass of the Vanadium, number of O2 atoms present, or the mole ratio of   1:1 , 3:2 , 2:1 , 5:2 . As Vanadium (V) and oxygen (O) form a series of compounds is given with masses of 76.10 67.98, 23.90 32.02, 33.28 2 39.94, etc.

Learn more about the Vanadium (V) and oxygen (O).  

brainly.com/question/2145642.

Answer:

filtering

Explanation:

you're pouring the mixture through muslin cloth to keep the particles and bigger peaces out of the soap.

Answer:

a. 2,3-dimethylbutane < 2-methylpentane < n-hexane

Explanation:

The boiling point of alkanes is highly affected by the degree of branching in the molecule. Branched alkanes generally have a lower boiling point than unbranched alkanes.

The reason for the higher boiling point of unbranched alkanes is because they have greater vanderwaals forces acting between their molecules due to their larger surface area. Recall that branched alkanes have a lesser surface area compared to unbranched alkanes.

n-hexane is an unbranched alkane hence it will have the highest boiling point followed by 2-methyl pentane and lastly 2,3-dimethyl butane. The boiling point continues to decrease as the extent of branching increases.

Answer:

4.32 is the ratio of f at the higher temperature to f at the lower temperature

Explanation:

Using the sum of Arrhenius equation you can obtain:

ln (f₂/f₁) = Eₐ / R ₓ (1/T₁ - 1/T₂)

Where f represents the rate constant of the reaction at T₁ and T₂ temperatures. Eₐ is the energy activation (155kJ / mol = 155000J/mol) and R is gas constant (8.314J/molK)

Replacing:

ln (f₂/f₁) = 155000J/mol / 8.314J/molK ₓ (1/495K - 1/515)

Where 2 represents the state with the higher temperature and 1 the lower temperature.

ln (f₂/f₁) = 155000J/mol / 8.314J/molK ₓ (1/495K - 1/515)

ln (f₂/f₁) = 1.4626

f₂/f₁ = 4.32

Answer:

185.49 grams of Zinc would react with 454g (1lb) of copper sulfate

Explanation:

Yo know the following balanced reaction:

CuSO₄(aq)+ Zn(s) →Cu(s) + ZnSO₄(aq)

You can see that by stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reagents and products are part of the reaction:

Being:

the molar mass of the compounds participating in the reaction is:

Then, by stoichiometry of the reaction, the following amounts of mass of reagent and product participate in the reaction:

Now you can apply the following rule of three: if 160 grams of CuSO₄ react with 65.37 grams of Zn by this reaction stoichiometry, 454 grams of CuSO₄ with how much mass of Zn will it react?

[tex]mass of Zn=\frac{454 grams of CuSO_{4} *65.37 grams of Zn}{160 grams of CuSO_{4}}[/tex]

mass of Zn= 185.49 grams

185.49 grams of Zinc would react with 454g (1lb) of copper sulfate