Using these metal ion/metal standard reduction potentials Cd2+(aq)|Cd(s) Zn2+(aq)|Zn(s) Ni2+(aq)|Ni(s) Cu2+(aq)/Cu(s) -0.40 V -0.76 V ‑0.25 V +0.34 V Calculate the standard cell potential for the cell whose reaction is Ni2+(aq) + Zn(s) →Zn2+(aq)+ Ni(s)

Answer:

c

Explanation:

it's found inside the atomic nucleus

Answer:

CO3^2-

Explanation:

In qualitative analysis, we try to use chemical reactions to determine the composition of an unknown substance. The addition of certain reagents to the unknown solution gives certain results that show the presence or absence of certain species from the unknown sample.

When dilute HCl is added to an unknown sample and effervescence is observed, then the unknown sample must contain CO3^2- or HCO3^-. The presence of these species is confirmed if the gas evolved is passed through limewater and the gas turns limewater milky.

Answer:

8.05 moles

Explanation:

5.50 / 2.954 = x / 4.325

x = 8.05

According to ideal gas equation, if the temperature and pressure stay constant during the process 0.520 moles have been added  so that the balloon expands to 4.325 L.

The ideal gas equation is a equation which is applicable in a hypothetical state of an ideal gas.It is a combination of Boyle's law, Charle's law,Avogadro's law and Gay-Lussac's law . It is given as, PV=nRT where R= gas constant whose value is 8.314.The law has several limitations.The law was proposed by Benoit Paul Emile Clapeyron in 1834.

In the given example if pressure and temperature are constant then V=nR substituting V=4.325 l and R=8.314  so n=V/R=4.325/8.314=0.520 moles.

Thus, 0.520 moles of helium are added if the temperature and pressure stay constant during this process.

Learn more about ideal gas equation,here:

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Explanation:

free your mind drink water and go outside take fresh air you will get answers

Answer:

The correct answer is 1332 KJ.

Explanation:

Based on the given information,  

ΔH°f of C2H4 is 51.9 KJ/mol, ΔH°O2 is 0.0 KJ/mol, ΔH°f of CO2 is -394 KJ/mol, and ΔH°f of H2O is -286 KJ/mol.  

Now the balanced equation is:  

C2H4 (g) + 3O2 (g) ⇔ 2CO2 (g) + 2H2O (l)

ΔH°rxn = 2 × ΔH°f CO2 + 2 × ΔH°fH2O - 1 × ΔH°fC2H4 - 3×ΔH°fO2

ΔH°rxn = 2 (-394) + 2(-286) - 1(51.9) - 3(0)

ΔH°rxn = -1411.9 KJ

Now, the given ΔS°f of C2H4 is 219.8 J/mol.K, ΔS°f of O2 is 205 J/mol.K, ΔS°f of CO2 is 213.6 J/mol.K, and ΔS°f of H2O is 69.96 J/mol.K.  

Now based on the balanced chemical reaction,  

ΔS°rxn = 2 × ΔS°fCO2 + 2 ΔS°fH2O - 1 × ΔS°f C2H4 - 3 ΔS°fO2

ΔS°rxn = 2 (213.6) + 2(69.96) - 1(219.8) -3(205)

ΔS°rxn = -267.68 J/K or -0.26768 KJ/K

T = 25 °C or 298 K

Now putting the values of ΔH, ΔS and T in the equation ΔG = ΔH-TΔS, we get

ΔG = -1411.9 - 298.0 × (-0.2677)

ΔG = -1332 KJ.  

Thus, the maximum work, which can obtained is 1332 kJ.  

Answer:

True

Explanation:

The percentage by mass of a substance in a solution can be calculated by dividing the mass of the substance dissolved in the solution by the total mass of the solution. This can be expressed mathematically as:

Percentage by mass = mass of substance in solution/mass of solution x 100

In this case;

mass of KBr = 4.34 grams

mass of water = 17.4 grams

mass of solution = mass of KBr + mass of water = 4.34 + 17.4 = 21.74

Percentage by mass of KBr = 4.34/21.74 x 100

                                              = 19.96 %

19.96 is approximately 20%.

Hence, the statement is true.

Answer:

Lime water, [tex]Ca(OH)_{2}_({aq} )[/tex] is formed.

Explanation:

Lime-water is a clear and colourless dilute solution of aqueous calcium hydroxide salt.

Small amounts of calcium hydroxide salt,  [tex]Ca(OH)_{2}_(s)[/tex]  is sparsely soluble at room temperature when dispersed vigorously. if in excess, a white suspension called 'milk of lime'is formed.

I hope this explanation is helpful.

Answer:

Pb + 2H2O --> PbO2 + 2H2

Explanation:

Products:

Solid metal; PbO2

Hydrogen; H

Reactants:

Metal; Pb

Steam; H2O

Reactants --> Products

Pb + H2O --> PbO2 + H2

Upon balancing we have;

Pb + 2H2O --> PbO2 + 2H2

Answer:12.9e-12g or in short 12.9pg

Explanation:as p=1e-12

Answer:

0.17 lb

Explanation:

78 g * (1 lb/454 g)=0.17 lb

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Answer:

ΔHrxn = -635.14kJ/mol

Explanation:

We can make algebraic operations of reactions until obtain the desire reaction and, ΔH of the reaction must be operated in the same way to obtain the ΔH of the desire reaction (Hess's law). Using the reactions:

(1)Ca(s) + 2 H+(aq) → Ca2+(aq) + H2(g) ΔH = 1925.9 kJ/mol

(2) 2H2(g) + O2 g) → 2 H2O(l) ΔH = −571.68 kJ/mole

(3) CaO(s) + 2 H+(aq) → Ca2+(aq) + H2O(l) ΔH = 2275.2 kJ/mole

Reaction (1) - (3) produce:

Ca(s) + H2O(l) → H2(g) + CaO(s)

ΔH = 1925.9kJ/mol - 2275.2kJ/mol = -349.3kJ/mol

Now this reaction + 1/2(2):

Ca(s) + ½ O2(g) → CaO(s)

ΔH = -349.3kJ/mol + 1/2 (-571.68kJ/mol)

Answer:

1.05 V

Explanation:

Since;

E°cell= E°cathode- E°anode

E°cathode= -0.40 V

E°anode= -1.45 V

E°cell= -0.40-(-1.45) = 1.05 V

Equation of the process;

2Zr(s) + 4Cd^2+(aq) ---->2Zr^4+(aq) + 4Cd(s)

n= 8 electrons transferred

From Nernst's equation;

Ecell = E°cell - 0.0592/n log Q

Ecell= E°cell - 0.0592/8 log [0.5]/[0.5]

Since log 1=0

Ecell= E°cell= 1.05 V

Answer:

0.0583g

Explanation:

The equation of the reaction is;

2HNO3(aq) + Mg(OH)2(aq) -------> Mg(NO3)2(aq) + 2H2O(l)

From the question, number of moles of HNO3 reacted= concentration × volume

Concentration of HNO3= 0.100 M

Volume of HNO3 = 20.00mL

Number of moles of HNO3= 0.100 × 20/1000

Number of moles of HNO3 = 2×10^-3 moles

From the reaction equation;

2 moles of HNO3 reacts with 1 mole of Mg(OH)2

2×10^-3 moles reacts with 2×10^-3 moles ×1/2 = 1 ×10^-3 moles of Mg(OH)2

But

n= m/M

Where;

n= number of moles of Mg(OH)2

m= mass of Mg(OH)2

M= molar mass of Mg(OH)2

m= n×M

m= 1×10^-3 moles × 58.3 gmol-1

m = 0.0583g

Answer:The concentration of the unknown HNO3 solution = 0.01809 M

Explanation:

For the acid-base reaction,  HNO3 + NaOH-----> NaN03 + H20

we have that

C1 V1 = C2 V2

Where ,

C1 = concentration of HNO3=?

V1 = volume of HNO3 = 25.00 mL,

V2 = volume of NaOH = 22.62 mL,  

C2 = concentration of NaOH = 0.02000 M

Therefore ,

25.00 mL x C1 = 22.62 mL x 0.02000 M    

 = (22.62 mL / 25.00 mL) x 0.02000 M = 0.01809 M

The concentration of the unknown HNO3 solution = 0.01809 M

Answer:

The amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure is 7.026 kJ.

Explanation:

A molar heat of vaporization of 40.66 kJ / mol means that 40.66 kJ of heat needs to be supplied to boil 1 mol of water at its normal boiling point.

To know the amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure, the number of moles represented by 3.11 g of water is necessary. Being:

the molar mass of water is:

H₂O= 2* 1 g/mole + 16 g/mole= 18 g/mole

So: if 18 grams of water are contained in 1 mole, 3.11 grams of water in how many moles are present?

[tex]moles of water=\frac{3.11 grams*1 mole}{18 gramos}[/tex]

moles of water= 0.1728

Finally, the following rule of three can be applied: if to boil 1 mole of water at its boiling point it is necessary to supply 40.66 kJ of heat, to boil 0.1728 moles of water, how much heat is necessary to supply?

[tex]heat=\frac{0.1728 moles*40.66 kJ}{1 mole}[/tex]

heat= 7.026 kJ

The amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure is 7.026 kJ.

Answer:

There are no forces holding the gas particles together.

Explanation:

Answer:

Due to the resonance structures

Explanation:

In the question:

"Explain why, when the guanidino group of arginine is protonated, the double-bonded nitrogen is the nitrogen that accepts the proton. There is a scheme of a reversible reaction, where one equivalent of the reactant reacts with two equivalents of H plus"

We have to take into account the structure of the amino acid arginine. In which, we have the amino and the carboxylic groups in the right and the guanidine group in the left.

In this group, we have a central carbon with three nitrogen atoms around and a double bond with the nitrogen on the top. This nitrogen on the top will accept the proton because the structure produced will have a positive charge on this nitrogen. Then, the double bond with the carbon can be delocalized into the nitrogen producing a positive charge in the carbon.

In this structure (the carbocation), we can have several resonance structures. In the blue option, we can produce a double bond with the nitrogen on the right. In the purple option, we can produce a double bond with the nitrogen on the left.

In conclusion, if the nitrogen in the top on the guanidine group accepts an hydrogen atom and we will have several resonance structures that can stabilize the molecule. Due to this, the nitrogen in the top its the best option to accept hydrogens.

See figure 1

I hope it helps!

Answer:

there is one atom of oxygen and two atoms of hydrogen

Explanation:

Answer:

NSAIDs are steroidal anti-inflammatories, their action is on the phospholipase A2 enzyme, this enzyme is responsible for breaking down the phospholipids of the membrane to trigger an inflammatory response. This is how steroidal anti-inflammatory drugs inhibit ALL inflammatory pathways (not like NSAIDs that they only inhibit the COX pathway).

These corticosteroid drugs cannot exceed the systemic mineralocorticoid value 1 in the body, since this corticosteroid hormone is also released by the adrenal cortex.

The NSAIDs generate: sporadic peaks in blood glucose, hypertension, fluid retention, increase in body fat mass, possible suppression of the adrenal cortex over time, inhibiting endogenous synthesis of corticosteroids.

On the other hand, naproxen and ibuprofen are NSAIDs, that is, non-steroidal anti-inflammatory drugs that do not influence both routes of inflammation, but only COX, this enzyme is abbreviated as COX but is called cyclooxygenase, and is responsible for a single route of inflammation.

NSAIDs such as naproxen and ibuprofen can cause gastric disorders such as ulcers or gastritis if they are consumed in a very repetitive manner.

In addition, both drugs are anti-inflammatory, analgesic and antipyretic. Although its two main functions are the first two, it was shown to have an effect in lowering body temperature.

That they are anti-inflammatory means that they inhibit the path of inflammation and analgesics the path of pain.

Explanation:

Both types of drugs generate the same effect but by different mechanisms.

Some are steroids and others are not, although steroids are considered to have a greater risk of benefit that is why they are administered against more systematically compromised instances such as anaphylactic shock.

NSAIDs such as naproxen and ibuprofen are the most prescribed today, since they have few risks and very good benefits, meaning that their adverse effects are not lethal or highly relevant and have a good effect on symptoms.

Both must be administered with care when treating a diabetic patient since corticosteroids generate glycemic peaks or increase in blood glucose, and NSAIDs compete for plasma protein with oral hypoglycemic agents, thus generating that these are in higher free concentrations. high diffusing better through the tissues and increases the potency of the adverse effects of these.

Answer:

A. Rank the following substances in order of decreasing standard molar entropy (S∘).

Rank the gases from largest to smallest standard molar entropy

I2(g)>Br2(g)>Cl2(g)>F2(g)

B. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.

H2O2(g)>H2S(g) >H2O(g)

C. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.

C(s, amorphous) >C(s, graphite)>C(s, diamond)

Explanation:

Hello,

In this case, we can apply the following principles to explain the order:

- The greater the molar mass, the larger the standard molar entropy.

- The greater the molar mass and the structural complexity, the larger the standard molar entropy.

- The greater the structural complexity, the larger the standard molar entropy.

A. Rank the following substances in order of decreasing standard molar entropy (S∘).

Rank the gases from largest to smallest standard molar entropy

I2(g)>Br2(g)>Cl2(g)>F2(g)

This is due to the fact that the greater the molar mass, the larger the standard molar entropy.

B. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.

H2O2(g)>H2S(g) >H2O(g)

This is due to the fact that the greater the molar mass and the structural complexity, the larger the standard molar entropy as the hydrogen peroxide has four bonds and weights 34 g/mol as well as hydrogen sulfide that has two bonds only.

C. Rank the gases from largest to smallest standard molar entropy. To rank items as equivalent, overlap them.

C(s, amorphous) >C(s, graphite)>C(s, diamond)

Since the molecular complexity is greater in the amorphous carbon (messy arrangement), mid in the graphite and lower in the diamond (well organized).

Regards.

Answer:

Half-life at 629K = 252.4min

Explanation:

Using Arrhenius equation:

[tex]ln\frac{K_1}{K_2} = \frac{Ea}{R} (\frac{1}{T_2} -\frac{1}{T_1})[/tex]

And as Half-life in a first order reaction is:

[tex]t_{1/2}=\frac{ln2}{K}[/tex]

We can convert the half-life of 58.0min to know K₁ adn replacing in Arrhenius equation find half-life at 629K:

[tex]58.0min=\frac{ln2}{K}[/tex]

K = 0.01195min⁻¹ = K₁

[tex]ln\frac{0.01195min^{-1}}{K_2} = \frac{218kJ/mol}{8.314x10^{-3}kJ/molK} (\frac{1}{629K} -\frac{1}{652K})[/tex]

[tex]ln\frac{0.01195min^{-1}}{K_2} =1.47[/tex]

[tex]\frac{0.01195min^{-1}}{K_2} =4.35[/tex]

K₂ = 2.75x10⁻³ min⁻¹

And, replacing again in Half-life expression:

[tex]t_{1/2}=\frac{ln2}{2.75x10^{-3}min^{-1}}[/tex]

The half-life of the first-order reaction of ethylene oxide decomposition at 629 K is 251.1 min when the half-life at 652 K is 58.0 min and the activation energy is 218 kJ/mol.   

The activation energy of a reaction is related to its rate constant as follows:  

[tex] k = Ae^{-\frac{E_{a}}{RT}} [/tex]   (1)

Where:

We can find the rate constant of the first-order reaction at 652 K with the half-life as follows:

[tex]k_{652} = \frac{ln(2)}{t_{1/2}_{(652)}}[/tex]   (2)

Where [tex]t_{1/2}_{(652)}[/tex] is the half-life at 652 K= 58.0 min

Hence, the rate constant at 652 K is:                            

[tex] k_{652} = \frac{ln(2)}{58.0 min} = 0.012 min^{-1} [/tex]

Now, from equation (1) we can find the pre-exponential factor (A):

[tex]A = \frac{k_{652}}{e^{(-\frac{E_{a}}{RT_{1}})}} = \frac{0.012 \:min{-1}}{e^{(-\frac{218\cdot 10^{3} \:J/mol}{8.314 \:J/(K*mol)*652 \:K})}} = 3.51 \cdot 10^{15} min^{-1}[/tex]  

With the pre-exponential factor we can calculate the rate constant at 629 K (eq 1):

[tex]k_{629} = 3.51 \cdot 10^{15} min^{-1}*e^{(-\frac{218 \cdot 10^{3} J/mol}{8.314 J/(K*mol)*629 K})} = 2.76 \cdot 10^{-3} min^{-1}[/tex]

Finally, the half-life at 629 K is (eq 2):

[tex] t_{1/2}_{629} = \frac{ln(2)}{2.76\cdot 10^{-3} min^{-1}} = 251.1 min [/tex]

Therefore, the half-life at 629 K is 251.1 min.

Find more about activation energy here:

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Answer:

A (Atomic mass number or Nucleon number)

Explanation:

The mass number is the total number of protons and nucleons in an atomic nucleus.

Hope this helps.

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Answer:

D. Malleable, conducts electricity, high melting point, giant structure, metallic lattice

Explanation:

Copper is a metal with an atomic number of 29. This metal is soft and reddish in color which explains why it is very malleable(beaten to form various shapes without breaking).

All metals are good conductors of electricity including copper which is also a metal. Metals generally are insoluble in water. Copper also has a high melting point which is a characteristic of metals due to their giant structure and metallic lattice which makes it difficult to be broken down.

Answer:

1. metals

2. atom

3. homogeneous

4. compounds

5. lustrous

6. saturated

7. colloidal

8. homogeneous

Explanation:

Answer: 2. Dalton Model

Explanation:

John Dalton proposed that atoms are indivisible spheres. Although his model of an atom was not entirely new to the scientific world since the ancient Greeks has made  a similar statement in the past ( all matter are made up of small indivisible particle called atom).

As of when Dalton proposed his model of an atom, electrons and nucleus where yet to be discovered.

Lakes, oceans, glaciers, clouds, etc. It categorizes all forms of water on earth.

hydro = water

Answer:

Lakes, streams, ground water, polar ice caps, glaciers, water vapor, and rivers!

Explanation:

The hydrosphere is made up of all the water on Earth. So anything that is water, like oceans, can be found in the hydrosphere:)

Answer:

0.8261 M.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

HC2H3O2 + NaOH —> NaC2H3O2 + H2O

From the balanced equation above, we obtained the following:

Mole ratio of the acid, HC2H3O2 (nA) = 1

Mole ratio of the base, NaOH (nB) = 1

Data obtained from the question include the following:

Volume of acid, HC2H3O2 (Va) = 10 mL

Molarity of acid, HC2H3O2 (Ma) =..?

Volume of base, NaOH (Vb) = 16.95 mL Molarity of base, NaOH (Mb) = 0.4874 M

Finally, we shall determine the molarity of the acid solution, as follow:

MaVa/MbVb = nA/nB

Ma x 10 / 0.4874 x 16.95 = 1

Cross multiply

Ma x 10 = 0.4874 x 16.95

Divide both side by 10

Ma = (0.4874 x 16.95) /10

Ma = 0.8261 M.

Therefore, the molarity of the vinegar solution (HC2H3O2) is 0.8261 M.

Answer:

Explanation:

we know that

ΔH=m C ΔT

where ΔH is the change in enthalpy (j)

m is the mass of the given substance which is water in this case

ΔT IS the change in temperature and c is the specific heat constant  

we know that given mass=2.9 g

ΔT=T2-T1 =98.9 °C-23.9°C=75°C

specific heat constant for water is 4.18 j/g°C

therefore ΔH=2.9 g*4.18 j/g°C*75°C

ΔH=909.15 j

Answer:

W = 278.64375 Joules

Explanation:

The information given in this problem are;

Initial volume = 0L

Final volume = 2.50L

ΔV = 2.50 - 0 = 2.50 L

External pressure, P =  1.10 atm

Work = ?

These parameters are related by the equation;

w = - P ΔV

W = - (1.10 )(2.50)

W = 2.75 L atm

Upon conversion to joules;

1 liter atmosphere is equal to 101.325 joule

W = 278.64375 Joules