Balance the following redox reaction if it occurs in acidic solution. What are the coefficients in front of Ni and H+ in the balanced reaction? Ni2+(aq) + NH4+(aq) → Ni(s) + NO3-(aq)

Answer:

this the partial removal of a substance from a solution or mixture by dissolving it in another immiscible solvent in which it is more soluble.

Answer:

landslide mudflows slump and creep

Explanation:

no explanation

This  problem provides the molar mass and radius of a metal that has an BCC unit cell and the density is required.

Firstly, we consider the formula that relates molar mass and also includes the Avogadro's number and the volume of the unit cell:

[tex]\rho =\frac{Z*M}{V*N_A}[/tex]

Whereas Z stands for the number of atoms in the unit cell, M the molar mass, V the volume and NA the Avogadro's number. Next, since BCC is able to hold 2 atoms and M and NA are given, we calculate the volume of the atom in the unit cell given the radius in meters:

[tex]V=a^3=(\frac{4R}{\sqrt{3} } )^3=(\frac{4*1.74x10^{-10}m}{\sqrt{3} } )^3=6.49x10^{-29}m^3[/tex]

And finally the required density in g/cm³:

[tex]\rho =\frac{2*341.1g/mol}{6.49x10^{-29}m^3\frac{m^3}{atom} *6.022x10^{23}\frac{atom}{mol} } =17455257.8g/m^3\\\\\rho=17.5g/cm^3[/tex]

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A sample of gas has Pi = 0.768 ATM, Vi = 10.5 L, and Ti = 300 K. What is the final pressure if VF = 7.85 L and T and f = 250 K?

Answer:

solution

Explanation:

solvent +solute =solution

A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.

A cylindrical weight with a mass (m) of 3 kg is dropped, that is, its initial velocity (u) is 0 m/s and travels 10 m (s). Assuming the acceleration (a) is that of gravity (9.8 m/s²). We can calculate the velocity (v) of the weight in the instant prior to the collision with the piston using the following kinematic equation.

[tex]v^{2} = u^{2} + 2as = 2 (9.8 m/s^{2} ) (10m) \\\\v = 14 m/s[/tex]

The object with a mass of 3 kg collides with the piston at 14 m/s, The kinetic energy (K) of the object at that moment is:

[tex]K = \frac{1}{2} m v^{2} = \frac{1}{2} (3kg) (14m/s)^{2} = 294 J[/tex]

The kinetic energy of the weight is completely converted into heat transferred into the gas cylinder. Thus, Q = 294 J.

Given all the process is at 250 K (T), we can calculate the change of entropy of the gas using the following expression.

[tex]\Delta S_{gas} = \frac{Q}{T} = \frac{294 J}{250K} = 1.18 J/K[/tex]

The change in the entropy of the environment, has the same value but opposite sign than the change in the entropy of the gas. Thus, [tex]\Delta S_{env} = -1.18 J/K[/tex]

A cylindrical weight with a mass of 3 kg is dropped onto the piston from a height of 10 m. The entropy of the gas is 1.18 J/K and the change in the entropy of the environment is -1.18 J/K.

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Answer:

Meiosis produces haploid cells (gametes), which contain single chromosomes, or on-half the number of chromosomes in diploid cells. When a sperm and an egg join, the single chromosomes pair up, which results in genetic diversity in the offspring

Answer:

umm

Explanation:

matter makes up everything. everything is matter

the more matter an object has the more mass It has

Answer:

This tells you that one mole of sulfur atoms, S , has a mass of 32 g

Answer: specialized body structures of Snakes

Explanation:

Snakes have many adaptations that allow them to move, hunt, eat and survive in their environments. Many animals have developed specific parts of the body adapted to survival in a certain environment. Among them are webbed feet, sharp claws, whiskers, sharp teeth, large beaks, wings, and hooves. In most aquatic animals, swimming is a must. To aid swimming, many animals have adapted and evolved with webbed feet.

The mole fraction of N₂ after the mixture of 4.0 L of N₂ at 15 atm with 4.0 L of H₂ at 7.0 atm is 0.68.

We can calculate the mole fraction of N₂ with the following equation:

[tex] X_{N_{2}} = \frac{n_{N_{2}}}{n_{t}} = \frac{n_{N_{2}}}{n_{N_{2}} + n_{H_{2}}} [/tex]   (1)

The number of moles of N₂ and H₂ can be found with the ideal gas law:

[tex] PV = nRT [/tex]

Where:

P: is the pressure

R: is the gas constant

T: is the temperature

V: is the volume

For nitrogen gas we have:

[tex] n_{N_{2}} = \frac{P_{N_{2}}V_{N_{2}}}{RT} [/tex]   (2)

And for hydrogen:

[tex] n_{H_{2}} = \frac{P_{H_{2}}V_{H_{2}}}{RT} [/tex]   (3)

After entering equations (2) and (3) into (1), we get:

[tex] X_{N_{2}} = \frac{\frac{P_{N_{2}}V_{N_{2}}}{RT}}{\frac{P_{N_{2}}V_{N_{2}}}{RT} + \frac{P_{H_{2}}V_{H_{2}}}{RT}} [/tex]  

Since RT are constants, we have:

[tex] X_{N_{2}} = \frac{P_{N_{2}}V_{N_{2}}}{P_{N_{2}}V_{N_{2}} + P_{H_{2}}V_{H_{2}}} [/tex]                

We know that:

[tex] P_{N_{2}} = 15 atm[/tex]                

[tex] V_{N_{2}} = 4.0 L[/tex]                

[tex] P_{H_{2}} = 7.0 atm[/tex]                

[tex] V_{H_{2}} = 4.0 L[/tex]          

so:

[tex] X_{N_{2}} = \frac{15 atm*4.0 L}{15 atm*4.0 L + 7.0 amt*4.0 L} = 0.68 [/tex]                

Therefore, the mole fraction of N₂ is 0.68.

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This graph shows a straight line.

Answer:

The graph is a straight line that angles upward.

Answer:BF3

Explanation:USA test prep

Fire extinguishment is referred to the stage in which flames are not visible

and the combustible item no longer generates heat.

This stage refers to when the fire has been put out through a fire

extinguisher or other compounds.   When fire is put out, there is no longer

flames present.

The absence of flames also means that there is no heat generation or

combustion products due to the absence of a heat source.

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Answer:

[tex]Molarity\,\,of\,\,the\,\,solution\,\,is\,\,S=0.039M[/tex]

Explanation:

[tex]W=2.18 g\\M=322g\\V=175mL=0.175L\\\\S=\frac{W}{MV} \\=>S=\frac{2.18}{322*0.175} \\So,S=0.039M[/tex]

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The reducing agent in the reaction 2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s) is lithium (Li).

The general reaction is:

2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s)   (1)

We can write the above reaction in two reactions, one for oxidation and the other for reduction:

Li⁰(s) → Li⁺(aq) + e⁻   (2)

Fe²⁺(aq) + 2e⁻ → Fe⁰(s)    (3)

We can see that Li⁰ is oxidizing to Li⁺ (by losing one electron) in the lithium acetate (reaction 2) and that Fe²⁺ in iron(II) acetate is reducing to Fe⁰ (by gaining two electrons) (reaction 3).  

We must remember that the reducing agent is the one that will be oxidized by reducing another element and that the oxidizing agent is the one that will be reduced by oxidizing another species.

In reaction (1), the reducing agent is Li (it is oxidizing to Li⁺), and the oxidizing agent is Fe(CH₃COO)₂ (it is reducing to Fe⁰).  

Therefore, the reducing agent in reaction (1) is lithium (Li).  

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Answer:

Many chemical reactions are reversible; that is, the products of the reaction can combine to re-form the reactants. An example of a reversible reaction is that of hydrogen with iodine to form hydrogen iodide:

H2(g) + I2(g) 2 HI(g)

We can study this reversible reaction by placing hydrogen and iodine in a reaction vessel and then measuring the concentrations of H2, I2, and HI at various times after the reactants are mixed. Figure 13.8 is a plot of the concentrations of reactants and products of this reaction versus time. The concentration of hydrogen iodide increases very rapidly at first, then more slowly, and finally, after the time indicated by the vertical line marked "Equilibrium," remains constant. Similarly, the concentrations of hydrogen and iodine are large at the start of the reaction but decrease, rapidly at first, and then more slowly. Finally, they, too, become constant.

If this reaction were not reversible, the concentrations of hydrogen and iodine would have continued to decrease and the concentration of hydrogen iodide to increase. This process does not happen. Instead, as soon as any molecules of hydrogen iodide are formed, some decompose into hydrogen and iodine. Two reactions are taking place simultaneously: the formation of hydrogen iodide and its decomposition. When the concentrations of all these components become constant (at the equilibrium point in Figure 13.8), the rate of the forward reaction (H2 + I2 2 HI) must be equal to the rate of the reverse reaction (2 HI H2 + I2). A state of dynamic chemical equilibrium has then been reached, one in which two opposing reactions are proceeding at equal rates, with no net changes in concentration.

PICTURE 13.8

FIGURE 13.8 Concentration changes during the reversible reaction

H2(g) + I2(g) 2 HI as it proceeds toward equilibrium.

We have encountered this criterion for equilibrium before. In the equilibrium between a liquid and its vapor, the rate of vaporization is equal to the rate of condensation. In the equilibrium of a saturated solution with undissolved solute, the rate of dissolution is equal to the rate of precipitation. In the equilibrium of a weak acid with its ions, the rate of dissociation is equal to the rate of recombination. Note that none of these reactions is static: Two opposing changes are occurring at equal rates.

B. The Characteristics of Chemical Equilibrium

1. Equal rates

At equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction.

2. Constant concentrations

At equilibrium, the concentrations of the substances participating in the equilibrium are constant. Although individual reactant molecules may be reacting to form product molecules and individual product molecules may be reacting to re-form the reactants, the concentrations of the reactants and the products remain constant.

3. No free energy change

At equilibrium, the free energy change is zero. Neither the forward nor the reverse reaction is spontaneous and neither is favored. Consider the ice-water change. Above 0°C, ice melts spontaneously to form liquid water; G for this change is negative. Below 0°C, the change from ice to water is not spontaneous; G is positive. At 0°C, the two states are in equilibrium. The rate of melting is equal to the rate of freezing: the amount of ice and water and the amount of liquid water present remain constant, and the free energy change is zero as long as no energy is added to or subtracted from the mixture.

C. The Equilibrium Constant

In Chapter 12, we introduced the mathematical relationship between the concentrations of the components of an equilibrium, known as the equilibrium constant, Keq. We said that, for the general equation of a reversible reaction

Explanation:

sorry(: hope to help

Answer:

A: Calcium is a group 2 element with two valence electrons. Therefore, it is very reactive and gives up electrons in chemical reactions. It is likely to react with an element with six valence electrons that “wants” to gain two electrons. This would be an element in group 6, such as oxygen.

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The major product obtained when 1-hexyne is treated with H2 and Pd is; 1-hexene.

The structure of 1-hexyne is such that it possesses a triple bond around its first Carbon in it's carbon chain.

On this note; hydrogenation by treatment with H2 and Paladium, Pd as catalyst yields 1-hexene as the major product.

PS: Paladium, Pd is the major constituent of the Lindlar's catalyst.

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Answer:

Explanation:

First you will find the mole from the molarity and then the desired mass from the mole.

Answer:

go get the stuff.

Explanation:

The genetic makeup of the majority of these organisms is either RNA or DNA. For instance, some viruses' genetic material may be RNA whereas others' genetic material may be DNA. RNA is present in the Human Immunodeficiency Virus (HIV), which after adhering to the host cell, transforms into DNA.

DNA is a collection of molecules that is in charge of transporting and passing genetic information from parents to children. A ribonucleic acid called RNA aids in the body's production of proteins. In the human body, new cells are created as a result of this nucleic acid.

Instead of thymine, uracil is present in RNA. All other bases are same as DNA like adenine, guanine and cytosine. The order of bases in RNA is:

UAACGUCG.

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Using melting and boiling temperature, hardness and electric current passing testing.

Ionic solids are materials that have a strong bond between their ions, thus producing well-defined shapes.

In addition, due to this strong attraction, the boiling and melting temperatures of these materials are very high, in addition to the resistance to breakage presented by them.

Finally, ionic solids are also excellent conductors of electricity.

So, their properties used to determine whether or not they are ionic solids are  melting and boiling temperature, hardness and electric current passing testing.

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Answer:

The answer is A and C.

Explanation:

The reactants in the process of cellular respiration are oxygen and glucose, respectively. It is ATP that serves as the primary product of cellular respiration, with carbon dioxide and water serving as waste products.

Sugar is a glucose.

Oxygen and glucose are the two  substances that are reactants in the chemical reactions of cellular respiration. Therefore, the correct options are options A, C.

Through the process of cellular respiration, organisms mix oxygen with food molecules, directing the chemical energy contained in these substances towards life-sustaining processes while excreting carbon dioxide and water as waste. Foods are broken down by microorganisms that do not require oxygen in a process known as fermentation.

Adenosine triphosphate (ATP), an energy-rich compound that absorbs the chemical energy generated by the decomposition of food molecules then releases it to power other cellular functions, is one goal of the breakdown of foodstuffs. ATP is created when the energy found inside chemical bonds is converted from one form to another. Oxygen and glucose are the two  substances that are reactants in the chemical reactions of cellular respiration.

Therefore, the correct options are options A, C.

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Answer:,   Correct option is    0.288m/s

Explanation:    

The relationship between the velocity of the wave, its wavelength and frequency is given by the formula

Wavelengthλ=

Frequency(ν)

Speed(v)

​

,

where, v - velocity of the wave

           λ - wavelength of the wave

          f - frequency of the wave.

In the question it is given that the frequency is 4.8 Hz and the wavelength is 6.0 cm, that is, 0.06 meters.

The velocity of the sound is calculated as follows.

v=f×λ=4.8 Hz×0.06 m=0.288 m/s

Hence, the speed of the water wave is 0.288 m/s.

This question is asking for a method for the determination of the freezing point in a solution that does not have a noticeable transition in the cooling curve, which is basically based on a linear fit method.

The first step, would be to understand that when the transition is well-defined as the one on the attached file, we can just identify the temperature by just reading the value on the graph, at the time the slope has a pronounced change. For instance, on the attached, the transition occurs after about 43 seconds and the freezing point will be about 4 °C.

However, when we cannot identify a pronounced change in the slope, it will be necessary to use a linear fit method (such as minimum squares) to figure out the equation for each segmented line having a significantly different slope and then equal them so that we can numerically solve for the intercept.

As an example, imagine two of the segmented lines have the following equations after applying the linear fit method:

[tex]y=-3.5 x + 25\\\\y=-0.52 x + 2[/tex]

First of all, we equal them to find the x-value, in this case the time at which the freezing point takes place:

[tex]-3.5 x + 25=-0.52 x + 2\\\\-3.5 x+0.52 x =2-25\\\\x=\frac{-23}{-2.98}=7.72[/tex]

Next, we plug it in in any of the trendlines to obtain the freezing point as the y-value:

[tex]y=-3.5 (7.72) + 25\\\\y = 1.84[/tex]

This means the freezing point takes place after 7.72 second of cooling and is about 1.84 °C. Now you can replicate it for any not well-defined cooling curve.

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