17.
In the reaction N2 + 3H2 → 2NH3 how many moles of ammonia will be produced from 1.30 mol
of hydrogen and excess nitrogen?

Answer:

go get the stuff.

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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This problem is describing the equilibrium whereby hydrofluoric acid decomposes to hydrogen and fluorine gases at 298 K whose equilibrium constant is 8.70x10⁻³, the equilibrium concentrations of all the reactants are both 1.33x10⁻³ M and asks for the initial concentration of hydrofluoric acid which turns out to be 2.86x10⁻³ M.

Then, we can write the following equilibrium expression for hydrofluoric acid once the change, [tex]x[/tex], has taken place:

[tex][HF]=[HF]_0-2x[/tex]

Now, since both products are 1.33x10⁻³ M we infer the reaction extent is also 1.33x10⁻³ M, and thus, we can calculate the equilibrium concentration of HF via the law of mass action (equilibrium expression):

[tex]8.70x10^{-3}=\frac{(1.33x10^{-3} M)^2}{[HF]} }[/tex]

[tex][HF]=\frac{(1.33x10^{-3} M)^2}{8.70x10^{-3}} }=2.03x10^{-4}M[/tex]

Finally, the initial concentration of HF is calculated as follows:

[tex][HF]_0=[HF]+2x=2.033x10^{-4}+2*(1.33x10^{-3})=2.86x10^{-3}M[/tex]

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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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The pH of the standard hydrogen electrode that has electrode potential of -0.128 V  is 4.3.

The equation of the hydrogen electrode is;

2H^+(aq) + 2e ⇄ H2(g)

The standard electrode potential of hydrogen is 0.00 V

Using the Nernst equation;

Ecell = E°cell - 0.0592/n log Q

Now;

E°cell = 0.00 V

n = 2

Q = 1/[H^+]

-0.128 = 0.00 - 0.0592/2 log  1/[H^+]

-0.128 = 0.00 - 0.0296 log 1/[H^+]

 -0.128 =  - 0.0296 log 1/[H^+]

-0.128/  - 0.0296  =  log 1/[H^+]

1/[H^+] = Antilog (4.32)

[H^+] = 4.79 × 10^-5

Now;

pH = -log[H^+]

pH = -log (4.79 × 10^-5)

pH = 4.3

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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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Answer:

composition of an element

Answer: Atomic number

Explanation:

I hope this helps you!

Answer:

Explanation:

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

To extract hydrogen from water, researchers insert two electrodes across the water and pass current, which can separate the hydrogen from water. The process called electrolysis of water. ... An electric field applied through the cobalt oxide to water molecules resulted in the electrolysis of water

Hydrogen gas is an environment-friendly fuel — it produces water on combustion in the presence of oxygen. ... To extract hydrogen from water, researchers insert two electrodes across the water and pass current, which can separate the hydrogen from water. The process called electrolysis of water.

Answer:

Explanation:

Thin-layer chromatography (TLC) is a chromatography technique used to separate non-volatile mixtures. ... After the sample has been applied on the plate, a solvent or solvent mixture (known as the mobile phase) is drawn up the plate via capillary action.

In this case, according to the given information about the oxidation numbers and the compounds given, it turns out possible to figure out the oxidation number of manganese in both MnI2, manganese (II) iodide and MnO2, manganese (IV) oxide, by using the concept of charge balance.

Thus, we can define the oxidation state of iodine and oxygen as -1 and -2, respectively, since the former needs one electron to complete the octet and the latter, two of them.

Next, we can write the following [tex]x[/tex], since manganese has five oxidation states, and it is necessary to calculate the appropriate ones:

[tex]Mn^xI_2^-\\\\Mn ^xO_2^{-2}[/tex]

Next, we multiply each anion's oxidation number by the subscript, to obtain the following:

[tex]Mn^xI_2^-\rightarrow x-2=0;x=+2\\\\Mn ^xO_2^{-2}\rightarrow x-4=0;x=+4[/tex]

Thus, the correct choice is Manganese has an oxidation number of +2 in Mnl2 and +4 in MnO2.

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Considering the definition of Kc, the equilibrium molar concentration of HCl at 500 K is 0.0894 [tex]\frac{mol}{L}[/tex].

The balanced reaction is:

H₂(g) +Cl₂(g) ⇆ 2 HCl(g)

Equilibrium is a state of a reactant system in which no changes are observed as time passes, despite the fact that the substances present continue to react with each other. In other words, reactants become products and products become reactants and they do so at the same rate.

In other words, chemical equilibrium is established when there are two opposite reactions that take place simultaneously at the same speed.

The concentration of reactants and products at equilibrium is related by the equilibrium constant Kc. Its value in a chemical reaction depends on the temperature and the expression of a generic reaction aA + bB ⇄ cC is

[tex]K_{c} =\frac{[C]^{c} x[D]^{d} }{[A]^{a} x[B]^{b} }[/tex]

That is, the constant Kc is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.

In this case, the constant Kc can be expressed as:

[tex]K_{c} =\frac{[HCl]^{2} }{[H_{2} ]x[Cl_{2} ] }[/tex]

You know that in an equilibrium mixture of HCl, Cl₂, and H₂:

Replacing in the expression for Kc:

[tex]4x10^{18} =\frac{[HCl]^{2} }{[1x10^{-11} ]x[2x10^{-10} ] }[/tex]

Solving:

[tex]4x10^{18} =\frac{[HCl]^{2} }{2x10^{-21} }[/tex]

[tex]4x10^{18} x 2x10^{-21}=[HCl]^{2}[/tex]

[tex]8x10^{-3} =[HCl]^{2}[/tex]

[tex]\sqrt[2]{8x10^{-3}} =[HCl][/tex]

0.0894 [tex]\frac{mol}{L}[/tex]= [HCl]

Finally, the equilibrium molar concentration of HCl at 500 K is 0.0894 [tex]\frac{mol}{L}[/tex].

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Answer:

247 ml

Explanation:

How many mL of 0.150 M HF solution are required to produce 0.0370 moles of HF 0.150 moles/ liter = 0.150/1000 moles/ml =0.000150 moles/ml

0.000150 x ? = 0.0370 moles HF

? = 0.0370/0.000150 = 247 ml

check

247 ml = 247/1000 L = 0.247

(0.247) x (0.150) =0.370 check

Answer:

›› FeBr2 molecular weight. Molar mass of FeBr2 = 215.653 g/mol. This compound is also known as Iron(II) Bromide. Convert grams FeBr2 to moles or moles FeBr2 to grams. Molecular weight calculation: 55.845 + 79.904*2 ›› Percent composition by element

Explanation:

If 38.6 grams of iron react with an excess of bromine gas, the mass of FeBr2 can form is 149 grams.

Mass is defined as a way to gauge how much matter there is in a substance or thing. The kilogram (kg) is the fundamental SI unit of mass, while lower masses can also be measured in grams (g). Atoms make up everyday matter. A majority of an atom's mass is contained in its nucleus.

Given Fe = 38.6 g.

Fe has a molar mass = 55.845 g/mol.

Given mass/molar mass equals 38.6g/55.845gmol-1, or 0.6912 moles of iron.

The reaction is described as Fe + Br2 FeBr2.

One mole Fe yields 1 mole of FeBr2.

FeBr2 would be produced from 0.6912 moles of Fe.

FeBr2 has a molar mass of 215.65 g/mol.

Moles of FeBr2 x Molar mass of FeBr2

= 215.65 g/mole x 0.6912 mole

= 149.06 g FeBr2 produced is the formula.

Thus, if 38.6 grams of iron react with an excess of bromine gas, the mass of FeBr2 can form is 149 grams.

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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.

Answer:   ionosphere

Explanation:       First it bounces off a top layer of the atmosphere called the ionosphere, then it bounces back to the Earth (this is reflection. It then bounces up again to the ionosphere, and continues bouncing back again until it reaches the radio receiver. This is called a skywave, which works around 3 to 30 MHz.

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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The solubility of nitrogen gas in water is 1.90 mL/dL at 1.00 atm and 13.3 mL/dL at 7.00 atm.

We want to relate the solubility of a gas with its partial pressure.

We can do so using Henry's law.

Henry's law states that the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid.

C = k × P

where,

The solubility of nitrogen gas is 1.90 mL/dL of blood at 1.00 atm.

Since the solvent is basically water, we can understand that the concentration of nitrogen gas is 1.90 mL/dL at 1.00 atm.

We can use this information to calculate Henry's Law constant.

k = C/P = (1.90 mL/dL)/1.00 atm = 1.90 mL/dL.atm

We want to calculate the solubility of nitrogen gas at a pressure of 7.00 atm.

We will use Henry's law.

C = k × P = (1.90 mL/dL.atm) × 7.00 atm = 13.3 mL/dL

The solubility of nitrogen gas in water is 1.90 mL/dL at 1.00 atm and 13.3 mL/dL at 7.00 atm.

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Answer: Chemical changes occur when a substance combines with another to form a new substance, called chemical synthesis or, alternatively, chemical decomposition into two or more different substances. These processes are called chemical reactions and, in general, are not reversible except by further chemical reactions.

A physical change is are changes affecting the form of a chemical substance, but not its chemical composition. Physical changes are used to separate mixtures into their component compounds, but can not usually be used to separate compounds into chemical elements or simpler compounds.

Answer:

13.6 is the correct answer written in standard form.

Explanation:

1.36, move the decimal once to the right to get 13.6

Answer:

13.6

Explanation:

The standard form is 13.6

Answer:

from the words below underline six example of rhetorical patterns

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 problem is describing the equilibrium whereby the formartion of a complex is attained when 1.47 g of nickel(II) chloride is dissolved in 100.0 mL of ammonia so that the latter's equilibium concentration is 0.20 M. Thus, it is asked to calculate the equilibrium concentrations of both nickel(II) ions and that of the complex.

Firstly, we can write out the chemical equation to be considered:

[tex]Ni^{2+}+6NH_3\rightleftharpoons Ni(NH_3)_6^{2+}[/tex]

Next, we can calculate the initial concentration of nickel(II) ions by using the concept of molarity:

[tex][Ni^{2+}]=\frac{1.47gNiCl_2*\frac{1molNiCl_2}{129.6g}*\frac{1molNi^{2+}}{1molNiCl_2} }{0.1000L}=0.113M[/tex]

Afterwards, we set up an equilibrium expression for this chemical reaction:

[tex]Kf=\frac{[Ni(NH_3)_6^{2+}]}{[Ni^{2+}][NH_3]^6}[/tex]

Which can be written in terms of the reaction extent, [tex]x[/tex]:

[tex]Kf=\frac{x}{(0.113-x)(0.2)^6}[/tex]

Now, for the calculation of [tex]x[/tex], we plug in Kf, and solve for it:

[tex]1.2x10^9=\frac{x}{(0.113-x)(0.2)^6}\\\\1.2x10^9=\frac{x}{(0.113-x)(6.4x10^{-5})}\\\\7.68x10^4(0.113-x)=x\\\\x=0.112999 M[/tex]

Which is about the same to the initial concentration of nickel(II) ions because the Kf is too large.

Thus, the required concentrations at equilibrium are about:

[tex][Ni(NH_3)_6^{2+}]=0.113M[/tex]

[tex][Ni^{2+}]=0M[/tex]

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Answer:

where is the question????????????

Explanation:

Carbon dioxide and water

I hope it helps

Answer:

The usual products of combustion reactions are carbon dioxide and water.

Explanation:

Combustion reaction is when a substance reacts with oxygen gas, resulting in a release of energy in the form of light and heat. Combustion reactions must have oxygen (O2) as one of the reactants.