How many stoms of oxygen in 4 molecules of HNO,?

Answer:

67.91 g of CuCl2; 32.09 g of Cu.

Explanation:

The two masses add to 100.0 g, the initial amount of starting material, demonstrating the law of conservation of matter.

Answer:

0.500 moles

Explanation:

In order to convert grams of any given substance into moles, we need the substance's molar mass:

We can find the molar masses of each element in the periodic table:

Now we can divide the given mass by the molar mass in order to calculate the number of moles:

Answer:

liquid, and of course

Explanation:

Answer:no

Explanation:

Answer:

Scandium is a transition metal

Explanation:

The more energy that particles have, the more they move.

Answer:

The activation energy is the amount of energy needed for a chemical reaction to occur.

Answer:

Activation energy, in chemistry, the minimum amount of energy that is required to activate atoms or molecules to a condition in which they can undergo chemical transformation or physical transport.

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Answer:

B

Explanation:

Vaccines prevent illness and disease

Explanation:

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Answer:

2.0kg

Explanation:

The mass of 7 red apples in kilograms is to be considered as the 1.16 kilograms.

Since the mass of a dozen apples is 2 kg.

we know that

1 dozen is 12 units

So the mass of 12 apples = 2kg

So mass of 1 apple = 2/12 = 1/6  kg

Now the mass of 7 apple is to be  

= 7/6 kg

= 1.16 kg

hence, The mass of 7 red apples in kilograms is to be considered as the 1.16 kilograms.

Learn more about mass here: https://brainly.com/question/24701836

Answer:

No, no precipitate is formed.

Explanation:

Hello there!

In this case, since the reaction between ammonium sulfide and aluminum nitrate is:

[tex]3(NH_4)_2S(aq)+2Al(NO_3)_3(aq)\rightarrow Al_2S_3(s)+6NH_4NO_3(aq)[/tex]

In such a way, we can calculate the concentration of aluminum and sulfide ions in the solution as shown below, and considering that the final total volume is 140.00 mL:

[tex][Al^3^+]=\frac{120.00mL*0.0082M}{140.00mL}=0.00703M[/tex]

[tex][S^2^-]=\frac{20.00mL*0.0090M}{140.00mL}=0.00129M[/tex]

In such a way, we can calculate the precipitation quotient by:

[tex]Q=[Al^3^+]^2[S^2^-]^3=(0.00703)^2(0.00129)^3=1.05x10^{-13}[/tex]

Which is smaller than Ksp and meaning that the precipitation does not occur.

Regards!

Answer:

See explanation

Explanation:

Hello there!

In this case, since the the concentrations are not given, and not even the Ksp, we can solve this problem by setting up the chemical equation, the equilibrium constant expression and the ICE table only:

[tex]AgCl(s)\rightleftharpoons Ag^+(aq)+Cl^-(aq)[/tex]

Next, the equilibrium expression according to the produced aqueous species as the solid silver chloride is not involved in there:

[tex]Ksp=[Ag^+][Cl^-][/tex]

And therefore, the ICE table, in which x stands for the molar solubility of the silver chloride:

       [tex]\ \ \ \ \ \ \ \ \ \ \ \ \ \ AgCl(s)\rightleftharpoons Ag^+(aq)+Cl^-(aq)[/tex]

I          -                   0             0

C        -                   +x           +x

E        -                    x             x

Which leads to the following modified equilibrium expression:

[tex]Ksp=x^2[/tex]

Unfortunately, values were not given, and they cannot be arbitrarily assigned or assumed.

Regards!

Answer:

+2

Explanation:

Answer:

2 electrons

Explanation:

In Iron (II) oxide, or Ferrous Oxide, or FeO, the Iron element (Fe) is bonded to the Oxygen, in the oxidation state of "2". This means that the Iron has accepted 2 electrons from the Oxygen.

Answer:

[tex]P_{H_2}=727mmHg[/tex]

Explanation:

Hello there!

In this case, according to the given data, it is possible to infer that the gas mixture lies on the 15.0 cm-high column of water, so that the total pressure or atmospheric pressure is given by:

[tex]P_{atm}=P_{water}+P_{H_2}[/tex]

Thus, since the atmospheric pressure is 745 mmHg and the vapor pressure of water is 18 mmHg, the pressure of hydrogen turns out to be:

[tex]P_{H_2}=P_{atm}-P_{water}\\\\P_{H_2}=745mmHg-18mmHg\\\\P_{H_2}=727mmHg[/tex]

Best regards!

Solution :

According to Chick's law

[tex]$\frac{N_t}{N_0}=e^{-k'C^n t}$[/tex]

where, t = contact time

            c = concentration of disinfectant

            k' = lethality coefficient = 4.71

            n = dilution coefficient = 1

            4 log removal = % removal = 99.99

[tex]$\frac{N_t}{N_0}=\frac{\text{bacteria remaining}}{\text{bacteria initailly present}}$[/tex]

      = 1 - R

      = 1 - 0.9999

Now for plug flow reactor contact time,

[tex]$\tau =\frac{V}{Q} =\frac{75000}{40 \times 10^6}$[/tex]

          = 0.01875 days

          = 27 minutes

For CSTR, [tex]$\tau =\frac{V}{Q} =\frac{150000}{40 \times 10^6}$[/tex]

                             [tex]$=3.75 \times 10^{-3}$[/tex] days

                            = 5.4 minute

There are 3 reactors, hence total contact time = 3 x 5.4

                                                                            = 16.2 minute

Or [tex]$\frac{N_t}{N_0}=e^{-k'C^n t}$[/tex]

or [tex]$(1-0.9999)=e^{-4.71 \times C \times t}$[/tex]

∴ C x t = 1.955

For PFR, [tex]$t_1 = 27 $[/tex] min

∴ C [tex]$=\frac{1.955}{27}$[/tex] = 0.072 mg/L

For CSIR, [tex]$t_2=16.2$[/tex] min

[tex]$C=\frac{1.955}{16.2} = 0.1206$[/tex] mg/L

∴ Chlorine required for PFR in kg/day

[tex]$=\frac{0.072 \times 40 \times 10^6 \times 3.785}{10^6}$[/tex]      (1 gallon = 3.785 L)

= 18.25 kg/day

Therefore we should go for PFR system.

Answer:

The correct answer is - transparent medium.

Explanation:

A transparent substance or medium is the substance that allows light to pass through it. Light moves through these substances as they do not absorb the light and do not reflect too.

The example of such substances is glass, air or water. These substances allow light to pass through them.

Thus, The correct answer is - transparent medium.

Answer:

Oxygen cycle, along with the carbon cycle and nitrogen cycle plays an essential role in the existence of life on the earth. The oxygen cycle is a biological process which helps in maintaining the oxygen level by moving through three main spheres of the earth which are:

Explanation:

The atmosphere is the layer of gases presents above the earth’s surface. The sum of Earth’s ecosystems makes a biosphere. Lithosphere is the solid outer section along with the earth’s crust and it is the largest reservoir of oxygen.

Answer:

Oxygen cycle, circulation of oxygen in various forms through nature. ... Free in the air and dissolved in water, oxygen is second only to nitrogen in abundance among uncombined elements in the atmosphere. Plants and animals use oxygen to respire and return it to the air and water as carbon dioxide (CO2).

Explanation:

Answer:

2200 L

Explanation:

There are 22.4 L in 1 mole of neon so if you have 98.2 L, then you have 98.2 mole x (22.4 L/mole) = 2200 L of neon.

Answer:

84672 J

Explanation:

From the question given above, the following data were obtained:

Height (h) = 72 m

Combined mass (m) = 120 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Energy (E) =?

We can obtain the energy by using the following formula:

E = mgh

Where

E => is the energy.

g => is the acceleration due to gravity

m => is the mass.

h => is the height.

E = 120 × 9.8 × 72

E = 84672 J

Thus, the energy is 84672 J

Answer:

Explanation:

From the given information:

The energy of photons can be determined by using the formula:

[tex]E = \dfrac{hc}{\lambda}[/tex]

where;

planck's constant (h) = [tex]6.63 \times 10^ {-34}[/tex]

speed oflight (c) = [tex]3.0 \times 10^8 \ m/s[/tex]

wavelength λ = 58.4 nm

[tex]E = \dfrac{6.63 \times 10^{-34} \ J.s \times 3.0 \times 10^8 \ m/s}{58.4 \times 10^{-9 } \ m}[/tex]

[tex]E =0.34 \times 10^{-17} \ J[/tex]

[tex]E = 3.40 \times 10^{-18 } \ J[/tex]

To convert the energy of photon to (eV), we have:

[tex]1 eV = 1.602 \times 10^{-19} \ J[/tex]

Hence

[tex]3.40 \times 10^{-18 } \ J = \dfrac{1 eV}{1.602 \times 10^{-19 } \ J }\times 3.40 \times 10^{-18 } \ J[/tex]

[tex]E = 2.12 \times 10 \ eV[/tex]

E = 21.2 eV

b)

The equation that illustrates the process relating to the first ionization is:

[tex]Hg_{(g)} \to Hg^+ _{(g)} + e^-[/tex]

c)

The 1st ionization energy (I.E) of Hg can be calculated as follows:

Recall that:

I.E  = Initial energy - Kinetic Energy

I₁ (eV) = 21.2 eV - 10.75 eV

I₁ (eV) = 10.45 eV

Since ;

[tex]1 eV = 1.602 \times 10^{-19} \ J[/tex]

∴

[tex]10.45 \ eV = \dfrac{1.602 \times 10^{-19 } \ J }{ 1 \ eV}\times 10.45 \ eV[/tex]

Hence; the 1st ionization energy of Hg atom = [tex]1.67 \times 10^{-18} \ J[/tex]

[tex]1.67 \times 10^{-21} \ kJ[/tex]

Finally;

[tex]I_1 \ of \ the \ Hg (kJ/mol) = \dfrac{1.67 \times 10^{-21 \ kJ} \times 6.02 \times 10^{23} \ Hg \ atom }{1 \ Kg \ atom }[/tex]

[tex]\mathbf{= 1.005 \times 10^3 \ kJ/mol}[/tex]

Answer:

[tex]n_{O_2}=0.866molSO_2\\\\n_{SO_2}=0.577molSO_2[/tex]

Explanation:

Hello there!

In this case, according to the given balanced chemical reaction by which ZnS reacts with O2, it is possible to calculate the moles of the latter that are consumed, not produced, according the 2:3 mole ratio between them and the following stoichiometric set up:

[tex]n_{O_2}=56.25gZnS*\frac{1molZnS}{97.47gZnS}*\frac{3molO_2}{2molZnS} \\\\n_{O_2}=0.866molO_2[/tex]

But also, we can compute the moles of SO2 that are produced via the the 2:2 mole ratio of ZnS to SO2:

[tex]n_{SO_2}=56.25gZnS*\frac{1molZnS}{97.47gZnS}*\frac{2molSO_2}{2molZnS} \\\\n_{SO_2}=0.577molSO_2[/tex]

Regards!

Answer:

c4 is an explosive..

contains RDX, DOS, DOA, and PIB.

Explanation:

Answer:

Pressure Affects the Boiling Point

Atmospheric pressure influences the boiling point of water. When atmospheric pressure increases, the boiling point becomes higher, and when atmospheric pressure decreases (as it does when elevation increases), the boiling point becomes lower.

Explanation:

i think it will help you

Answer: 8%

Explanation:

Can't you just do 4/50= 8%?  If not, just remember that for empirical formula, do percent to grams, and grams to moles, divide by smallest and multiply to whole!

Answer:

Ionization Energy = 9 eV

Explanation:

If we apply the law of conservation of energy to the given situation, we will get the following equation:

[tex]Energy\ of\ Photon = Ionization\ Energy + Kinetic\ Energy\ of\ Electron\\[/tex]

where,

Energy of Photon = 10 eV

Ionization Energy = ?

Kinetic Energy of Electrons = 1 eV

Therefore,

[tex]10\ eV = Ionization\ Energy + 1\ eV\\Ionization\ Energy = 10\ eV - 1\ eV[/tex]

Ionization Energy = 9 eV