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Answer:

2,1,2

1,2,1,1

the answer

Answer:

i dont know

Explanation:

its simple i dont know

Answer:

Explanation:Red

According to the Avogadro's number, there are 3.011×10²³ atoms in 0.5 mole of aluminium.

Avogadro's number is defined as a proportionality factor which relates number of constituent particles with the amount of substance which is present in the sample.

It has a SI unit of reciprocal mole whose numeric value is expressed in reciprocal mole which is a dimensionless number and is called as Avogadro's constant.It relates the volume of a substance with it's average volume occupied by one of it's particles .

According to the definitions, Avogadro's number depend on determined value of mass of one atom of those elements.It bridges the gap between macroscopic and microscopic world by relating amount of substance with number of particles.

Number of atoms can be calculated using Avogadro's number as follows: mass/molar mass×Avogadro's number, on substituting values in formula, 0.5×6.023×10²³=3.011×10²³ atoms.

Thus, there are 3.011×10²³ atoms in 0.5 mole of aluminium.

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Answer:

A

Explanation:

I had this in a test

Answer is in the photo. I can only upload it to a file hosting service. link below!

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Answer:

The responses to the given points can be defined as follows:

Explanation:

In point 1:

The answer is "No".

In point 2:

The mass of solute = The mass of precipitate [tex]= 0.56\ kg =0.56\ \times 1000 = 560 \ g[/tex]

Calculating the solubility:

[tex]= \frac{\text{mass of solute in g}}{\text{volume of solution in ml}}\\\\ = \frac{560}{ 4000} \\\\ = 0.14 \ \frac{g}{ml}[/tex]

Answer:

Splits into

Explanation:

Because fission reaction is divided into two or more pieces.

Answer:

i don't know but I just want to have coins

Answer:

Follows are the solution to this question:

Explanation:

Methane inlet humidity content of [tex]=7.00\%[/tex]

Methane moisture outlet content [tex]=0.05\%[/tex]

Zeolite absorption humidity [tex]= 6.95\%[/tex]

Dry zeolite 1 kg will accommodate water[tex]= 0.1000 \ kg[/tex]

One kilogram of Dry Zeolite will carry water from [tex]=0.1000\ kg[/tex]

The water can contain 1000 kg of zeolite [tex]= 100 \ kg[/tex]

Methane which would be made [tex]=\frac{100}{6.95\%}= 1,439 \ kg[/tex]

That's why it will be producing 1439 kg of methane.

The amount of methane that can be produced before the regeneration of zeolite is 1,439 kg.

What is zeolite?

Zeolite belongs from the family of hydrated aluminosilicate minerals, in which alkali and alkaline earth metals are present.

In the question, it is given that:

Capacity of 1kg of zeolite to hold water = 0.100kg

Capacity of 1000kg of zeolite to hold water = 100kg

Inlet moisture content of the methane = 7.00 % (by mass)

Outlet moisture content of the methane = 0.05% (by mass)

Capacity of zeolite to absorb methane content = 6.95 %

Amount of methane before the regeneration of zeolite = 100kg / 6.95% = 14.39% of kg = 1,439 kg

Hence, 1,439 kg methane will be produced before the zeolite must be regenerated.

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Answer:

[tex]\huge\boxed{\sf Q = 13.7\ Joules}[/tex]

Explanation:

Given Data:

Mass = m = 96 g = 0.096 kg

[tex]T_1[/tex] = 124 °C

[tex]T_2[/tex] = 158 °C

Change in Temp. = ΔT = 158 - 124 = 34 °C

Specific Heat Constant = c = 4.186 J/g °C

Required:

Specific Heat Capacity = Q = ?

Formula:

Q = mcΔT

Solution:

Q = (0.096)(4.186)(34)

Q = 13.7 Joules

[tex]\rule[225]{225}{2}[/tex]

Hope this helped!

Answer:

[tex]\Large \boxed{\sf 13600 \ J}[/tex]

Explanation:

Use formula

[tex]\displaystyle \sf Heat \ (J)=mass \ (g) \times specific \ heat \ capacity \ (Jg^{-1}\°C^{-1}) \times change \ in \ temperature \ (\°C)[/tex]

Specific heat capacity of water is 4.18 J/(g °C)

Substitute the values in formula and evaluate

[tex]\displaystyle \sf Heat \ (J)=96 \ g \times 4.18 \ Jg^{-1}\°C^{-1} \times (158\°C-124 \°C)[/tex]

[tex]\displaystyle Q=96 \times 4.18 \times (158-124 )=13643.52[/tex]

Answer:

–132.55 °C

Explanation:

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

Initial volume (V₁) = 30.8 L

Initial temperature (T₁) = –67 °C

Final volume (V₂) = 21.0 L

Final temperature (T₂) =?

Next, we shall convert –67 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T₁) = –67 °C

Initial temperature (T₁) = –67 °C + 273

Initial temperature (T₁) = 206 K

Next, we shall determine the final temperature of the gas. This can be obtained as follow:

Initial volume (V₁) = 30.8 L

Initial temperature (T₁) = 206 K

Final volume (V₂) = 21.0 L

Final temperature (T₂) =?

V₁/T₁ = V₂/T₂

30.8 / 206 = 21 / T₂

Cross multiply

30.8 × T₂ = 206 × 21

30.8 × T₂ = 4326

Divide both side by 30.8

T₂ = 4326 / 30.8

T₂ = 140.45 K

Finally, we shall convert 140.45 K to celsius temperature. This can be obtained as follow:

T(°C) = T(K) – 273

T₂ = 140.45 K

T₂ = 140.45 K – 273

T₂ = –132.55 °C

Thus, the new temperature of the gas is –132.55 °C

Answer:

Explanation:

First we convert the given masses of each element into moles, using their respective molar masses:

Now we divide them in order to calculate the molar ratio of S to Fe:

Meaning that for each 1 Fe mol, there's 1.5 S moles. We can write that as Fe₁S₁.₅

Finally we double those subscripts so that they become the lowest possible integers: Fe₂S₃.

Answer:

0.06moles/litres

Explanation:

molarity = no of moles / volume in litres = 3.00*10^-2/(450/1000)=0.06 moles/litres

Answer:

0.07mol/dm^3

Explanation:

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Answer:  An electrochemical cell that produces electricity from a spontaneous oxidation-reduction reaction

Explanation:

Galvanic cell is a device which is used for the conversion of the chemical energy produced in a redox reaction into the electrical energy.

Galvanic cell consists of two half cells known as anode and cathode. Oxidation takes place at anode and reduction takes place at cathode. The two half cells are connected by a salt bridge which contains an electrolyte which maintains the electrical neutrality.

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Answer:

C. Resource use by humans has gone up and down over the years is a false statement

Explanation:

In the graph you can make an observation on how the increase of human population is directly positive.

The resource use either has a positive or negative correlation, making it impossible for the usage to fluctuate over the years.

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Answer:

4.80×10³ years

Explanation:

Let the original amount (N₀) of ²²⁶Rn = 1 g

Therefore,

12.5% of the original amount = 12.5% × 1 = 12.5/100 × 1 = 0.125 g

Next, we shall determine the number of half-lives that has elapse. This can be obtained as follow:

Original amount (N₀) = 1

Amount remaining (N) = 0.125 g

Number of half-lives (n) =?

N = 1/2ⁿ × N₀

0.125 = 1/2ⁿ × 1

0.125 = 1/2ⁿ

Cross multiply

0.125 × 2ⁿ = 1

Divide both side by 0.125

2ⁿ = 1/0.125

2ⁿ = 8

Express 8 in index form with 2 as the base

2ⁿ = 2³

n = 3

Thus, 3 half-lives has elapsed.

Finally, we shall determine the time taken for only 12.5% of the original sample of ²²⁶Rn to remain.

This can be obtained as follow:

Half-life (t½) = 1.60×10³ years

Number of half-lives (n) = 3

Time (t) =?

t = n × t½

t = 3 × 1.60×10³

t = 4.80×10³ years.

Thus, it will take 4.80×10³ years for 12.5% of the original sample of ²²⁶Rn to remain.

Answer:

its a option neutralization takes place between acid and base plz mark me branliest

Answer:

1 mole = 28.0101

Hence,

82.4 = around 2.8 mole

Have a great day <3

Answer:

Explanation:

From the information given:

left half-cell = 3.00 mM     M(NO)

right half-cell = 3.00 M       M(NO)

Since the electrode are the same in both cells, then the concentration for the cell are also the same.

Negative electrode = Anode = lower concentration = 3.00 mM

Positive electrode = cathode = higher concentration = 3.00 M

Thus, right half cell will be postive electrode.

To determine the concentration cell:

[tex]Ecell =\Big( \dfrac{2.303\times R\times T}{nF} \Big)log\Big(\dfrac{[cathode]}{[anode]}\Big)[/tex]

SInce [Cathode] > [anode],

[tex]R = 8.314 J/K/mol, \\ \\ T = 273+20 = 293 K \\ \\ Faraday's constant (F)= 96500 C/mol[/tex]

n = 3

[tex]Ecell ={ \dfrac{2.303\times 8.314\times 293 }{(3\times 96500)}} log\dfrac{3}{0.003}[/tex]

[tex]\mathbf{E_{cell} = 0.0582 V } \\ \\ \mathbf{E_{cell} = 0.06 V}[/tex]

Answer: Igneous rocks may be simply classified according to their chemical/mineral composition as felsic, intermediate, mafic, and ultramafic, and by texture or grain size: intrusive rocks are course grained (all crystals are visible to the naked eye) while extrusive rocks may be fine-grained (microscopic crystals) or glass.

Explanation: Hope this helped! :)

Answer:

Igneous rocks may be simply classified according to their chemical/mineral composition as felsic, intermediate, mafic, and ultramafic, and by texture or grain size: intrusive rocks are course grained (all crystals are visible to the naked eye) while extrusive rocks may be fine-grained (microscopic crystals) or glass ( ...

Answer:

There are three types of hydropower facilities: impoundment, diversion, and pumped storage. Some hydropower plants use dams and some do not. The images below show both types of hydropower plants.

Explanation:

Explanation:

heyyyyyyyyy how are you sure

Answer: sorta

Explanation:

In fact, they do not flow at all: they simply vibrate back and forth. Because the atoms in a solid are so tightly packed, solid matter holds its shape and cannot be easily compressed

Answer:

True

Explanation:

A chemical change is otherwise known as a chemical reaction. A chemical change is characterized by the formation of new substances, evolution of heat and change in chemical composition of substances. The change in the composition of substances during a chemical change is the result of the making and breaking of chemical bonds during a chemical change.

A chemical change is not easily reversible because it involves changes in chemical composition of substances.

An example of a chemical change is the combination of sodium hydroxide and hydrochloric acid according to the reaction equation;

HCl(aq) + NaOH (aq) -----> NaCl(aq) + H2O (l)

We easily notice the change in the chemical composition of each of the substances involved in the reaction.

The equilibrium concentration of N₂O₄, given that the concentration of NO₂ is 1.78×10⁻² is 5.42×10⁻²

Keq = [Product] / [Reactant]

Keq = [NO₂]² / [N₂O₄]

5.85×10⁻³ = [1.78×10⁻²]² / [N₂O₄]

Cross multiply

5.85×10⁻³ × [N₂O₄] = [1.78×10⁻²]²

Divide both sides by 5.85×10⁻³

[N₂O₄] = [1.78×10⁻²]² / 5.85×10⁻³

[N₂O₄] = 5.42×10⁻²

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Answer:

First answer is "The equilibrium lies to the left", and the second one is 5.42 x 10^-2 M.

Explanation:

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