1°C if the pressure is 149.3 kPa? How many moles of hydrogen sulfide H2S are contained in a 327.3 mL bulb at 48.

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]

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:

Hope i helped yall :D

Answer:

peace of vassals

Explanation:

I think i spelled it worng

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]

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:

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: The molarity of [tex]Na_2S_2O_3[/tex] is 0.108 M

Explanation:

[tex]KIO_3+5KI+3H_2SO_4\rightarrow 3K_2SO_4+3H_2O+3I_2[/tex]

[tex]2Na_2S_2O_3+I_2\rightarrow Na_2S_4O_6+2NaI[/tex]

To calculate the number of moles for given molarity, we use the equation:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]    

[tex]\text{Moles of }KIO_3=\frac{0.0131mol/L\times 21.55}{1000}=2.8\times 10^{-4}mol[/tex]

1 mole of [tex]KIO_3[/tex]  produces = 3 moles of   [tex]I_2[/tex]

[tex]2.8\times 10^{-4}[/tex] moles of [tex]KIO_3[/tex] produces = [tex]\frac{3}{1}\times 2.8\times 10^{-4}=8.4\times 10^{-4}[/tex] moles of [tex]I_2[/tex]  

Now 1 mole of [tex]I_2[/tex] uses = 2 moles of [tex]Na_2S_2O_3[/tex]

[tex]8.4\times 10^{-4}[/tex] moles of [tex]I_2[/tex] uses =  [tex]\frac{2}{1}\times 8.4\times 10^{-4}=1.69\times 10^{-3}[/tex]  moles of [tex]Na_2S_2O_3[/tex]

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution in L}}=\frac{1.69\times 10^{-3}\times 1000}{15.65}=0.108M[/tex]

The molarity of [tex]Na_2S_2O_3[/tex] is 0.108 M  

Answer:

C) Mercury based thermometer

Explanation:

Answer:

Mercury based thermometer

Explanation:

When you touch a lightbulb it is warm. Hairdryer produces heat to dry your hair. Stove heats up to warm food and liquids up.

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:

i dont know

Explanation:

its simple i dont know

Answer:

Explanation:Red

Answer:

2,1,2

1,2,1,1

the answer

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:

Answer:

Approximately [tex]279\; \rm mg[/tex] of fluorine would be produced.

Explanation:

Look up the relative atomic mass of these calcium and fluorine on a modern periodic table:

In other words, the mass of each mole of calcium atoms is (approximately) [tex]40.078\; \rm g[/tex]. Similarly, the mass of each mole of fluorine atoms is (approximately) [tex]18.998\; \rm g[/tex].

Calculate the number of moles of calcium atom that were produced:

[tex]\begin{aligned} &n({\rm Ca}) \\ &= \frac{m({\rm Ca})}{M({\rm Ca})} \\ &= \frac{294\; \rm mg}{40.078\; \rm g \cdot mol^{-1}} \\ &= \frac{0.294\; \rm mg}{40.08 \; \rm g \cdot mol^{-1}}\\ & \approx 7.3357 \times 10^{-3}\; \rm mol\end{aligned}[/tex].

Every formula unit of calcium fluoride, [tex]\rm CaF_2[/tex], contains twice as many fluorine atoms as calcium atoms.

Hence, if the sample contained approximately [tex]7.33570 \times 10^{-3}\; \rm mol[/tex] calcium atoms, it would contain twice as many fluorine atoms:

[tex]\begin{aligned}n({\rm F}) &= 2\, n({\rm Ca}) \\ &\approx 2 \times 7.33570\times 10^{-3}\; \rm mol \\ &\approx 1.4671 \times 10^{-2}\; \rm mol\end{aligned}[/tex].

Calculate the mass of that many fluorine atoms:

[tex]\begin{aligned}& m({\rm F}) \\ &= n({\rm F}) \cdot M({\rm F}) \\ &\approx 1.4671 \times 10^{-2}\; \rm mol \times 18.998\; \rm g \cdot mol^{-1} \\ &\approx 278.72 \times 10^{-3}\; \rm g \\&\approx 279\; \rm mg\end{aligned}[/tex].

279 mg of Fluorine was formed in the decomposition reaction

The equation of the reaction is;

CaF2 -----> Ca + 2F

Mass of calcium formed = 294 mg

Number of moles of Ca in 294 mg = 294 × [tex]10^-3[/tex] g/40 g/mol = 0.00735 moles

From the equation, the products are in ration 1 : 2

Since 1 mole of Ca is to 2 moles of F

0.00735 moles of Ca is to 0.00735 moles × 2 moles/1 mole

= 0.0147 moles of F

Mass of F = 0.0147 moles × 19 g/mol

= 0.279g or 279 mg of F

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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:

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

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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:

Fractional distillation of crude oil in laboratory

Explanation:

In general, we can sort them according to their boiling point.

We should first place a ball of mineral wool soaked with petroleum into a boiling tube and heat it with a Bunsen flame. At the beginning of the experiment, heat the mineral with a gentle flame, and then more strongly later. The fraction with a lower boiling range will distil out first. Since petrol is quite volatile, it should be condensed by cooling with water or an ice-water bath.

Petrol will be collected first, as it has a lower boiling point than kerosene. It will turn into vapour and condense again in the receive tub in the beaker.

Answer:

Splits into

Explanation:

Because fission reaction is divided into two or more pieces.

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:

_c+_so2-_cs2+_co= Toxic

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:

294.87 gm  CaCl_2

Explanation:

The computation of the  mass of calcium chloride is shown below:

But before that following calculations need to be done

Number of moles of chlorine atom is

= 3.20 × 10^24 ÷ 6.022 × 10^23

= 5.314 moles

As we know that

1 mole CaCl_2 have the 2 moles of chlorine atoms

Now 5.341 mole chloride atoms would be

= 1 ÷ 2 × 5.314

= 2.657 moles

Now

Mass of CaCl_2 = Number of moles × molar mass of  CaCl_2

= 2.657 moles × 110.98 g/mol

= 294.87 gm  CaCl_2

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:

Please mark as brainliest

Answer:

0.75mol/LNaOH

Explanation:

Data: Soln:

M= ? M= n/ V(l)

n= 3moles. M= 3mol/ 4L

V= 4L. M = 0.75 mol/LNaOH

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:

The readings of equilibrium constant is incorrect.

Explanation:

There is a mistake in the value of equilibrium constant if the burette contains traces of distilled water because the water will be present in the solution from which equilibrium constant is calculated. The spectrophotometer did not give the correct reading because its calibration is incorrect. If we want to gain correct readings from the spectrophotometer, we must have to calibrate the instrument correctly.

Incorrect value of equilibrium constant is obtained.

Equilibrium constant of any reaction tells about the relative ration of the concentration of products to the reactants.

Value of equilibrium constant can be calculated in the aqueous medium where concentration of water also present. If we don't rinse the burette with the reaction solution then the already present traces of water affect the concentration of water during the calculation of equilibrium constant.

As the absorbance of any solution is also directly proportional to the concentration, so if the spectrometer is not calibrated accurately then incorrect value of absorbance also affects the value of equilibrium constant.

Hence, we get incorrect value of equilibrium constant for each of the given mistakes.

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