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Answer:

The answer is "[tex]\bold{4.97 \times 10^{-2}}[/tex]"

Explanation:

Please find the complete question in the attached file.

Equation:

[tex]2SO_2+O_2 \leftrightharpoons 2SO_3[/tex]

at [tex]t=0 3.3 \ \ \ \ \ \ \ \ \ \ 0.79[/tex]

at equilibrium [tex]3.3-p \ \ \ \ \ \ \ \ \ \ 0.79 - \frac{P}{2} \ \ \ \ \ \ \ \ \ \ \ \ P[/tex]

[tex]p= 0.47 \ \ atm\\\\SO_2=3.3-0.47 = 2.83 \ \ atm\\\\O_2= 0.74 -\frac{0.47}{2}=0.74-0.235=0.555 \ atm\\\\K_P=\frac{[PSO_3]^2}{[PSO_2]^2[PO_2]}\\\\[/tex]

     [tex]=\frac{0.47^2}{2.83^2\times 0.555}\\\\=4.97 \times 10^{-2}[/tex]

Answer:

12.26%

Explanation:

MM K = 39.1 g/mol

MM K2CO3 . 10H2O = 2x39.1 + 12.01 + 3 x 15.99 + 10 x 18.02= 318.38 g/mol

% by weight = (MM K/MM K2CO3.10H2O ) x 100= (39.1/318.8)  x 100=12.26 %

The percent composition by weight of potassium in potassium carbonate decahydrate is 13.98%

Let's represent the chemical formula of the compound as follows;

potassium carbonate decahydrate = CH₂0KO₁₃

The molecular mass of the compound can be calculated as follows:

mass of potassium = 39 g

Therefore,

% mass of potassium = 39 / 279 × 100

% mass of potassium = 3900 / 279

% mass of potassium = 13.9784946237

% mass of potassium = 13.98 %

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Answer: arterioles

Explanation:

Answer:

The mass of 0.063*10⁻⁴ moles of aluminum sulphate  is 2.15*10⁻³ grams.

Explanation:

Aluminum sulfate Al₂(SO₄)₃ has a molar mass of 342.15 g/mol.

Molar mass is the amount of mass that a substance contains in one mole of a substance, which can be an element or a compound.

So in this case you can apply the following rule of three: if 342.15 grams are present in 1 mole of aluminum sulfate, how much mass is present in 0.063*10⁻⁴ moles of the compound?

[tex]mass of aluminum sulphate=\frac{0.063*10^{-4}moles*342.15 grams }{1 mole}[/tex]

mass of aluminum sulphate= 2.15*10⁻³ grams

The mass of 0.063*10⁻⁴ moles of aluminum sulphate  is 2.15*10⁻³ grams.

Answer:

letting the car fall down the ramp i got that right.

Explanation:

Answer:

T = 4.062V

Explanation:

from PV = nRT => T = PV/RT

P = 1 atm

V = Final Volume

n = 3 moles

R = 0.08206 L·atm/mol·K

T = ?

T = 1 atm · V(Liters)/(3 moles)(0.08206L·atm/mol·K) = 4.062·V(final) Kelvin

The temperature of the balloon is 406 K

We'll begin by listing out what was given from the question. This is shown below:

Volume (V) = 100 L

Mole of oxygen (n) = 3 moles

Pressure (P) = 1 atm

We can obtain the temperature of the balloon by using the ideal gas equation as shown below:

P is the pressure.

V is the volume.

n is the number of mole

R is the gas constant (0.0821 atm.L/Kmol)

T is the temperature.

Applying the ideal gas equation, we have:

PV = nRT

1 × 100 = 3 × 0.0821 × T

100 = 0.2463 × T

Divide both side by 0.2463

T = 100 / 0.2463

Therefore, the temperature of the balloon is 406 K

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Answer:

32.7

Explanation:

I just did it and got it right

Answer:

A liquid that evaporates rapidly at low temperatures

Explanation:

Answer:

A liquid that evaporates rapidly at low temperatures.

Explanation:

A volatile liquid such as gasoline evaporates easily. Its chemicals aren't held together as strongly, and room temperature is often enough to break intermolecular bonds and evaporate them. 

PF

Answer:

V₁ = 0.025 Liters = 25 ml

Explanation:

P₁ = 6 Atm      P₂ = 7800mm/760mm/Atm = 0.01 Atm

V₁ = ?              V₂ = 29 Liters

T₁ = 115 K        T₂ = 225 K

P₁V₁/T₁ = P₂V₂/T₂ => V₁ = P₂V₂T₁/P₁T₂

V₁ = (0.01 Atm)(29L)(115K)/(6 Atm)(225K) = 0.025 Liters = 25ml

The question is incomplete, the complete question is;

We obviously cannot wait for 4 million years to determine the half-life of technetium, or even 87.7 years to determine the half-life of plutonium. The half-life of a substance is determined using a Geiger counter, which is essentially a tube filled with an inert gas. When a particle decays and produces radiation, it briefly ionizes the gas and produces a detectable electric pulse, which is amplified and counted by a detector (before digital displays were available, you would hear clicks and see a needle register on a dial). Determine the formula for the half-life of a substance with an initial count of No pulses and a count of N1 pulses M minutes later.

Answer:

M1/2 = 0.693/k

Explanation:

Given that the initial count rate is No pulses

The count rate after M minutes is N1

Then from;

A = Aoe^-kt

A= N1

Ao=No

I can now write;

N1 = Noe^-kM

The half life is the time taken for the activity of the radioactive nuclide to reach half its initial value. Hence;

N1/No =e^-kM = 1/2

Taking natural logarithm of both sides as shown below;

-kM1/2 = ln(1/2)

M1/2 = -  (ln(1/2)/k)

M1/2 = 0.693/k

Answer:

C. 21%oxygen

Explanation:

Dry air contains 78.09% nitrogen, 20.95%oxygen, 0.93% argon, 0.04% carbon dioxide, and small amounts of other gases.

Answer:

molar mass of H2CO3 = 72.0228

Answer:

62.02 g/mol

Explanation:

Took the test and got it right. if you have enough faith in brainly youll choose the right answer

Answer:

Explanation:

The formula for determining the energy of state [tex]m_l[/tex] can be computed by using the formula:

[tex]Em_i = \dfrac{m_1^2h^2}{2I}[/tex]

Also, the momentum is:

[tex]l_2 = m_i h[/tex]

There are 22 electrons with two electrons in each of the lowest II energy levels so that the highest occupied states are [tex]m_1 = \pm 5[/tex]

The moment of inertia of an electron on a ring of radius 440 ppm is:

[tex]I = mR^2 \\ \\ = 9.109 \times 10^{31} \ kg (440 \times 10^{-12}) ^2 \\ \\ = 1.76 \times 10^{-49} \ kgm^2[/tex]

[tex]E_{\pm 5 } = \dfrac{25h^2}{2I} \\ \\ = \mathbf{7.89 \times 10^{-19} \ J}[/tex]

The angular  momentum is:

[tex]l_2 = \pm 5h \\ \\ = \pm 5 \tiems ( 1.05457 \times 10^{-34} \ Js)[/tex]

[tex]= \mathbf{5.275 \times 10^{-34} \ Js}[/tex]

B) Let's recall that:

The lowest occupied energy level is [tex]m_1 = \pm 6[/tex] which implies that the energy  [tex]E_{\pm 6}= 1.14 \times 10^{-18} \ J[/tex]

Thus;

[tex]\Delta E = E_{\pm 6} - E_{\pm 5} \\ \\ = 1.14 \times 10^{-18 \ J} - 0.79 \times 10^{-18} \ J \\ \\ = 0.35 \times 10^{-18} \ J \\ \\ 0.35 \times 10^{-18} \ J = hv \\ \\ 0.35 \times 10^{-18} \ J = h \dfrac{c}{\lambda}[/tex]

Hence, the radiation which would induce a transition that relates to the wavelength of about 570nm, a wavelength of visible light.

Answer:

The Florian predict the metamorphic rocks because it changes when subjected to heat and pressure.

Explanation:

i hope this helps

Answer:

632 g

Explanation:

First we convert 579 g of CO₂ into moles, using its molar mass:

Then we convert CO₂ moles into O₂ moles, using the stoichiometric coefficients:

Finally we convert O₂ moles into grams, using its molar mass:

Answer:

There are two dots are present on the left side of calcium atom which represents the unpaired electrons present in the outermost shell. This structure is presented by Lewis in order to show the number of unpaired electrons in the atom.

Explanation:

Answer: 22.3 *10^23 S atoms

Explanation:

Answer:

513.975g

Explanation:

I'm assuming that the question has Aluminium and Chlorine as the question is not clear.

1. Balance the equation

  2 Al + 3 Cl2  → 2AlCl3

2. Use the mole = mass/mr equation and molar ratio to find the mass of AlCl3

   133.5 x 3.85 = 513.975g

Answer: Polarity, Boiling and Freezing Points, Specific Heat Capacity, Density, Surface Tension, Heat of Vaporization and Vapor Pressure, Capillary Action, Solid State (Ice), Liquid State (Liquid Water), Gas State (Steam),  

Explanation:  Water molecules are polar, so they form hydrogen bonds. This gives water unique properties, such as a relatively high boiling point, high specific heat, cohesion, adhesion, and density.

Answer:

Explanation:

Lithium is in group 1 so there's 1 outermost shell electron.

Oxygen is in group 6 so there's 6 outermost shell electron.

To react lithium with oxygen, we need 2 lithium and 1 oxygen

Each lithium transfer 1 electron to the oxygen and oxygen gain 2 electrons.

[tex]Li_{2}O\\[/tex] formed.

If there's 4 lithium oxide, 4 molecules will be there.

If you mean how many ATOMS are there, there will be 3×4=12 atoms.

Answer:

[tex]5.77x10^{47}particles.[/tex]

Explanation:

Hello!

In this case, since we are given 9.58x10²³ moles of potassium sulfate, it is possible to compute the particles according to the Avogadro's number:

6.022x10²³

As shown below:

[tex]9.58x10^{23}mol*\frac{6.022x10^{23}}{1mol}\\\\5.77x10^{47}particles.[/tex]

Best regards!

Answer:

Kc = 1.09x10⁻⁴

Explanation:

HF = 1.62g

H₂O = 516g

F⁻ = 0.163g

H₃O⁺ = 0.110g

To solve this question we need to find the moles of each reactant in order to solve the molar concentration of each reactan and replacing in the Kc expression. For the reaction, the Kc is:

Kc = [H₃O⁺] [F⁻] / [HF]

Because Kc is defined as the ratio between concentrations of products over reactants powered to its reaction coefficient. Pure liquids as water are not taken into account in Kc expression:

[H₃O⁺] = 0.110g * (1mol /19.01g) = 0.00579moles / 5.6L = 1.03x10⁻³M

[F⁻] = 0.163g * (1mol /19.0g) = 0.00858moles / 5.6L = 1.53x10⁻³M

[HF] = 1.62g * (1mol /20g) = 0.081moles / 5.6L = 0.0145M

Kc = [1.03x10⁻³M] [1.53x10⁻³M] / [0.0145M]

Answer:

should be D. Hope this helpsss!

Answer:

3. 116.5 V

4. 119.6 V

Explanation:

3. Determination of the voltage.

Resistance (R) = 25 Ω

Current (I) = 4.66 A

Voltage (V) =?

V = IR

V = 4.66 × 25

V = 116.5 V

Thus, the voltage is 116.5 V

4. Determination of the voltage.

Current (I) = 9.80 A

Resistance (R) = 12.2 Ω

Voltage (V) =?

V = IR

V = 9.80 × 12.2

V = 119.6 V

Thus, the voltage is 119.6 V