I2(g) + Br2(g) ↔ 2 IBr(g) What is the equilibrium concentration of IBr (in M to 4 decimal places) if 0.200 mol of I2 and 0.100 mol Br2 in a 5.00 L vessel if Kc = 29 at 150°C?

Answer:

5 g

Explanation:

From the question, we have,

Molarity of FeIII solution= 0.245 M

Volume of solution = 125 ml

From

number of moles= concentration × volume

We have;

Number of moles= 0.245 M × 125/1000

Number of moles = 0.031 moles

Molar mass of Fe III = 162.5g/moles

Mass of iron III = number of moles× molar mass = 0.031 × 162.5= 5 g

n

Answer: It’s D

Explanation:

Cuz I said so

Answer: d. Shearing

Explanation:

Answer:

3

Explanation:

I wish you the best, its three or 2

Answer:

2

Explanation:

Edge 2021

Answer:

C₃H₆

Explanation:

We'll begin by calculating the empirical formula for cyclopropane. This is illustrated below:

Carbon (C) = 85.7%

Hydrogen (H) = 14.3%

Divide by their molar mass

C = 85.7/12 = 7.14

H = 14.3/1 = 14.3

Divide by the smallest:

C = 7.14/7.14 = 1

H = 14.3/7.14 = 2

Therefore, the empirical formula for the cyclopropane is CH₂

Next, we shall determine the number of mole of cyclopropane.

This can be obtained as follow:

Temperature (T) = 50 °C = 50 °C + 273 = 323 K

Pressure (P) = 0.984 atm

Volume (V) = 1 L

Gas constant (R) = 0.0821 atm.L/Kmol

Number of mole (n) of cyclopropane =.?

PV = nRT

0.984 × 1 = n × 0.0821 × 323

Divide both side by 0.0821 × 323

n = 0.984 / (0.0821 × 323)

n = 0.0371 mole

Next, we shall determine the molar mass of cyclopropane. This can be obtained as follow:

Number of mole of cyclopropane = 0.0371 mole

Mass of cyclopropane = 1.56 g

Molar mass of cyclopropane =.?

Mole = mass /Molar mass

0.0371 = 1.56/Molar mass

Cross multiply

0.0371 × Molar mass = 1.56

Divide both side by 0.0371

Molar mass = 1.56 /0.0371

Molar mass of cyclopropane = 42.05 g/mol.

Finally, we shall determine the molecular formula for cyclopropane. This can be obtained as follow:

[CH₂]ₙ = 42.05

[12 + (2×1)]ₙ = 42.05

[12 + 2]ₙ = 42.05

14n = 42.05

Divide both side by 14

n = 42.05/ 14

n = 3

Thus,

[CH₂]ₙ => [CH₂]₃ => C₃H₆

Therefore, the molecular formula for cyclopropane is C₃H₆

The molecular formula of cyclopropane ([tex]CH_2[/tex]) is equal to [tex]C_3H_6[/tex]

Given the following data:

Scientific data:

To determine the molecular formula of cyclopropane:

First of all, we would determine the molar mass and number of moles of cyclopropane by using the ideal gas law equation;

[tex]PV=\frac{M}{MM} RT[/tex]

Where;

Making MM the subject of formula, we have:

[tex]MM = \frac{MRT}{PV} \\\\MM = \frac{1.56\;\times \;0.0821\times \;323}{0.984\;\times \;1}\\\\MM = \frac{41.3686}{0.984}[/tex]

Molar mass, MM = 42.04 g/mol.

For number of moles:

[tex]Number\;of\;moles = \frac{mass}{molar\;mass}\\\\Number\;of\;moles = \frac{1.56}{42.04}[/tex]

Number of moles = 0.0371 moles

Next, we would determine the empirical formula of cyclopropane:

For carbon (C):

[tex]C = \frac{85.7}{12}[/tex]

Carbon (C) = 7.1417

For hydrogen (H):

[tex]H = \frac{14.3}{1}[/tex]

Hydrogen (H) = 14.3

Simplest whole number ratio:

[tex]C = \frac{7.1417 }{7.1417 } =1[/tex]

[tex]H= \frac{14.3 }{7.1417 } =2[/tex]

Empirical formula of cyclopropane = [tex]CH_2[/tex]

Now, we can determine the molecular formula of cyclopropane:

[tex](CH_2)n = 42.04\\\\(12 + 1\times2)n=42.04\\\\14n=42.04\\\\n=\frac{42.04}{14}[/tex]

n = 3

Molecular formula of cyclopropane ([tex]CH_2[/tex]) = [tex](CH_2)_3 =C_3H_6[/tex]

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The pressure can be increased by a factor of 3 by decreasing the volume by a factor of 3.

Given that the temperature of the system is held constant. So it implies that it is an isothermal process.

Now we know that the ideal gas equation is given by:

PV = nRT

where, P is the pressure

V is the volume

n is the number of moles

R is the gas constant, and

T is the temperature

Assuming n is constant, R is universal as constant, so if T is also constant, then:

PV = constant

So if P becomes 3P. that is the pressure increased by a factor of 3, then V must become V/3, so that: (3P)(V/3) = PV = constant.

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Answer:

The ΔH for the reaction is -456.5 KJ

Explanation:

Here we want to determine ΔH for the reaction;

Mathematically;

ΔH = ΔH(product) - ΔH(reactant)

In the case of the first reaction;

ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3)  ...........................(*)

From the other reactions, we can get the respective ΔH for the individual molecule in the reaction

In second reaction;

Kindly note that for elements, molecule of gases, ΔH = 0

What this means is that throughout the solution;

ΔH(Ca)  = 0 KJ

ΔH(O2) = 0 KJ

ΔH(C) = 0 KJ

Thus, in writing the equation for the subsequent chemical reactions, we shall need to write and equate the overall ΔH for the reaction to that of the product alone

So in the second reaction

ΔH = 2ΔH(CaCO3)

Thus;

-2414/2 = ΔH(CaCO3)

ΔH(CaCO3) = -1,207  KJ

Moving to the third reaction, we have;

ΔH = ΔH(CO2)

Hence ΔH(CO2) = -393.5 KJ

For the last reaction;

ΔH = ΔH(CaO)

Hence ΔH(CaO) = -1270 KJ

Going back to equation *

ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3)

Using the values of the ΔH  of the respective molecules given above,

ΔH  = -1270 + (-393.5) - (-1207)

ΔH  = -456.5 KJ

Answer:

XH₂ = 0.4885

XN₂ = 0.4290

XAr = 0.0825

Explanation:

Step 1: Given data

Step 2: Calculate the total pressure (P)

The total pressure is equal to the sum of the partial pressures of all the gases.

P = pH₂ + pN₂ + pAr = 441.0 Torr + 387.3 Torr + 74.5 Torr = 902.8 Torr

Step 3: Calculate the mole fraction (X) of each gas

We will use the following expression.

Xi = pi / P

where,

Xi: mole fraction of the gas i

pi: partial pressure of the gas i

P: total pressure

XH₂ = pH₂ / P = 441.0 Torr / 902.8 Torr = 0.4885

XN₂ = pN₂ / P = 387.3 Torr / 902.8 Torr = 0.4290

XAr = pAr / P = 74.5 Torr / 902.8 Torr = 0.0825

Answer:

O2

Explanation:

Diatomic molecules are molecules that have 2 of either H,N,F,O,I,Cl, or Br. In this case, 2H is not correct because the 2 is in front of the element name. Diatomic molecules will have the two after which means that the molecule has these two elements together.

Answer:

Explanation:

a) O₂(g) + 2H₂(g) = 2H₂O

b) H₂O(l) = H₂O(s)

c) NH₃(g) + HCl(g) =  NH₄Cl(g)

d) 2NH₄NO₃(s) =  2N₂(g) + O₂(g) + 4H₂O(g)

e) N₂O₄(g) =  2NO₂(g)

ΔS is positive when there is increase in disorderliness. It happens when there is increase in volume .

Increase is volume is maximus in the following reaction.

d) 2NH₄NO₃(s) =  2N₂(g) + O₂(g) + 4H₂O(g)

in this reaction solid NH₄NO₃ is changed to 7 x 22.4 L of gases so there is maximum increase in volume . Hence maximum increase in entropy . Hence ΔS is most positive .

Answer:

1) easy compressed

2) fills its container

3) far more space

Explanation:

Answer:c

Explanation: more expensive so people use nonrenewed energy more.

Answer:

Will be more acidic

Explanation:

The equilibrium of NH3 in water is:

NH3(aq) + H2O(l) ⇄ NH4⁺(aq) + OH⁻(aq).

Where equilibrium constant, Kb, is:

Kb = 1.85x10⁻⁵ = [NH4⁺] [OH⁻] / [NH3]

From 0.10M NH3, the reaction will produce X of NH4⁺ and X of OH⁻ and Kb will be:

1.85x10⁻⁵ = [X] [X] / [0.10M]

1.8x10⁻⁶ = X²

X = 1.34x10⁻³ = [OH⁻]

As pOH = -log[OH⁻] = 2.87

And as pH = 14 - pOH

pH of the 0.10M NH3 is 11.13

Now, to find the pH of the NH4Cl and NH3 we need to use H-H equation for bases:

pOH = pKb + log [NH4⁺] / [NH3]

Where pKb is -log Kb = 4.74 and [] are moles of both compounds.

Moles of [NH4⁺] = [NH3] = 60mL, 0.060L*0.22M = 0.0132moles:

pOH = 4.74 + log [0.0132] / [0.0132]

pOH = 4.74

pH = 14 - 4.74 = 9.26

That means the pH of the resulting solution will be more acidic

Answer: B

Explanation: a calcium ion has a charge of +2 because it has 2 more protons than electrons giving it a positive charge instead of neutral or negative.

Answer:

K = 0.0505

Explanation:

Based on the equilibrium:

N2O3 ⇄ NO2 + NO

K, equilibrium constant, is defined as:

K = [NO2] [NO] / [N2O3]

Where [] are the equilibrium concentration of each species in the mixture.

The initial molarity of N2O3 is:

1.191mol / 2.00L = 0.5955M

In equilibrium, 0.5955M of N2O3 reacts producing X Molar of NO2 and X Molar of NO:

[N2O3] = 0.5955M - X

[NO2] = X

[NO] = X

As equilibrium concentration of NO2 is 0.300mol/2.00L = 0.15M; X = 0.15M:

[N2O3] = 0.5955M - 0.15M = 0.4455M

[NO2] = 0.15M

[NO] = 0.15M

And K is:

K = [0.15M] [0.15M] / [0.4455M]

Answer:

The moon blocks the Sun's light from hitting the surface of the earth

Explanation:

Answer:

B. are electrical signals.

Answer:

The energy of each transition is approximately [tex]1.98\times 10^{-19}\; \rm J[/tex].

The frequency of photons released in such transitions is approximately [tex]3.00\times 10^{14}\; \rm Hz[/tex].

Explanation:

The Rydberg Equation gives the wavelength (in vacuum) of photons released when the electron of a hydrogen atom transitions from one main energy level to a lower one.

The Rydberg Equation gives the following relation:

[tex]\displaystyle \frac{1}{\lambda_\text{vac}} = R_\text{H} \cdot \left(\frac{1}{{n_2}^2}} -\frac{1}{{n_1}^2}\right)[/tex].

Rearrange to obtain and expression for [tex]\lambda_\text{vac}[/tex]:

[tex]\displaystyle \lambda_\text{vac} = \frac{1}{\displaystyle R_\text{H}\cdot \left(\frac{1}{{n_2}^2} - \frac{1}{{n_1}^2}\right)}[/tex].

In this question, [tex]n_1 = 7[/tex] while [tex]n_2 = 3[/tex]. Therefore:

[tex]\begin{aligned} \lambda_\text{vac} &= \frac{1}{\displaystyle R_\text{H}\cdot \left(\frac{1}{{n_2}^2} - \frac{1}{{n_1}^2}\right)} \\ &\approx \frac{1}{\displaystyle 1.09678 \times 10^{7}\; \rm m^{-1} \cdot \left(\frac{1}{3^2} - \frac{1}{7^2}\right)} \approx 1.0 \times 10^{-6}\; \rm m \end{aligned}[/tex].

Note, that [tex]1.0\times 10^{-6}\; \rm m[/tex] is equivalent to [tex]1000\; \rm nm[/tex]. That is: [tex]1.0\times 10^{-6}\; \rm m = 1000\; \rm nm[/tex].

Look up the speed of light in vacuum: [tex]c \approx 3.00\times 10^{8}\; \rm m \cdot s^{-1}[/tex]. Calculate the frequency of this photon:

[tex]\begin{aligned} f &= \frac{c}{\lambda_\text{vac}} \\ &\approx \frac{3.00\times 10^{8}\; \rm m\cdot s^{-1}}{1.0\times 10^{-6}\; \rm m} \approx 3.00 \times 10^{14}\; \rm Hz\end{aligned}[/tex].

Let [tex]h[/tex] represent Planck constant. The energy of a photon of wavelength [tex]f[/tex] would be [tex]E = h \cdot f[/tex].

Look up the Planck constant: [tex]h \approx 6.62607 \times 10^{-34}\; \rm J \cdot s[/tex]. With a frequency of [tex]3.00\times 10^{14}\; \rm Hz[/tex] ([tex]1\; \rm Hz = 1\; \rm s^{-1}[/tex],) the energy of each photon released in this transition would be:

[tex]\begin{aligned}E &= h \cdot f \\ &\approx 6.62607 \times 10^{-34}\; \rm J\cdot s^{-1} \times 3.00 \times 10^{14}\; \rm s^{-1} \\ &\approx 1.98 \times 10^{-19}\; \rm J\end{aligned}[/tex].

The energy of the transition between n = 7 and n = 3  is 1.96 × 10^-19 J while the frequency is 3 × 10^14 Hz.

Using the Rydberg Equation for energy;

ΔE = -RH(1/n^2final - 1/n^2initial)

Given that;

nfinal = 3

ninitial = 7

RH = 2.18 × 10^-18 J

ΔE = - 2.18 × 10^-18(1/3^2 - 1/7^2)

ΔE = - 2.18 × 10^-18(0.11 - 0.02)

ΔE = - 1.96 × 10^-19 J

For the second part;

Since the wavelength is 1000nm, we have;

λ = 1000nm

c = 3 × 10^8 m/s

f = ?

c = λf

f = c/λ

f = 3 × 10^8 m/s/1000 × 10^-9 m

f = 3 × 10^8 m/s/ 1 × 10^-6 m

f = 3 × 10^14 Hz

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Answer:

A) enthalpy of reaction

Explanation:

The region C signifies the enthalpy of reaction.

This diagram is the energy profile of an endothermic reaction. In such reaction, heat is absorbed from the surrounding. At the end of the reaction, the heat of product is lesser than that of the reactants.

it is indeed A) enthalpy of reaction

Answer:

See explanation

Explanation:

A drying tube prevents moisture from contaminating the reactants. The drying tube contains CaCl2 a hygroscopic material whose function is to absorb the moisture so that it does not react with the Grignard reagent. Without the desiccating action of CaCl2, moisture will enter the reaction chamber,contaminating the reactants.

If water reacts with the Grignard reagent, an alkane is formed. The mechanism of this reaction is shown in the image attached. R here represents the alkyl moiety of the Grignard reagent.

Answer:

C. Hydrosphere for 7

B. Biosphere for 8

Explanation: I am not sure about the rest of them.

Answer:

0.150 M

Explanation:

The molarity of the solution can be calculated using the formula;

C1V1 = C2V2

Where; C1 = initial concentration of solution

C2 = final concentration of solution

V1 = volume of initial solution

V2 = volume of final solution

According to this question, C1 = 0.200M, C2 = ?, V1 = 75mL, V2 = 100.0mL

Hence,

C1V1 = C2V2

0.200 × 75 = C2 × 100

15 = 100C2

C2 = 15/100

C2 = 0.150M

Therefore, the molarity of the final KCl solution is 0.150M

Answer: the type of cup

Explanation: the independent variable is what you change in your experiment, to find out the different results, in this experiment the factor that they are changing in order to see which outcome is best, is the type of cup.

Answer:

NH₂O

Explanation:

Given compound:

       N₄H₈O₄

Unknown:

The empirical formula of the compound  = ?

Solution:

The empirical formula of a compound is its simplest formula. It expresses the composition of a the compound in the simplest whole ratio of atoms of the different elements present in the compound.

 For the given compound:

                                  N₄H₈O₄

  Number of moles of N = 4

                                      H = 8

                                      O = 4

 the highest common factor is 4 and we simply divide through by this number;

                                      N = 1

                                      H = 2

                                      O = 1

So, the empirical formula of compound is NH₂O

Answer:

i wish i could help but t know either so i wish you luck for an answer once again truly sorry don'

Explanation:

i am sorry but i do not know :(

Answer:

The actual concentration of the hydrochloric acid is 1.2M

Explanation:

The formula to be used here is that of concentration (popularly used during titration);

CₐVₐ/CbVb =nₐ/nb

where Cₐ is the concentration of acid (supposed to be 1.0 M but unsure)

Vₐ is the volume of acid (100 ml)

Cb is the concentration of base (1.3 M)

Vb is the volume of base (89.13 ml)

nₐ is the volume of acid

nb is the volume of base

The equation for the reaction described in the question is

HCl + NaOH ⇒ NaCl + H₂O

we can see from the above equation the ratio of the number of moles for both the acid and the base is 1:1

Thus;

Cₐ × 100/1.3 × 89.13 =1/1

Cₐ = 1.3 × 89.13/100

Cₐ = 1.2M

The actual concentration of the hydrochloric acid is 1.2M

Answer:

1.63cm

Explanation:

Given parameters:

Density of silicon = 2.13g/cm³

Mass of Cr₂O₃ = 88.3g

Width of the block = 2.65cm

Length of the block  = 2.65cm

Unknown:

Thickness of the block  = ?

Solution:

Reaction equation:

          3Si   +   2Cr₂O₃   →  3SiO₂ + 4Cr

Let us find the mass of Si,

   So,

    Number of moles  = [tex]\frac{mass}{molar mass}[/tex]

  we find the number of moles of Cr₂O₃ ;

Molar mass of Cr₂O₃ = 2(52) + 3(16) = 152g/mol

  Number of moles of Cr₂O₃ = [tex]\frac{88.3}{152}[/tex]   = 0.58mole

Then;

  we know that;

              2 moles of Cr₂O₃ reacted with 3 moles of Cr₂O₃

              0.58moles of Cr₂O₃ will react with [tex]\frac{0.58 x 3}{2}[/tex]   = 0.87mole

So,

    Mass of Si  = number of moles x molar mass

          molar mass of Si = 28g/mol

   Mass of Si  = 0.87 x 28  = 24.4g

Since density is the mass per unit volume of a compound;

     Mass of Si  = density of Si x volume of Si block

  Volume of Si block  = length x width x thickness = 2.65 x 2.65 x thickness

   Volume of Si block = 7.02 x thickness

Mass of Si  = 2.13 x 7.02 x thickness

    24.4  = 15 x thickness

         thickness = [tex]\frac{24.4}{15}[/tex] = 1.63cm

Answer:

The mass of N a H C O 3 present is 2.431 g

Explanation:

The sample contains 57.2 % N a H C O 3  by mass.

To find the mass of N a H C O 3  in the sample, we need to find what the equivalent of 57.2 %.

Mass of N a H C O 3  = Percentage Composition * Mass of sample

Mass of N a H C O 3  = 57.2 / 100     * 4.25

Mass of N a H C O 3   = 2.431 g

The mass of N a H C O 3 present is 2.431 g

The mass of the compound (NaHCO₃) contained in 4.25 g is 2.431 g.

The given parameters;

The mass of the compound (NaHCO₃) contained in 4.25 g is calculated by the finding its equivalent in the given percent composition as follows;

[tex]mass \ = \frac{57.2}{100} \times 4.25\\\\mass = 2.431 \ g[/tex]

Thus, the mass of the compound (NaHCO₃) contained in 4.25 g is 2.431 g.

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