Ethane, C2H6, burns in air according to the equation
2 C2H6(g) + 7 O2(g) 4 CO2(s) + 6 H2O(g)
How many liters of O2 are required for complete reaction with 4.0 L of C2H6?

Answer:

The decrease of decomposers. The decrease of decomposers led to a change in carbon dioxide in biodome files.

Answer:

See explanation below

Explanation:

When the bromine (Br₂) reacts in presence of FeBr₃ with the isopropylbenzene, the isopropylbenzene is a ring activator, therefore, it will promote the substitution in the ortho and para positions. However, as the Isopropyl is voluminic, the ortho position would have steric hindrance and hence, this product is not formed in greater proportions. Instead, the para position becomes more available to reaction, and this product is formed in majority.

The mechanism of reaction can be seen in the following picture, along with the products of the reaction

Hope this helps

The second option 2NO(g) + O2(g) - 2NO2(g) could be reaction 2 in the overall reaction 2O(g) + 2NO(g) - 2NO2 + 2O3(g).

The overall balanced  equation involves the combination of two reactants to form two products. The reactants in reaction 2 must include O2 and NO, which combine to form NO2. The balanced equation for this reaction is 2NO(g) + O2(g) → 2NO2(g). This equation matches the stoichiometry of the overall reaction, and the enthalpies of reactions 1 and 2 can be added to determine the enthalpy of the overall reaction.

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Answer:

23.5 %

Explanation:

La fórmula del ácido sulfúrico es H₂SO₄, mientras que su masa molar es de 98 g/mol.

Siendo la densidad 1.4 g/mL, quiere decir que en 1 mL de solución tenemos 1.4 g de solución. Como la concentración es de 330 g/L entendemos que la solución contiene 330 gramos de soluto en 1L de solución.

Entonces en 1000 mL de solución (lo que equivale a 1 litro), tendremos:

1.4 g/mL = masa de solución / 1000 mL

masa de solución = 1.4 g/mL . 1000 mL =  1400 g

Como % en masa es la concentración que define la masa de soluto en 100 gramos  de solución, para esta solución de H₂SO₄ es:

(330 g / 1400 g) . 100 = 23.5 %

A represents the central atom, X represents the terminal atom, E represents the non-bonding electron group (usually lone pairs), and n represents the number of bonding electron pairs.

We describe each term as follows:

A: Central atom represents the atom in the center of the molecule to which other atoms are bonded.

X: Terminal atom represents the atoms bonded to the central atom.

E: Non-bonding electron group represents the valence electron group that is not involved in bonding and usually exists as lone pairs on the central atom.

n: Number of bonding electron pairs represents the number of pairs of electrons shared between the central atom and the terminal atoms.

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Answer:

Ionization energy

Electron affinity

Electronegativity

Explanation:

Answer: D:wavelenght

Explanation: Students will understand that shorter wavelengths have higher frequency and energy.

Answer:

C I believe... I might be wrong though

Explanation:

Answer:

79

Explanation:

Gold has 79 electrons, which makes it equal to 79 protons

Answer:

Superposition results in adding the two waves together. Constructive interference is when two waves superimpose and the resulting wave has a higher amplitude than the previous waves. Destructive interference is when two waves superimpose and cancel each other out, leading to a lower amplitude.

Explanation:

I hope I hlep :)

Catalysts are correctly characterized by each of the following statements except one. Catalysts are characterized by being able to exist in various states (solids, liquids, or gases),
The correct answer is (e) no correct response.

All of the statements are correct and accurately describe catalysts.
(a) Catalysts can exist in any state, whether they are solids, liquids or gases.
(b) One of the primary functions of a catalyst is to lower the activation energy required for a reaction to occur. This allows the reaction to proceed more quickly and efficiently.
(c) Catalysts do not participate in the reaction itself, but rather facilitate it.
(d) Catalysts are not consumed during a reaction and can be reused multiple times.

Catalysts are characterized by being able to exist in various states (solids, liquids, or gases), lowering the activation energy for a reaction, and not being consumed in a reaction. However, they do actively participate in a reaction by providing an alternative reaction pathway and forming temporary intermediate compounds, which ultimately helps increase the reaction rate.

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Answer:

Calculate the number of moles of oxygen required to react exactly with 4.30 ... How many moles of NH3 can be made from 2.00 moles of Nz if you have excess Hz? ... 10. How many grams of water are made when 3.00 moles of butanol is combusted? ... If 50.0 g of calcium oxide were produced, how many grams of calcium ...

Answer:

41.44 g

Explanation:

First of all, we must put down the equation of the reaction;

\(CaO + H2O ----->Ca(OH)2\)

Number of moles of CaO = 33g/56 g/mol = 0.59 moles

Number of moles of H20 = 10g/18 g/mol = 0.56 moles

Since the reaction is in 1:1 mole ration, H2O is the limiting reactant

Hence;

mass of Ca(OH)2 produced = 0.56 moles * 74 g/mol = 41.44 g

The pressure : 800 kPa

Boyle's law and Gay Lussac's law  

\(\tt \dfrac{P_1.V_1}{T_1}=\dfrac{P_2.V_2}{T_2}\)

P1 = initial gas pressure (N/m² or Pa)  

V1 = initial gas volume (m³)  

P2 = gas end pressure  

V2 = the final volume of gas  

T1 = initial gas temperature (K)  

T2 = gas end temperature  

The pressure after the gas is compressed :

\(\tt \dfrac{100\times 6.4}{300}=\dfrac{P_2\times 2.4}{900}\\\\P_2=800~KPa\)

Here is a basic procedure for creating an improvised musical instrument:

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The volume of 8.8g of carbon dioxide at STP is 4.38 L.

1 mole of any gas will take up 22.4 L of space at standard temperature and pressure (STP). A balanced chemical equation and the Ideal Gas Law can be used to determine the amount or mass of gas consumed or created in a chemical process.

n = m/M

where m is the molar mass of carbon dioxide and M is its mass in terms of molecules.

Considering that the molar mass of carbon dioxide is 44.01 g/mol:

n = 8.8 g / 44.01 g/mol

n = 0.1998 mol

Next, we can plug in the values of n, R, P, and T into the ideal gas law and solve for V:

V = (nRT)/P

V = (0.1998 mol x 0.0821 L·atm/(mol·K) x 273.15 K) / 1 atm

V = 4.38 L

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Answer:

the method is fractional distillation

1) a) The apparatus that can be used to separate a mixture of ethanol and water is called a fractional distillation apparatus. It consists of the following components:

b) The method of separation used in this apparatus is called fractional distillation.

2) In fractional distillation, the fractionating column provides additional surfaces for the vaporized components to condense and revalorize.

Distillation flask: This is a round-bottomed flask where the mixture of ethanol and water is initially placed.

Fractionating column: A long column with several glass beads or plates. It provides a large surface area for the vaporized components to condense and revalorize, aiding in the separation process.

Thermometer: It is placed at the top of the fractionating column to monitor the temperature during the distillation process.

Condenser: It is a coiled glass tube connected to the fractionating column. Cold water flows through the condenser, causing the vaporized components to condense back into liquid form.

Receiver flask: This is where the separated components are collected. The receiver flask is placed at the end of the condenser.

b) The method of separation used in this apparatus is called fractional distillation. Fractional distillation is employed when the components of a mixture have similar boiling points. In the case of ethanol and water, they form an azeotropic mixture with a boiling point of around 78.2°C. Simple distillation would not effectively separate these two components because they would boil together and vaporize simultaneously.

In fractional distillation, the fractionating column provides additional surfaces for the vaporized components to condense and revaporize. This repeated condensation and revalorization process allows for more efficient separation. The higher surface area in the fractionating column helps to achieve better separation of the ethanol and water, resulting in a more concentrated sample of ethanol in the distillate collected in the receiver flask.

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Answer:

The first thing that you need to do here is to figure out the mass of the sample.

To do that, you can use its volume and the fact that aluminium is said to have a density of

2.702 g cm

−

3

, which implies that every

1 cm

3

of aluminium has a mass of

2.702 g

.

Explanation:

Answer:

a) 48KJ

b) -48KJ

Explanation:

Given that;

ln(K2/K1) = ΔH°/R(1/T2 - 1/T1)

K2= equilibrium constant at T2

K1 = equilibrum constant at T1

R = gas constant

T1 = initial temperature

T2 = final temperature

When we double the equilibrium constant K1; K2 = 2K1

T1 = 310 K

T2 = 310 + 15 = 325 K

ln(2K1/K1) =- ΔH°/R(1/T2 - 1/T1)

ln2 = -ΔH°/8.314(1/325 - 1/310)

0.693 = -ΔH°/8.314(3.08 * 10^-3 - 3.2 * 10^-3)

0.693 = -ΔH°/8.314 (-0.00012)

0.693 = 0.00012ΔH°/8.314

0.693 * 8.314 = 0.00012ΔH°

ΔH° = 0.693 * 8.314/0.00012

ΔH° = 48KJ

b) K2 =K1/2

ln(K1/2/K1) =- ΔH°/R(1/T2 - 1/T1)

ln (1/2) = -ΔH°/8.314 (1/325 - 1/310)

-0.693 = -ΔH°/8.314  (-0.00012)

-0.693 = 0.00012ΔH°/8.314

-0.693 * 8.314 = 0.00012ΔH°

ΔH°= -0.693 * 8.314/0.00012

ΔH°= -48KJ

Answer:

R, the gas law constant, is a constant,

Solubility of gas is inversely proportional to temperature.

Solubility of gas ∝ 1/Temperature

That means as temperature increases the solubility gas goes on decreasing and vice versa.

But R is inversely proportional to temperature .

So Solubility ∝ R. (1/T ∝ R)

That means solubility of gas goes on increasing as R increases.

Step 2: Result

Result:

So from above conclusion the solubility of Oxygen in water increases that result into increasing R value.

Explanation:

The maximum number of moles of sodium oxide produced is 2.00 moles

The balanced chemical equation for the reaction between sodium and oxygen to form sodium oxide is:

4Na + O2 → 2Na2O

From the equation, we can see that 4 moles of sodium react with 1 mole of oxygen to produce 2 moles of sodium oxide.

To find the maximum number of moles of sodium oxide produced, we need to determine which reactant is limiting.

The molar mass of Na is 22.99 g/mol and the molar mass of O2 is 32.00 g/mol.

Using the given masses, we can calculate the number of moles of each reactant:

moles of Na = 46 g Na / 22.99 g/mol = 2.00 mol Na

moles of O2 = 32 g O2 / 32.00 g/mol = 1.00 mol O2

According to the balanced equation, 4 moles of Na react with 1 mole of O2. Therefore, since we have only 1.00 mol of O2, it is the limiting reactant.

From the equation, we see that 1 mole of O2 reacts to produce 2 moles of Na2O. Therefore, the maximum number of moles of Na2O that can be produced is:

moles of Na2O = (1.00 mol O2) × (2 mol Na2O/1 mol O2) = 2.00 mol Na2O

Therefore, the maximum number of moles of sodium oxide produced is 2.00 moles

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Answer:

Yes

Explanation:

3NaOH + H3PO4

3Na + 3O + 3H + 3H + P + 4O

3Na + 7O + 6H + P

Na3PO4 + 3H2O

3Na + P + 4O + 6H + 3O

3Na + 7O + 6H + P

Answer:

(c) 15.39 kg of copper present in 1tonne of ore.

Explanation:

We are given that ore contains 3% of copper carbonate

1tonne = 1000kg

3% of copper carbonate in 1000kg of ore will be

= \(\frac{3}{100}*1000\) = 30kg

30 kg of copper carbonate is present in ore

CuCO3 has 63.5g of cupper present in it

molar mass of CuCO3 = 123.5

so the percentage of copper present in CuCO3

= \(\frac{63.5}{123.5}*100\) = 51.3% of copper present per kg CuCO3

Now

amount of copper present in 30kg of CuCO3

= \(\frac{51.3}{100}* 30\) = 15.39kg

15.39 kg of copper present in 1tonne of ore

The mass of copper obtained from 1 tonne of the ore is 15.4 kg

The copper ore has 3% of copper carbonate by mass.

The mass of copper carbonate in 1 tonne of the ore can be calculated below.

1000 kg = 1 tonne

Therefore,

mass of copper carbonate = 3 /100 × 1000 = 30 kg

atomic mass of copper = 63.5 g

molar mass of CuCO₃ = 123.55 g

123.55 g of CuCO₃ gives 63.5 g of copper  

30, 000g of CuCO₃ will give ? of copper

cross multiply

mass of copper = 30,000 × 63.5 / 123.55

mass of copper = 1905000 / 123.55 = 15418.8587616 g

mass of copper = 15418.8587616 / 1000 ≈ 15. 4 kg  

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To shift the potential of the lead-zinc cell upward, toward 0.70 V, the following actions would give the desired result:

Add some concentrated Pb(C2H3O2)2 solution to the Pb2+/Pb couple

Adding some concentrated Zn(C2H3O2)2 solution to the Zn2+/Zn couple

Any action that makes [Zn2+] larger than [Pb2+]

A lead-zinc cell is a galvanic cell consisting of lead and zinc electrodes immersed in a suitable electrolyte. The potential of the lead-zinc cell is the difference in electrode potential between the lead and zinc electrodes, as determined by their concentration gradient.

Therefore, to shift the potential of the lead-zinc cell upward, toward 0.70 V, one should add some concentrated Pb(C2H3O2)2 solution to the Pb2+/Pb couple, add some concentrated Zn(C2H3O2)2 solution to the Zn2+/Zn couple, and any action that makes [Zn2+] larger than [Pb2+].

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The color of phenolphthalein indicator in a solution of hydrochloric acid will be colorless.

Phenolphthalein is a weak acid that changes color over a pH range of 8.2 to 10.0, going from colorless to pink as the pH increases. In a solution of hydrochloric acid, the pH is less than 8.2, so the phenolphthalein will not be pink, but will remain colorless. This indicates that the solution is acidic, with a pH lower than 8.2.

Phenolphthalein is often used in laboratory experiments, especially in the field of chemistry, to determine the endpoint of titrations and to monitor the pH changes during a reaction.

The color change of phenolphthalein occurs because the pink form of the indicator is a weak base and the colorless form is a weak acid. In an acidic solution, phenolphthalein will exist in its colorless, acidic form, and in a basic solution, it will exist in its pink, basic form. The exact pH range over which the color change occurs depends on the concentration of the phenolphthalein solution, but it is typically between 8.2 and 10.0.

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Answer:

A car speeds up from 0 km/hr to 100 km/hr in 6 seconds.

Explanation:

The main reservoirs of the carbon cycle are the atmosphere, oceans, land (including vegetation and soils), and fossil fuels. In these reservoirs, carbon exists in both inorganic and organic forms.

The inorganic carbon cycle involves the exchange of carbon dioxide (CO2) between the atmosphere and oceans through processes like photosynthesis and respiration.

Organic carbon, on the other hand, is found in living organisms, dead organic matter, and soil organic matter. It is cycled through processes such as decomposition and consumption by organisms. The interactions between the inorganic and organic carbon cycles occur primarily in the biosphere, where photosynthesis converts inorganic carbon into organic carbon compounds. While inorganic carbon is primarily in the form of CO2, organic carbon is present in complex organic molecules. Both forms of carbon play crucial roles in energy transfer, nutrient cycling, and climate regulation.

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The system adjust to reestablish equilibrium at D. The reaction shifts to the right products and the concentration of HI increases.

When \(H_{2}\) gas is added to the system at equilibrium, Le Chatelier's principle states that the system will respond by shifting the equilibrium position in the direction that consumes or reduces the excess reactant. In this case, the excess \(H_{2}\) gas is consumed to reestablish equilibrium.

Since the reaction is written in the forward direction, an increase in the concentration of \(H_{2}\) gas will drive the reaction towards the product side, leading to an increase in the concentration of HI gas. At the same time, the concentrations of \(H_{2}\) and \(I_{2}\) gases will decrease as they are consumed in the forward reaction.

Therefore, the system will adjust by shifting to the right, favoring the formation of HI gas and increasing its concentration, while decreasing the concentrations of \(H_{2}\) and \(I_{2}\) gases. This shift helps to reestablish equilibrium in the system. Therefore, Option D is correct.

The question was incomplete. find the full content below:

\(H_{2}\) gas is added to the system at

equilibrium below. How does the

system adjust to reestablish

equilibrium?

51.8kJ + H_{2}(g) +l 2 (g)

A. The reaction shifts to the left reactants and the concentration of HI increases

B. The reaction shifts to the right products and the concentrations of \(H_{2}\) and \(I_{2}\) increase

C. The reaction shifts to the left reactants and the concentration of \(H_{2}\) and \(I_{2}\) increase

D. The reaction shifts to the right products and the concentration of HI increases

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Answer:

5 is the answer since I did this and got it rigt

Answer:

Outer planets are significantly colder than other planets! The answer is B.

Explanation:

The main difference between inner and outer planets is that inner planets have a high temperature compared to outer planets.