CHEMISTRY
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Answer:

reaction between oxygen and hydrocarbon is combustion

The compound would most likely be the reactants in a combustion reaction is oxygen and hydrocarbon.

Those reaction in which a fuel will change into a water molecule and carbon dioxide gas, in the presence of oxygen is known as combustion reaction.

Combustion reaction will represent as:

CₓHₙ + O₂ → H₂O + CO₂, where

CₓHₙ is a hydrocarbon which gets oxidized in the presence of oxygen oxidant.

Hence option (2) is correct i.e. oxygen and hydrocarbon.

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Answer: B. It increases quickly

Explanation:

As the bromine gas in a container is heated over a flame, the average kinetic energy increases quickly.

The temperature of gas molecules is the measure of average kinetic energy of the gas molecules.

The increase in temperature increases the heat gained by the gas molecules which in turn increases the average kinetic energy of the individual gas molecules.

Thus, as the bromine gas in a container is heated over a flame, the average kinetic energy increases quickly.

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The substance begins to boil at 2750⁰C, the substance begins to melt at 1500⁰C, the temperature at which the substance is both a liquid and a gas at 2750⁰C, temperature is the substance both a solid and a liquid​ at 1500⁰C.

Heating curves are the graphical correlations between heat added to a substance. When viewed from a cooling perspective, ie. loss of heat, it is the cooling curve.

The gradient of the cooling curve is related to the heat capacity, the thermal conductivity of the substance, and the external temperature. The more heat is required to change the temperature of the substance, the slower it cools, so the smaller the gradient of the curve. The higher the thermal conductivity, the faster heat is transferred, so the faster the substance cools.

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Answer:

Instrumental methods

surface analysis

high performance liquid chromatography

atomic spectroscopy

potentiometry

Classical methods

precipitation titration

gravimetric analysis

Explanation:

Instrumental methods of analysis are those analytical methods in which the responsibility of detection has been removed from human beings and placed on automated instruments while classical methods are those analytical methods in which the responsibility of detection remains the responsibility of human beings.

Many instrumental methods such as HPLC rely on computer screens as readout devices.

Answer:

A neutron is a subatomic particle found in the nucleus of every atom except that of simple hydrogen.

Explanation:

Answer:

With the exception of simple hydrogen, every atom has a neutron in its nucleus. The fact that the particle has no electrical charge it is neutral gives it its name. Neutrons have a very high density.

_________________

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Composite boards can be recycled and need not be sealed against  effect of moisture .Recyclable materials do not cause harm to environment as the do not leave any toxins.

Recyclable materials help in protecting the environment through the conservation of the natural resources.It helps in improving the energy efficiency as it promotes the three R's of waste management which are reduce ,reuse and recycle.

Recycling helps in reducing the amount of the waste that gets collected in the landfills.Solid waste is burnt for electricity is efficient but leads to increases amounts of carbon dioxide in the environment.

It helps in conservation of the limited resources such as  oil and precious metals.Recycling helps in improving energy efficiency as it helps in reduction of energy costs.

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Answer:

see explanations

Explanation:

The graphic is the heating curve for water. Note that it is divided into 5 distinct heat flow segments. The segments with changing slopes are single phase segments with changes in temperature values. From left to right segment A is solid ice being warmed to it's melting point. Segment B is the melting segment in which 2 phases are in contact (solid + liquid). Note that addition of heat does not change the temperature. Segment C is warming of the liquid (single phase) up to its boiling point. At the boiling point the liquid begins to pass into the gas phase and again 2 phases are in contact; i.e., liquid & gas. Note again when two phases are in contact no temperature change occurs. Finally, segment E is the heating of the pure, single phase gas.

In summary ...

Segment A => heating single phase (solid) ice up to melting pt.

Segment B => melting of ice => 2 phases in contact (s & l) ΔT = ∅.

Segment C => heating single phase (liquid) water up to boiling pt.

Segment D => boiling of liquid => 2 phases in contact (l & g). ΔT = ∅.

Segment E => heating single phase (steam) up to desired temperature.

For what it's worth, the equation for the segments that show increasing temperature values is q = mcΔT (m= mass, c = specific heat & ΔT temp change.

The segments with zero slopes (horizontal lines) are defined by equations  q = m·ΔHₓ where m = mass & ΔHₓ = heat of fusion (a constant = 335 j/g). The same is true for the line at 100°C where q = m·ΔH(v) where m = mass & ΔH(v) is the heat of vaporization (a constant = 2259 j/g.

Calculations involve calculating the amount heat transfer for each segment individually and then adding the heat values to obtain the total heat transfer.

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Answer:

Picture attached has answers

Explanation:

Answer:

4.81 g / mol

Explanation:

Given :

Titanium has an HCP unit cell

Radius of titanium, R = 0.1445 nm

Unit cell volume, \($V_c= 6R^2 C\sqrt3$\)

But for Ti, c/a = 1.58

           So, c  = 1.58 a

And a = 2R or c = 3.16 R

\($V_c = 6R^2 \times 3.16 R \times \sqrt3$\)

\($V_c=6 \times 3.16 \times \sqrt3 \times (1.445 \times 10^{-8})^3 $\)

\($V_c= 9.91 \times 10^{-23}\ cm^3 / unit \ cell $\)

Density of Ti (theoretical), \($\rho=\frac{n. A_{Ti}}{V_C.N_A} $\)

For HCP, n = 6 atoms per unit cell and atomic mass = 47.87 g/mol

\($\rho=\frac{6 \times 47.87}{9.91 \times 10^{-23} \times 6.023 \times 10^{23}} $\)

  = 4.81 g/mol

This is the theoretical density of titanium.

The value given in literature is 4.51 g/mol

Answer: 78.59 lb/ft

Explanation:

use stoichiometry to solve

1.26g/cm x ( 1pound/ 454 grams) X (28317 Cm/1 ft) =  78.59 pounds/ft

Answer:

If n = 4, then the possible values of 1 and m depend on the equation or expression being used. Without more information, it is impossible to determine what the possible values of 1 and m might be. Can you please provide more context or information about the problem you are trying to solve?

Answer:

newton-meters or (N-m)

Explanation:

This is challenging to provide an exact prediction for the melting and boiling points. Experimental factors, impurities, and other variables can influence these values.

To predict the melting and boiling points for methylamine (CH3NH2), we can consider the differences in molecular structure and intermolecular forces compared to ethane. Methylamine has a nitrogen atom in place of one carbon atom in ethane.

Methylamine exhibits hydrogen bonding due to the presence of the nitrogen atom, which can form hydrogen bonds with other methylamine molecules. Ethane, on the other hand, lacks a hydrogen atom bonded to an electronegative atom, so it does not participate in hydrogen bonding.

Hydrogen bonding is stronger than the London dispersion forces present in ethane. Consequently, methylamine is expected to have stronger intermolecular forces and thus higher melting and boiling points compared to ethane.

Considering this, we can predict that the melting point of methylamine will be higher than -183 °C, the melting point of ethane. Similarly, the boiling point of methylamine will be higher than -89 °C, the boiling point of ethane.

However, without specific experimental data or a detailed analysis of methylamine's properties.

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At equilibrium, the number of moles of \(Cl_2\) (g) will be 0.2025 mol.

1: Write the balanced chemical equation:

\(C_O\)(g) + \(Cl_2\)(g) ⟶ \(C_OCl_2\)(g)

2: Set up an ICE table to track the changes in moles of the substances involved in the reaction.

Initial:

\(C_O\)(g) = 0.3500 mol

\(Cl_2\)(g) = 0.05500 mol

\(C_OCl_2\)(g) = 0 mol

Change:

\(C_O\)(g) = -x

\(Cl_2\)(g) = -x

\(C_OCl_2\)(g) = +x

Equilibrium:

\(C_O\)(g) = 0.3500 - x mol

\(Cl_2\)(g) = 0.05500 - x mol

\(C_OCl_2\)(g) = x mol

3: Write the expression for the equilibrium constant (Kc) using the concentrations of the species involved:

Kc = [\(C_OCl_2\)(g)] / [\(C_O\)(g)] * [\(Cl_2\)(g)]

4: Substitute the given equilibrium constant (Kc) value into the expression:

1.2 x \(10^3\) = x / (0.3500 - x) * (0.05500 - x)

5: Solve the equation for x. Rearrange the equation to obtain a quadratic equation:

1.2 x \(10^3\) * (0.3500 - x) * (0.05500 - x) = x

6: Simplify and solve the quadratic equation. This can be done by multiplying out the terms, rearranging the equation to standard quadratic form, and then using the quadratic formula.

7: After solving the quadratic equation, you will find two possible values for x. However, since the number of moles cannot be negative, we discard the negative solution.

8: The positive value of x represents the number of moles of \(Cl_2\)(g) at equilibrium. Substitute the value of x into the expression for \(Cl_2\)(g):

\(Cl_2\)(g) = 0.05500 - x

9: Calculate the value of \(Cl_2\)(g) at equilibrium:

\(Cl_2\)(g) = 0.05500 - x

\(Cl_2\)(g) = 0.05500 - (positive value of x)

10: Calculate the final value of \(Cl_2\) (g) at equilibrium to get the answer.

Therefore, at equilibrium, the number of moles of \(Cl_2\) (g) will be 0.2025 mol.

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Answer:

Volume of the reaction vessel is increased - shift to the left

The reaction is cooled down - shift to the right

H2 is added to the system - shift to the right

The pressure of the system is decreased - shift to the left

A catalyst is added to the system - no change

Water is removed from the system - shift to the right

Explanation:

When a constraint such as a change in temperature, pressure or volume is imposed on a reaction system in equilibrium, the equilibrium position will shift in such a way as to annul the constraint.

When the volume of a reaction system is increased, the equilibrium position shifts in the direction in which there is the highest total volume. This is the left hand side.

Since the reaction is exothermic (heat is given out) when the reaction is cooled down, the forward reaction is favoured.

Adding of reactants shifts the equilibrium position to the right hand side hence when H2 is added, the equilibrium position shifts to the right.

Decreasing the pressure shifts the equilibrium position to the direction of higher total volume hence the equilibrium shifts to the left when pressure is decreased.

A catalyst has no effect on the equilibrium position. It increases the rate of forward and reverse reaction to the same extent hence the equilibrium position is unaffected.

Removal of water from the system increases the rate of forward reaction since a product is being removed from the reaction system.

The equation C2H4(g) + H2(g) + C2H6(g) → Caco3(s) + Cao(s) + CO2(g) shows an increase in entropy due to the formation of a gas as a product. Option A

In this equation, the reactants on the left-hand side consist of gases (C2H4 and H2), while the products on the right-hand side include a solid (Caco3) and a gas (CO2).

When a reaction involves a change from gaseous to solid or liquid states, there is typically a decrease in entropy because the particles become more ordered and constrained in the solid or liquid phase.

Conversely, when a reaction involves the formation of gases, there is generally an increase in entropy because gases have higher degrees of molecular motion and greater freedom of movement compared to solids or liquids.

In the given equation, the reactants include three gaseous compounds (C2H4, H2, and C2H6), and one of the products is a gas (CO2). Therefore, the overall entropy of the system increases during this reaction.

The equation Fe(s) + S(s) → FeS(s) does not show an increase in entropy. Both the reactants (Fe and S) and the product (FeS) are solids. Since solids have lower entropy compared to gases or liquids, the entropy of the system does not increase in this reaction. Option A

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Answer:

Q1 D

Explanation:

Q2 C

Global warming brought on by rising carbon dioxide concentrations in the atmosphere is changing weather patterns and ocean temperatures. Changes in precipitation patterns are being brought on by the warming of the atmosphere.

Because it is one of the gases in the atmosphere that causes the greenhouse effect, which causes the Earth to warm, carbon dioxide is known as a greenhouse gas. Long-wavelength infrared radiation (heat) from the Earth is absorbed by carbon dioxide molecules in the atmosphere, and some of it is then radiated back downward.

Similarly, as air temperatures rise in response to rising carbon dioxide concentrations, more water vapor escapes into  the atmosphere—which then amplifies greenhouse heating

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Answer:

1.84

Explanation: just input the value into the formula pv=nrt

p*30=2*0.08206*337 then divide both sides by 30 and  get 1.84

Answer: chlorine 17Cl35 has 18 neutrons, and 17Cl37 has 20 neutrons

Explanation: 17 is the atomic number of chlorine which tells us what is the number of protons in chlorine nucleus. 35 and 37 is the mass number of two isotopes of chlorine, it tells us what is the total number of protons and neutrons in atomic nucleus of two chlorine isotopes .

Now to know how many neutrons are in both isotopes we have to substract the number of protons from the mass number of each isotope

For 17Cl35 it will be 35- 17 what makes 18 and for 17 Cl 37 it will be 20 neutrons

Aqueous acids open the epoxide rings of the provided epoxide, forming 2-methyl-2,3-hexanediol as a result.

For the ring opening of an epoxide, a wide range of basic nucleophiles can be employed, including amines, hydrides, Grignard reagents, acetylide anions, and hydride. SN2 mechanisms are typically used to open these rings.

A heterocyclic organic compound called an epoxide has oxygen as a ring component.

The opening of the ring by the acid's protonation of the oxygen is the first stage of this reaction. This now produces a carbocation, which a water molecule quickly attacks.

The product is produced and the acid catalyst is renewed when a proton is lost.

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Answer:

decreases

Explanation: decreases,as shelter is a limiting factor becasue the trees for example were there home now it is covered in melaleuca

hope this helps

Answer:

a should be the correct answer

Explanation:

there will not be many shelters that suit some animals needs

The change in entropy for the crystallization of 1.25 moles of butane can be calculated using the formula ΔS = ΔH_fusion / T, where ΔS is the change in entropy, ΔH_fusion is the enthalpy of fusion, and T is the temperature in Kelvin.

In the given problem, the enthalpy of fusion of butane is 24.3 kJ/mol. To convert this to J/mol, we multiply it by 1000, resulting in 24,300 J/mol. The melting point of butane is -0.5°C, which is equivalent to 272.65 K.

Using the formula, we can calculate the change in entropy as follows:

ΔS = (24,300 J/mol) / 272.65 K

    ≈ 89.25 J/(mol·K)

Therefore, the change in entropy for the crystallization of 1.25 moles of butane is approximately 89.25 J/(mol·K).

The change in entropy (ΔS) for the crystallization of a substance can be determined using the equation ΔS = ΔH_fusion / T, where ΔH_fusion is the enthalpy of fusion and T is the temperature in Kelvin. In this case, the enthalpy of fusion of butane is given as 24.3 kJ/mol, which is converted to 24,300 J/mol. The melting point of butane is -0.5°C, which is equivalent to 272.65 K.

By substituting the values into the equation, we find that the change in entropy is approximately 89.25 J/(mol·K). This means that for the crystallization of 1.25 moles of butane, the entropy decreases by 89.25 J/(mol·K).

Entropy is a measure of the disorder or randomness of a system. In this case, the crystallization process involves the transition from a disordered liquid state to an ordered solid state, resulting in a decrease in entropy. The magnitude of the entropy change depends on the enthalpy of fusion and the temperature. A higher enthalpy of fusion or a lower temperature leads to a larger change in entropy.

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When the conductivity is at a minimum, the true thing about the amount of Ba(OH)₂ is that the amount of the two reactants are the same when conductivity is low.

The ability of an aqueous solution to carry an electric current is measured by its conductivity. Ionic strength is measured by conductivity monitors. In automated chromatography systems, conductivity is the main input control parameter used to enable the generation of salt gradients or to control buffer dilution or in-line buffer preparation.

Conductivity monitors can be used to automate and monitor cleaning and equilibration procedures. It's important to remember that temperature affects measurements of conductivity and pH.

When conductivity is low, the amounts of the two reactants are equal. Between the reactants, there are no excesses. The aqueous solution cannot dissolve the precipitate BaSO₄. Since there are no ions, the conductivity is at its lowest.

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Complete question

The Ba(OH)₂ dissociates as Ba+2 + 2 OH-. H₂SO₄ dissociates as 2 H+ + SO₄-2. When the conductivity is at a minimum, what must be true about the amount of Ba(OH)₂ compared to H₂SO₄

Answer:

First, we need to determine the molar mass of HCl.

The molar mass of HCl = the mass of hydrogen (1.008 g/mol) + the mass of chlorine (35.45 g/mol) = 36.45 g/mol.

Next, we can use Avogadro's number (6.02 x 10^23 molecules/mol) to convert the number of molecules to moles:

6.02 x 10^24 molecules / 6.02 x 10^23 molecules/mol = 10 moles

Finally, we can use the molar mass to convert moles to grams:

10 moles x 36.45 g/mol = 364.5 grams

Therefore, the mass of 6.02 x 10^24 molecules of HCl is 364.5 grams.

Answer:

the reactivity and the valence electrons

Explanation:

the reactivity of the elements would have played a significant role in why such elements were grouped together. the number of valence electrons dictates how reactive an element is - the less valence electrons the more reactive it is. the column, group 1 in which these elements are put together in, show that each of the elements have 1 valence electrons and are therefore reactive.

you can go on to further explain what valence electrons are, explain what the group numbers are associated with the valence electrons and how valence electrons effect reactivity. further this, talk about how the three elements have the same number of valence electrons and therefore were grouped together

The heat of combustion of the substance is obtained as -6154 kJ/mol.

The heat of combustion is the heat that is evolved when one mole of sucrose is burnt in oxygen. Let us now try to find the heat of combustion of the substance.

Number of moles of sucrose = 1.151 grams /342.3 g/mol = 0.00336 moles

Heat capacity of the calorimeter = 4.900 kJ/°C

Temperature change = 26.63°C -  22.41°C = 4.22 °C

Now we can find the heat of combustion of the sucrose knowing that heat is evolved.

H = -(4.900 kJ/°C) * 4.22 °C/ 0.00336 moles

H = -6154 kJ/mol

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If there is sufficient fossil fuel, we will cope by utilizing it in a responsible and sustainable manner while actively transitioning towards alternative energy sources. Coping with the abundance of fossil fuels requires a multi-faceted approach that considers environmental, economic, and social aspects.

To cope effectively, we can:

1. Promote energy efficiency: Invest in technologies and practices that minimize energy waste and maximize efficiency in all sectors, including transportation, industries, and buildings.

2. Transition to renewable energy: Increase the adoption of renewable energy sources such as solar, wind, hydro, and geothermal power. This reduces our reliance on fossil fuels and mitigates environmental impacts.

3. Implement policy measures: Enact policies that incentivize the use of renewable energy, discourage excessive fossil fuel consumption, and promote sustainable practices.

4. Invest in research and development: Support and fund research efforts aimed at developing cleaner and more sustainable energy technologies, such as advanced battery storage, hydrogen fuel cells, and carbon capture and storage.

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