Can anyone answer this question please it is very important


Which statement about bases is true?

(a) They are all alkalis

(b) They can neutralize acids

(c) They are all soluble


please can anyone explain what the question is about please brother/sister?

The name of the compound given in the question above is 4-ethyl-5,5-dimethyl-2-hexyne

To know the correct name of the compound illustrated in the diagram above, we shall obtain the IUPAC name of the compound.

The IUPAC nomenclature gives the standard for naming compounds. This is illustrated below:

Thus, with the above information, the name of the compound is 4-ethyl-5,5-dimethyl-2-hexyne

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Answer:

he oxygen ion you have with 8 protons attracting 10 electrons is larger, ... As a result, the 3Al3+ ion will be smaller than the Si element, which has ... O<Al3+<Si<Al<O2-<Ag<Rb<Cs ... It is fairly easy to arrange the atoms according to atomic size. ... I would put it between Al and Ag .

Explanation:

By using pH formula, the concentration of hydroxyl ion [OH-] will be 5.3 ×\(10^{-4}\) M.

What is pH?

The pH scale is a measurement of hydrogen ion concentration that is used to determine the acidity as well as alkalinity of a solution.

Calculation of [OH-] by using formula of pH.

Given data:

[H+] = 1.87 x \(10^{-11} M\)

[OH-] = ?

Calculation of [OH-] is shown below:

It can be calculated by using the formula:

pH = -log [H+]

Put the value of concentration of hydrogen ion [H+]  in above formula.

pH = -log (1.87 x \(10^{-11} M\))

pH = -log (1.87) -log ( \(10^{-11} M\))

pH = -0.27 +11

pH = 10.73

Now, use the formula, pH +pOH = 14

pOH = 14 - pH

pOH = 14 - 10.73

pOH = 3.27

Now, use formula pOH = -log [OH-]

[OH-] = \(10^{-pOH}\)

[OH-] = \(10^{-3.27}\)

[OH-] = 5.3 ×\(10^{-4}\) M

Therefore, the concentration of hydroxyl ion [OH-] will be 5.3 ×\(10^{-4}\) M.

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If we add the chromate as an oxidizing agent, a green Cr³⁺ solution be formed :

The oxidizing agent is the compound which itself rets reduce and oxidizes the other compounds. The primary alcohols get oxidized to the carboxylic acid. The secondary alcohols will convert to the ketones. The tertiary alcohols do not undergoes this reaction. The reactions are as follows :

1) CH₃ - CH₂ -CH - CH₂ CH₃   --------->   CH₃ - CH₂ - C - CH₂ - CH₃

                      |                                                             ||

                    OH                                                          O

2) 2-methyl-2-butonol ----->  no reaction

3) CH₃ - CH₂ - CH₂ - OH ------>CH₃ - CH₂ - CHO --->  CH₃ - CH₂ - COOH

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Answer:

thermal shock

Explanation:

the temperatures inside the glass jar should have continued to increase over time. Internal stresses due to uneven heating. This is also known as “thermal shock”.

In general, the thicker the glass, the more prone it will be to breaking due to the immediate differences in temperature across the thickness of glass.

Borosilicate glass is more tolerant of this, as it has a higher elasticity than standard silicon glass.

You may also note that laboratory test tubes and flasks are made with thinner walls, and of borosilicate glass, when designated for heating.

The fugacity in atm for pure ethane at 310 K and 20.4 atm is given by the equation: f = 20.4 exp (-Δg1/RT). The change in chemical potential between this state and a second state of ethane where the temperature is constant but the pressure is 24 atm is -0.0911RT.

Fugacity is a measure of the escaping tendency of a component in a mixture, which is defined as the pressure that the component would have if it obeyed ideal gas laws. It is used as a correction factor in the calculation of equilibrium constants and thermodynamic properties such as chemical potential. Here we need to determine the fugacity in atm for pure ethane at 310 K and 20.4 atm and the change in the chemical potential between this state and a second state of ethane where the temperature is constant but the pressure is 24 atm. So, using the formula of fugacity: f = P.exp(Δu/RT) Where P is the pressure of the system, R is the gas constant, T is the temperature of the system, Δu is the change in chemical potential of the system.  Δu = RT ln (f / P)The chemical potential at the initial state can be calculated using the ideal gas equation as: PV = nRT    

=>  P

= nRT/V

=> 20.4 atm

= nRT/V

=> n/V

= 20.4/RT The chemical potential of the system at the initial state is:

Δu1 = RT ln (f/P)

= RT ln (f/20.4) Also, we know that for a pure substance,

Δu = Δg. So,

Δg1 = Δu1 The change in pressure is 24 atm – 20.4 atm

= 3.6 atm At the second state, the pressure is 24 atm.

Using the ideal gas equation, n/V = 24/RT The chemical potential of the system at the second state is: Δu2 = RT ln (f/24) = RT ln (f/24) The change in chemical potential is Δu2 – Δu1 The change in chemical potential is

Δu2 – Δu1 = RT ln (f/24) – RT ln (f/20.4)

= RT ln [(f/24)/(f/20.4)]

= RT ln (20.4/24)

= - 0.0911 RT Therefore, the fugacity in atm for pure ethane at 310 K and 20.4 atm is:

f = P.exp(Δu/RT)

=> f

= 20.4 exp (-Δu1/RT)

=> f

= 20.4 exp (-Δg1/RT) And, the change in the chemical potential between this state and a second state of ethane where the temperature is constant but pressure is 24 atm is -0.0911RT. Therefore, the fugacity in atm for pure ethane at 310 K and 20.4 atm is given by the equation: f = 20.4 exp (-Δg1/RT). The change in chemical potential between this state and a second state of ethane where the temperature is constant but the pressure is 24 atm is -0.0911RT.

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Carbonyl groups is bonded to sp2 hybridized carbon atom.

The interaction within a carbonyl group, when both the oxygen and carbon atoms are sp2 hybridized, is explained by the same hypothesis.

When a carbon atom bonds with one s-orbital and two p orbitals, it becomes sp2 hybridized. Between the three atoms, 2 individual connections and a double bond are formed. The hybridization orbitals are arranged in a triangle with bonds at 120° angles.

A carbon atom is the double to an oxygen atom to form a carbonyl group, which is a chemical organic functional group. Aldehydes and ketones, which are typically joined to another carbon compound, are the simplest carbonyl groups. Many aromatic chemicals have these structures, which contribute to flavor and fragrance.

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Answer:

Pizza, salad, tacos, burgers, etc.

Explanation:

:D

Answer:

=> 2.8554 g/mL

Explanation:

To determine the formula to use in solving such a problem, you have to consider what you have been given.

We have;

mass (m)     = 16.59 g

Volume (v) =  5.81 mL

From our question, we are to determine the density (rho) of the rock.

The formula:

\(p = \frac{m}{v}\)

Substitute the values into the formula:

​\(p = \frac{16.59 g}{5.81 mL} \\ = 2.8554 g/mL\)

= 2.8554 g/mL

Therefore, the density (rho) of the rock is 2.8554 g/mL.

The chest electrode placed on the fifth intercostal space on the mid-clavicular line is known as the V4 electrode. V4 is part of the precordial electrodes used in a standard 12-lead electrocardiogram (ECG) to assess the electrical activity of the heart.

The placement of these electrodes is crucial for obtaining an accurate ECG, which helps healthcare professionals diagnose and monitor various cardiac conditions.
The V4 electrode is positioned on the left side of the chest in line with the middle of the clavicle (collarbone), at the level of the fifth intercostal space. To locate this space, one should palpate the ribcage and count down from the first rib until the fifth rib space is found. Proper electrode placement ensures that the ECG will accurately capture the electrical signals originating from the heart.
Other precordial electrodes, such as V1, V2, V3, V5, and V6, are also strategically placed on specific areas of the chest to obtain a comprehensive view of the heart's electrical activity. Collectively, the information from all 12 leads provides a detailed picture of the heart's functioning, assisting in the diagnosis of heart conditions and guiding treatment decisions.
In conclusion, the V4 electrode plays a vital role in electrocardiography, as its placement on the fifth intercostal space on the mid-clavicular line enables healthcare professionals to monitor the heart's electrical activity and make informed decisions about patient care.

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To determine the vapor pressure of the substance at 29°C, we can use the Clausius-Clapeyron equation, which relates the vapor pressure (P) of a substance at a given temperature (T) to its normal boiling point (T_boiling), and the enthalpy of vaporization (ΔH_vap):

ln(P/P_boiling) = -ΔH_vap/R * (1/T - 1/T_boiling)

Where P_boiling is the vapor pressure at the boiling point, and R is the ideal gas constant (8.314 J/(mol·K)).

First, we need to convert the given temperatures to Kelvin:

T = 29°C + 273.15 = 302.15 K

T_boiling = 76°C + 273.15 = 349.15 K

Substituting the values into the equation:

ln(P/P_boiling) = -38.7 kJ/mol / (8.314 J/(mol·K)) * (1/302.15 K - 1/349.15 K)

Simplifying the equation further, we can solve for ln(P/P_boiling):

ln(P/P_boiling) = -4.6616 * (0.003312 - 0.002862)

ln(P/P_boiling) = -4.6616 * 0.00045

ln(P/P_boiling) = -0.002097

To find P/P_boiling, we take the exponential of both sides:

P/P_boiling = e^(-0.002097)

Finally, we can solve for P (vapor pressure) by multiplying P_boiling to both sides of the equation:

P = P_boiling * e^(-0.002097)

Explanation:

The Clausius-Clapeyron equation is a useful tool for calculating the vapor pressure of a substance at a temperature different from its boiling point. It is based on the relationship between temperature, vapor pressure, and enthalpy of vaporization. By using the equation, we can determine the vapor pressure at a given temperature using known values.

In this case, we are given the normal boiling point of the substance (76°C) and its enthalpy of vaporization (38.7 kJ/mol). The boiling point represents the temperature at which the vapor pressure is equal to the atmospheric pressure. By plugging the values into the Clausius-Clapeyron equation and solving for the vapor pressure (P), we can obtain the desired result.

It is important to note that the equation assumes ideal gas behavior and relies on the assumption that the enthalpy of vaporization remains constant over the temperature range. Additionally, the ideal gas constant (R) is used to convert the units of enthalpy from kJ/mol to J/(mol·K).

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Answer:

B is the answer

Explanation:

ik this because I take the class

Answer:

chloroplast

Explanation:

Answer:

6:4

Explanation:

The coefficients of a balanced chemical equation give us the mole ratios. The coefficient of carbon dioxide here is 4 and water is 6.

Answer:

0.0074 moles

Explanation:

We'll have to use this forumla: PV = nRT

P = Pressure (0.724)

V = Volume in liters (0.25)

n = moles

R = 0.0821

T = temperature in kelvins (25 + 273 = 298)

Our equation will look like this:

(0.724)(.25) = n(0.0821)(298)

0.181 = 24.4658n

Divide both sides by 24.4658 to isolate n

0.181/24.4658 = 24.4658n/24.4658

n = 0.00739808

Option a is false, option b is true and option is false. The second law of thermodynamics does not state that entropy is conserved.

(a) False. The second law of thermodynamics states that the total entropy of an isolated system can only increase or remain constant over time, but not decrease. Entropy is not conserved, but tends to increase. Other options are incorrect because they may misinterpret the concept of entropy conservation.

(b) True. In a reversible process, the system's entropy increase is balanced by the surroundings' entropy decrease by the same amount. This ensures that the total entropy change of the universe remains zero, which is a characteristic of a reversible process. Other options are incorrect because they may not account for the reversibility condition.

(c) False. In a spontaneous process, the total entropy change of the universe must be positive. While the system may undergo an entropy change of 4.2 J/K, the surroundings' entropy change should be greater than 4.2 J/K to ensure a positive total entropy change. Other options are incorrect because they may assume that the entropy changes of the system and surroundings must be equal and opposite, which is not the case for spontaneous processes.

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To reset the simulation, click the orange circle in the bottom right. Then, check the white box with the energy symbol.

By clicking the orange circle in the bottom right, you will reset the simulation to its initial state, allowing you to start over. This is useful when you want to undo any changes or modifications you have made and return to the original setup. Once you have reset the simulation, make sure to check the white box with the energy symbol again. This box likely contains important information or controls related to the energy aspect of the simulation, which is crucial for understanding and manipulating the system accurately.

The orange circle in the bottom right. Resetting the simulation can be done by clicking on the orange circle located in the bottom right corner. This action reverts the simulation back to its original state, undoing any changes or adjustments that have been made. It provides a way to start over and ensures a clean slate for further experimentation or analysis. Additionally, after resetting the simulation, it is recommended to check the white box with the energy symbol.

This box likely contains relevant information, controls, or data related to the energy aspect of the simulation. Understanding and monitoring the energy dynamics are essential for accurately interpreting and manipulating the simulation's behavior. Therefore, regularly referring to the energy symbol box can provide valuable insights and enable effective adjustments to achieve desired outcomes.

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Answer:  The electron transition from 4p→2s  would produce light with a longer wavelength.

Explanation: The energy of an electron in a hydrogen atom is higher when it is at a higher energy level. When an electron undergoes transition it releases energy in the form of electromagnetic radiation or light. The higher the transition, the greater the energy.

Now we know that the energy released in the 4p→1s transition is greater than the energy released in the 4p→2s transition. And since energy is inversely proportional to wavelength, the wavelength of light in the latter case is more.

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Answer:

\(K_{c} =\frac{[O_{2} ] [SO_{2} ]^{2} }{[SO_{3} ]^{2} }\)

Explanation:

To construct the equilibrium constant, you need the balanced equation:

2 SO₃(g) -----> O₂(g) + 2 SO₂(g)

The equilibrium constant compares the concentrations of the products and the reactants.

The given expression follows this structure:

aA(g) ----> bB(g) + cC(g)

In this equation, the uppercase letters symbolize the molecules and the lowercase letters symbolize their corresponding coefficients in the balanced equation.

The general equilibrium expression looks like this:

\(K_{c} =\frac{[B]^{b} [C]^{c} }{[A]^{a} }\)

To be clear, the concentrations in the numerator represent the gaseous products and the concentrations in the denominator represent the gaseous reactants.

Therefore, the equilibrium expression for this equation is:

\(K_{c} =\frac{[O_{2} ] [SO_{2} ]^{2} }{[SO_{3} ]^{2} }\)

Answer:

its not inclusive fitness which is confusing but

Explanation:

ill give the answer when I finish the test

Answer:

phenotypic plasticity

Explanation:

Answer:

a

Explanation:

becuase it make cents

Ne gas has a partial pressure of 0.297 atm.

We must first translate the mass of each gas into moles in order to determine the partial pressure of Ne:

1.38 mol n He is equal to 5.50 g / 4.00 g/mol.

15.0 g/20.18 g/mol n Ne = 0.747 mol

n Kr is equal to 35.0 g/83.8 g/mol, or 0.416 mol.

The total number of moles in the combination is then determined:

He, Ne, and Kr together make up n total, which is 1.38 + 0.747 + 0.416 = 2.53 mol.

We can now determine the mole fraction of Ne.

n Ne = 0.747 / 2.53 = 0.297; X Ne = n total / n Ne

The total pressure is 1 atm since the combination is at standard temperature and pressure (STP). Thus, the following formula can be used to determine the partial pressure of Ne:

P Ne is equal to X Ne * P total, or 0.297 atm * 1 atm.

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Answer:

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

\(volume = \frac{mass}{density} \\ \)

From the question

mass of gold = 200 g

density = 20.5 g/mL

It's volume is

\(volume = \frac{200}{20.5} \\ = 9.7560975609...\)

We have the final answer as

Hope this helps you

Answer:

INTENSIVE PROPERTY OF MATTER

Explanation:

DENSITY IS AN INTENSIVE PROPERTY OF MATTER THAT ILLUSTRATES HOW MUCH MASS A SUBSTANCE HAS IN A GIVEN AMOUNT OF VALUE,

Answer:

Mg (NO3)2+ K2CO3 ⇒ Mg CO3 + KNO3

Mg=1 ,N =2, O =6 +3 9 ,K =2 ,C=1 ,  (reactant side)

Mg =1, C = 1  ,O = 3+3 = 6 , K= 1., N =1 (product side)

Mg(NO3)2 + K2CO3  ⇒ Mg CO3 + 2KNO3

Mg =1 , C = 1, O =3+6=9, K = 2 IN product side

ANS:

Mg (NO3)2 +K2CO3 ⇒ MgCO3 +2 KNO3

Explanation:

Answer:

A thermodynamic quantity equivalent to the capacity of a system to do work.

Explanation:

Answer:

Sulfer

Explanation:

We just took the lesson lol

The product of electrolysis : Al at cathode and Cl₂(chlorine) at anode

Given

Aluminum chloride compound

Required

The product of electrolysis

Solution

The rule :

The reaction at the cathode(the negative pole) :

1. the reduced active metal is water, other than that the metal will be reduced

2. H⁺ of the acid will be reduced

The reaction at the anode((the positive pole) :

1. if the electrodes are not inert then the metal is oxidized

2. If inert then:

a. OH⁻ from the base will be oxidized

b. The halogen metal will oxidize

In the electrolysis of molten AlCl₃ with an inert electrode, the cation will be reduced at the cathode and the anion will be oxidized at the anode

Cathode : Al³⁺ + 3e⁻ ⇒ Al(s)

Anode : 2Cl⁻⇒Cl₂(g) + 2e⁻