Which of the following elements is in the same group as Sulfur (S)?

Answer:

A) CH3CH2CH2CH2CH2CH2OH

Explanation:

For this question, we have the following answer options:

A) CH3CH2CH2CH2CH2CH2OH

B) (CH3CH2)2CH(OH)CH2CH3

C) (CH3CH2)2CHOHCH3

D) (CH3CH2)3COH

E) (CH3CH2)2C(CH3)OH

We have to remember the reaction mechanism of the substitution reaction with [tex]PBr_3[/tex]. The idea is to generate a better leaving group in order to add a "Br" atom.

The [tex]PBr_3[/tex] attacks the "OH" generation new a bond to P (O-P bonds are very strong), due to this new bond we will have a better leaving group that can remove the oxygen an allow the attack of the Br atom to generating a new C-Br bond. This is made by an Sn2 reaction. Therefore we will have a faster reaction with primary substrates. In this case, the only primary substrate is molecule A. So, "CH3CH2CH2CH2CH2CH2OH" will react faster.

See figure 1

I hope it helps!

Answer:

The final temperature of the entire water sample after all the ice is melted, is 12,9°C. We should realize that if there is no loss of heat in our system, the sum of lost or gained heat is 0.  It is logical to say that the temperature has decreased because the ice gave the water "heat" and cooled it

Thats all i know

Answer:

CO3^2-

Explanation:

In qualitative analysis, we try to use chemical reactions to determine the composition of an unknown substance. The addition of certain reagents to the unknown solution gives certain results that show the presence or absence of certain species from the unknown sample.

When dilute HCl is added to an unknown sample and effervescence is observed, then the unknown sample must contain CO3^2- or HCO3^-. The presence of these species is confirmed if the gas evolved is passed through limewater and the gas turns limewater milky.

Answer:

an increase in 1-butene was observed when t-butoxide was used

Explanation:

When a base reacts with an alkyl halide, an elimination product is formed. This reaction is an E2 reaction.

Here we are to compare the reaction of two different bases with one substrate; 2-bromobutane. Both reactions occur by the E2 mechanism but follow different transition states due to the size of the base.

The Saytzeff product, 2-butene, is obtained when the methoxide is used while the non Saytzeff product, 1-butene, is obtained when t-butoxide is used.

The Saytzeff rule is reliable in predicting the major products of simple elimination reactions of alkyl halides given the fact that a small/strong bases is used for the elimination reaction. Therefore hydroxide, methoxide and ethoxide bases give similar results for the same alkyl halide substrate. Bulky bases such as tert-butoxide tend to yield a higher percentage of the non Saytzeff product and this is usually attributed to steric hindrance.

Answer:

a. the Kb of NO₂⁻ to find the hydroxide concentration.

Explanation:

When sodium nitrite is dissolved in water, it dissociates in sodium cation and nitrite anion according to the following equation.

NaNO₂(s) ⇒ Na⁺(aq) + NO₂⁻(aq)

Na⁺ comes from NaOH (strong base) so it doesn't react with water.

NO₂⁻ comes from HNO₂ (weak acid) so it reacts with water according to the following equation.

NO₂⁻(aq) + H₂O(l) ⇄ HNO₂(aq) + OH⁻(aq)

This is the basic reaction of nitrite ion, so we need the Kb of NO₂⁻ to  find the hydroxide concentration.

Answer:

[tex]T_b=-88.48\°C[/tex]

Explanation:

Hello,

In this case, since the entropy of vaporization is defined in terms of the enthalpy of vaporization and the boiling point of the given substance, nitrous oxide, as shown below:

[tex]\Delta _{vap}S=\frac{\Delta _{vap}}{T_b}[/tex]

Solving for the boiling point of nitrous oxide, we obtain:

[tex]T_b=\frac{\Delta _{vap}H}{\Delta _{vap}S}=\frac{16.53\frac{kJ}{mol}*\frac{1000J}{1kJ} }{89.51\frac{J}{mol} } \\ \\T_b=184.67K[/tex]

Which in degree Celsius is also:

[tex]Tb=184.67-273.15\\\\T_b=-88.48\°C[/tex]

Best regards.

Answer:

PLEASE LOOK INN TO THE FILE YOU WILL GET ANSWER AND ALSO SUMMARY THANKS FOR ASKING QUESTION.

Explanation:

Answer:

35 hrs

Explanation:

half life of the substance [tex]t_{1/2 }[/tex] = 8.1 hr

initial amount [tex]N_{0}[/tex] = 75 g

The final amount [tex]N[/tex] = 3.9 g

The time elapsed [tex]t[/tex] = ?

we use the relationship

[tex]N[/tex] = [tex]N_{0}[/tex] [tex](\frac{1}{2} )^{\frac{t}{t_{1/2} } }[/tex]

substituting values, we have

3.9 = 75 x [tex]\frac{1}{2}^{\frac{t}{8.1} }[/tex]

0.052 = [tex]\frac{1}{2}^{\frac{t}{8.1} }[/tex]

take the log of both side

log 0.052 = log  [tex]\frac{1}{2}^{\frac{t}{8.1} }[/tex]

log 0.052 = [tex]\frac{t}{8.1}[/tex] log 1/2

-1.284 =  [tex]\frac{t}{8.1}[/tex]  x -0.301

1.284 = 0.301t/8.1 =

1.284 = 0.0372t

t = 1.284/0.037 = 34.5 ≅ 35 hrs

Answer:

105/22*(1.251-0.620)

105/22*0.631

4.772*0.631

3.011132

Hope it helps

Answer:

3.0

Explanation:

First, complete the operations inside the parenthesis according to the normal rules for significant figures. Because there are subsequent calculations, keep at least one extra significant figure when possible: (4.7727) × (0.631).

The final product will be rounded to two significant figures because it can’t be more precise than the least precise number in the problem, 22. The final product is 3.0.

Answer:

Explanation:

Given that:

The argon atoms are excited into an excited state before emitting the 488.0 nm laser.

the energy of the first ionization energy of argon is 1520 kJ mol-1.

SInce 1 eV = 96.49 kJ/mol

Therefore, the energy of the first ionization energy of argon in eV is = ( 1520/ 96.49) eV

=  15.75 eV

To find where  the energy level of the excited state lies below the vacuum energy level, let's first determine, the energy liberated by using planck expression.

[tex]E = \dfrac{hc}{\lambda}[/tex]

[tex]E = \dfrac{6.6 \times 10^{-34} \times 3 \times 10^8}{488 \times 10^{-9}}[/tex]

[tex]E = \dfrac{1.98 \times 10^{-25}}{488 \times 10^{-9}}[/tex]

[tex]E = \dfrac{1.98 \times 10^{-25}}{488 \times 10^{-9}}[/tex]

[tex]E =4.057 \times 10^{-19} \ J[/tex]

Converting Joules (J) to eV ; we get,

[tex]E =\dfrac{4.057 \times 10^{-19}}{1.6 \times 10^{-19}}[/tex]

E = 2.53 eV

The energy levels of the first exited state = -13.223 eV

Answer:

The answer is "[tex]\bold{\log \frac{[0] mole}{[R]mole}}[/tex]"

Explanation:

[tex]E_{cell} =E_{cell}^{\circ} - \frac{0.0591}{n}= \log\frac{[0]}{[R]}\\[/tex]

In the above-given equation, we can see from [tex]E_{ceu}[/tex], of both oxidant [tex]conc^n[/tex]as well as the reactant were connected. however, weight decreases oxidant and reduction component concentration only with volume and the both of the half cells by the very same factor  and each other suspend

[tex]\to \log \frac{\frac{\text{oxidating moles}}{25 \ ml}}{\frac{\text{moles of reduction}}{25 ml}} \ \ = \ \ \log \frac{\frac{\text{oxidating moles}}{30 \ ml}}{\frac{\text{moles of reduction}}{30 ml}} \\\\\\[/tex]

[tex]\to {\log \frac{[0] mole}{[R]mole}}[/tex]

The cell potential of the electrochemical reaction has been the same when the volume has been reduced from 30 mL to 25 mL in each half cells.

The cell potential has been given as the difference in the potential of the two half cells in the electrochemical reaction.

The two cells has been set with the concentration of solutions in the oxidation and reduction half cells.

The cell potential has been changed when there has been a change in the potential of the half cells.

The volume of 30 mL to the solution has been, resulting in the cell potential difference of x.

With the volume of 25 mL, there has been the difference in the potential being similar to the 30 mL solution, i.e. x.

Thus, the cell potential of the electrochemical reaction has been the same when the volume has been reduced from 30 mL to 25 mL in each half cells.

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Answer:

H+/PO-4^-2

Explanation:

hydrogen has dissolved completely

In the given reaction conjugate base is H₂PO₄⁻, it also behave as a weak acid.

An acid and conjugate base pairs are those pairs in which they are differentiated by the one atom of hydrogen atom.

Given chemical reaction is:

H₃PO₄ + H₂O → H₂PO₄⁻ + H₃O⁺

In the above reaction H₃PO₄ is an acid as it gives H⁺ ion to the solution and formed H₂PO₄⁻, which is a conjugate base of H₃PO₄ acid. H₂PO₄⁻ will also behave as an acid because it have H⁺ ion to gives in the solution but nature of this acid is weak as they not readily dissociates.

Hence, H₂PO₄⁻ is a conjugate base.

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Answer:

All chemical reactions involve energy. Energy is used to break bonds in reactants, and energy is released when new bonds form in products. Endothermic reactions absorb energy, and exothermic reactions release energy. The law of conservation of energy states that matter cannot be created or destroyed.

Answer:

Corrective lense or just lens.

Explanation:

Answer:

123.96 g Na₂O

Explanation:

4 Na  +  O₂  ⇒  2 Na₂O

You first need to find the limiting reagent.  Convert the reactants to moles and see which produces the least amount of product using the mole ratios in the chemical equation.

(91.96 g Na)/(22.99 g/mol Na) = 4 mol Na

(4 mol Na) × (2 mol Na₂O/4 mol Na) = 2 mol Na₂O

(32.0 g O₂)/(32.0 g/mol) = 1 mol O₂

(1 mol O₂) × (2 mol Na₂O/1 mol O₂) = 2 mol Na₂O

Since they both produce the same amount of product, you don't need to pick a limiting reagent.  Now, convert moles of Na₂O to grams.

(2 mol Na₂O) × (61.98 g/mol Na₂O) = 123.96 g Na₂O

Answer:

Chemistry is one of the causes for global warming, and in some cases it can even cause certain illnesses.

Answer:

Chemists have both hurt the environment and helped the environment by their actions.

Explanation:

<3

Answer:

I would say c) a closed circuit.

Hope I was right.

Answer:

Diferentes tipos de arcilla

ARCILLA DE LADRILLOS. Contiene muchas impurezas. ...

ARCILLA DE ALFARERO. Llamada también barro rojo y utilizada en alfarería y para modelar. ...

ARCILLA DE GRES. Es una arcilla con gran contenido de feldespato. ...

ARCILLAS “BALL CLAY” O DE BOLA. ...

CAOLIN. ...

ARCILLA REFRACTARIA. ...

BENTONITA.

Explanation:

Answer:

Coil method and the slab method.

Explanation:

Answer:

Sodium, magnesium, aluminium, silicon, phosphorus, sulfur, chlorine, and argon are the elements of third period.

Explanation:

There are three energy levels in sodium atom. It has 11 electrons revolving around the nucleus. the atomic radius of sodium atom is 227 ppm. Magnesium, aluminium, silicon, phosphorus, sulfur, chlorine, and argon has also three energy levels like sodium because all these elements belongs to third period. There are 12 electrons present in magnesium, 13 in aluminium, 14 in silicon, 15 in phosphorus, 16 in sulfur, 17 in chlorine, and 18 electrons in argon. The atomic radius of magnesium atom is 173 ppm.  The atomic radius of aluminium atom is 143 ppm.  The atomic radius of silicon atom is 111 ppm.  The atomic radius of phosphorus atom is 98 ppm.  The atomic radius of sulfur atom is 87 ppm. The atomic radius of chlorine atom is 79 ppm and the atomic radius of argon atom is 71 ppm.

Answer:

[tex]C_{C_4H_6}=0.179M[/tex]

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

[tex]2C_4H_6\rightarrow C_8H_{12}[/tex]

And the rate law is:

[tex]\frac{dC_{C_4H_6}}{dt}=kC_{C_4H_6}^2[/tex]

Which integrated is:

[tex]\frac{1}{C_{C_4H_6}} =\frac{1}{C_{C_4H_6}^0}+kt[/tex]

In such a way, the concentration after 10.0 min is:

[tex]\frac{1}{C_{C_4H_6}} =\frac{1}{0.200M}}+5.76x10^{-2}\frac{L}{mol*min}*10.0min\\ \\\frac{1}{C_{C_4H_6}}=5.58\frac{L}{mol} \\\\C_{C_4H_6}=\frac{1}{5.58\frac{L}{mol} } \\\\C_{C_4H_6}=0.179M[/tex]

Regards.

Answer:

b) dime

Explanation:

a dime is approximately 2.2g

half of this is 1.1g, which can be rounded down to one gram.

hope this helps

Answer:I think it's Geosphere

Explanation:

Answer:

A

Explanation:

Geo means rock, or earth. Hydro means water, Atmosphere is space, and Bio global ecosystem composed of living organisms

Answer:

Bisphenol A (BPA)

Explanation:

Bisphenol A (BPA) is a chemical additive commonly found in resins and plastics, such as water bottles or food containers. It can also be found in household electronics, medical devices, dental fillings and sales receipts, just to name a few other applications.

Answer:

I. dipole-dipole

III. dispersion

IV. hydrogen bonding

Explanation:

Intermolecular forces are weak attraction force joining nonpolar and polar molecules together.

London Dispersion Forces are weak attraction force joining non-polar and polar molecules together. e.g O₂, H₂,N₂,Cl₂ and noble gases. The attractions here can be attributed to the fact that a non -polar molecule sometimes becomes polar because the constant motion of its electrons may lead to an uneven charge distribution at an instant.

Dispersion forces are the weakest of all electrical forces that act between atoms and molecules. The force is responsible for liquefaction or solidification of non-polar substances such as noble gas an halogen at low temperatures.

Dipole-Dipole Attractions are forces of attraction existing between polar molecules ( unsymmetrical molecules) i.e molecules that have permanent dipoles such as HCl, CH3NH2 . Such molecules line up such that the positive pole of one molecule attracts the negative pole of another.

Dipole - Dipole attractions are more stronger than the London dispersion forces but weaker than the attraction between full charges carried by ions in ionic crystal lattice.

Hydrogen Bonding is a dipole-dipole intermolecular attraction which occurs when hydrogen is covalently bonded to highly electronegative elements such as nitrogen, oxygen or fluorine. The highly electronegative elements have very strong affinity for electrons. Hence, they attracts the shared pair of electrons in the covalent bonds towards themselves, leaving a partial positive charge on the hydrogen atom and a partial negative charge on the electronegative atom ( nitrogen in the case of CH3NH2 ) . This attractive force is know as hydrogen bonding.

Answer:

The intermolecular forces present in CH_3NH_2 includes

Explanation:

The intermolecular forces present in CH_3NH_2 includes II. (ion-dipole) and IV. (hydrogen bonding)

It is a polar molecule due to NH polar bond and it can form Hydrogen bond also due to NH bond.

Interaction will be dipole- dipole and Hydrogen dispersion forces can always be taken into account.

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Answer:

7.50 L

Explanation:

The balloon has a volume of 1.20 L (V₁) when the pressure at the sea floor is 6.25 atm (P₁). When it reaches the surface, the pressure is that of the atmosphere, that is, 1.00 atm (P₂). If we consider the gas to behave as an ideal gas and the temperature to be constant, we can calculate the final volume (V₂) using Boyle's law.

P₁ × V₁ = P₂ × V₂

V₂ = P₁ × V₁ / P₂

V₂ = 6.25 atm × 1.20 L / 1.00 atm

V₂ = 7.50 L

Answer:

B. NH3 < PH3< CH4

Explanation:

Hello,

In this case, taking into account that the boiling point of ammonia, methane and phosphorous trihydrate are -33.34 °C , -161.5 °C  and -87.7 °C , clearly, methane has the lowest boiling point (most negative) and ammonia the greatest boiling point (least negative), therefore, ranking is:

B. NH3 < PH3< CH4

Best regards.

Answer:

A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as 5. molar mass decreases.

B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as 3. molar mass and structure complexity decreases.

C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as 4. structure complexity decreases.

Explanation:

Hello.

In this case, we can understand a higher entropy when more disorder is present and a lower entropy when less disorder is present, thus:

A (I_2(g), Br_2 (g), Cl_2 (g), F_2 (B): The ranking can best be explained by the trend entropy decreases as 5. molar mass decreases since iodine has the greatest molar mass (254 g/mol) and fluorine the least molar mass (38 g/mol).

B (H_2O_2 (g), H_2S(g), H_2O(g): The ranking can best be explained by the decreases a trend entropy decreases as 3. molar mass and structure complexity decreases since hydrogen peroxide weights 34 g/mol as well as hydrogen sulfide but the peroxide has more bonds (more complex, higher entropy).

C. (C(s, amorphous), C(s, graphite), C(s, diamond): The ranking can best be explained by the trend entropy decreases as 4. structure complexity decreases since diamond has a well-ordered structure and amorphous carbon has a very disordered one.

Best regards.

Answer:

\text{0.30 cm}^{3} \times \left (\dfrac{10^{-2}\text{ m}}{\text{1 cm}}\right )^{3} = 3.0 \times 10^{-7} \text{ m}^{3}  

Explanation:

0.030 cm³ × ? = x m³

You want to convert cubic centimetres to cubic metres, so you multiply the cubic centimetres by a conversion factor.

For example, you know that centi means "× 10⁻²", so  

1 cm = 10⁻² m

If we divide each side by 1 cm, we get 1 = (10⁻² m/1 cm).

If we divide each side by 10⁻² m, we get (1 cm/10⁻² m) = 1.

So, we can use either (10⁻² m/1 cm) or (1 cm/10⁻² m) as a conversion factor, because each fraction equals one.

We choose the former because it has the desired units on top.

The "cm" is cubed, so we must cube the conversion factor.

The calculation becomes

[tex]\text{0.30 cm}^{3} \times \left (\dfrac{10^{-2}\text{ m}}{\text{1 cm}}\right )^{3} = 0.30 \times 10^{-6}\text{ m}^{3} = \mathbf{3.0 \times 10^{-7}} \textbf{ m}^{\mathbf{3}}\\\\\textbf{0.30 cm}^{\mathbf{3}} \times \left (\dfrac{\mathbf{10^{-2}}\textbf{ m}}{\textbf{1 cm}}\right )^{\mathbf{3}} = \mathbf{3.0 \times 10^{-7}} \textbf{ m}^{\mathbf{3}}[/tex]

Answer:

That means that if you are calculating entropy change, you must multiply the enthalpy change value by 1000. So if, say, you have an enthalpy change of -92.2 kJ mol-1, the value you must put into the equation is -92200 J mol-1

The entropy change in the surroundings associated with this reaction occurring at 25 degree C is calculated as ΔS = -ΔH/T J/K.

Entropy is a quantity which gives idea about the randomness or arrangement of atoms or molecules present in any sample.

Entropy change will be calculated as:
ΔS = -ΔH/T, where

ΔH = chnage in enthalpy (J/mole)

T = temperature (K)

So to calculate the entropy change first we have to know about the value of enthalpy in joules and then divide it by the temperature.

Hence the unit of entropy is joule per kelvin.

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