In the picture this is my last question pls.

Answer:

5–bromo–9–chlorodecane

Explanation:

To name the compound given above, the following must be obtained:

1. The longest continuous carbon chain. This gives the parent name of the compound.

2. The substituent group attached to the compound.

3. Position of the substituent group.

4. Combine the above to obtain the name.

Now, we shall determine the name of the compound as follow:

1. The longest continuous carbon chain is 10. Thus, the parent name of the compound is decane.

2. The substituent groups attached to the compound are:

I. Bromine (Br) => Bromo

II. Chlorine (Cl) => Chloro

3. The position of the substituent groups are:

I. Br => carbon 5

II. Cl => carbon 9

NOTE: numbering is done alphabetically.

4. Therefore, the name of the compound is:

5–bromo–9–chlorodecane

Answer:

A.

Explanation:

I chose this answer and it was correct ‍♀️

Answer:

Decrease temperature or reduce pressure

Answer:

Explanation:

Hope this helps

Answer:

See explanation.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to realize that the NaCl solution must react with an aluminum-containing substance, say the hydroxide or any other salt, so that the following equation will take place:

[tex]3NaCl(aq)+Al^{3+}\rightarrow 3Na^+(aq)+AlCl_3(s)[/tex]

In such a way, given the volume of the NaCl solution, it must be necessary to know its concentration, in order to get moles of this salt, further use the 3:1 mole ratio of NaCl to AlCl3 and the molar mass of the latter (133.34 g/mol) in order to solve an stoichiometric setup like the following:

[tex]m_{AlCl_3}=V_{NaCl}*M_{NaCl}*\frac{1molAlCl_3}{3molNaCl} *\frac{133.34gAlCl_3}{1molAlCl_3}[/tex]

Besides, you must make sure the volume is in liters.

Best regards!

Equilibrium shifts to the right.

OPTION A

Answer:

ee = 96%

Explanation:

Enantiomeric excess, ee, is a way to express a mixture that is not enantiomerically pure. It is defined as 100 times the ratio between the  differences of amounts of enantiomers and the total amunt. that is:

ee = |A-B|/ A+B * 100

ee = |98%-2%| / 98+2 * 100

Answer:

Research that releases a poisonous gas into the air.

Explanation:

Since I don't know the options I will guess it is ^

Answer:

6.33×10¯²² g

Explanation:

From the question given above, the following data were obtained:

Number of atoms = 6 atoms

Mass of copper (Cu) =?

From Avogadro's hypothesis, we understood that:

6.02×10²³ atoms = 1 mole of Cu

But 1 mole of Cu = 63.5 g

Thus,

6.02×10²³ atoms = 63.5 g of Cu

Finally, we shall determine the mass of 6 atoms of copper. This can be obtained as illustrated below:

6.02×10²³ atoms = 63.5 g of Cu

Therefore,

6 atoms = (6 × 63.5) / 6.02×10²³

6 atoms = 6.33×10¯²² g of Cu

Therefore, the mass of 6 atoms of copper is 6.33×10¯²² g.

Answer:

Phenols do not exhibit the same pka values as other alcohols;

They are generally more acidic.

Using the knowledge that hydrogen acidity is directly related to the stability of the anion formed, explain why phenol is more acidic than cyclohexane.

Explanation:

According to Bromsted=Lowry acid-base theory,

an acid is a substance that can release [tex]H^{+}[/tex] ions when dissolved in water.

So, acid is a proton donor.

If the conjugate base of an acid is more stable then, that acid is a strong acid.

In the case of phenol,

the phenoxide ion formed is stabilized by resonance.

[tex]C_6H_5OH -> C_6H_5O^- +H^+[/tex]

The resonance in phenoxide ion is shown below:

Whereas in the case of cyclohexanol resonance is not possible.

So, cyclohexanol is a weak acid compared to phenol.

Answer:

A sound wave is the pattern of disturbance caused by the movement of energy traveling through a medium (such as air, water, or any other liquid or solid matter) as it propagates away from the source of the sound. The source is some object that causes a vibration, such as a ringing telephone, or a person's vocal chords.

HEY. HOPE THIS HELPS♡

Answer:

4.7 g

Explanation:

The reaction that takes place is:

First we convert 4.8 g of CaCl₂ into moles, using its molar mass:

Then we convert CaCl₂ moles into Na₃PO₄ moles, using the stochiometric coefficients of the balanced reaction:

Finally we convert 0.0288 moles of Na₃PO₄ into grams, using its molar mass:

Answer:

Examples. Glucose (C6H12O6), ribose (C5H10O5), Acetic acid (C2H4O2), and formaldehyde (CH2O) all have different molecular formulas but the same empirical formula: CH2O.

Explanation:In chemistry, the empirical formula of a chemical compound is the simplest positive integer ratio of atoms present in a compound.

Answer:

Equation 1 - nuclear fission

Equation 2 - nuclear fusion

Explanation:

Nuclear fission is a reaction in which a large nucleus is split into smaller nuclei when it is bombarded by neutrons. The process produces more neutrons to continue the chain reaction. This is clearly depicted in equation 1 as shown in the question.

Nuclear fusion is a reaction in which two light nuclei combine in order to form a larger nuclei. This is clearly depicted in equation 2 as shown in the question.

In the first reaction, a neutron is released, and in the second a helium atom is released. The given two equations represent nuclear fission and fusion.

A nuclear reaction is a reaction that involves the nuclei of the atom and the absorption and release of energy. In the first reaction, a big nucleus is split by the neutron bombardment into smaller nuclei.

In the second reaction the process of nuclear fusion, two nuclei combine into a single larger nucleus that is shown as:

₁¹H+ ²₁H → ³₂He

Therefore, nuclear fission and fusion are represented by each of these equations.

Learn more about nuclear reactions here:

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answer is 2546 h

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Answer:

Light can be treated like particles because it is made of chunks like things called protons.

Answer:

68.97g of milk must be added

Explanation:

To solve this problem we need to use the equation:

Q = C*m*ΔT -Coffee cup calorimeter equation-

This equation relates mass and change in temperature with heat of solution, Q.

The energy that decreases in the milk is the same that increases in the milk. That is:

Q(milk) = Q(tea)

Using the equation:

4.184J/molK* Mass milk * (65.0°C - 7.00°C) = 4.184J /molK * 200g (Tea) * (85.0°C - 65.0°C)

Mass milk * 58.0°C = 200g * 20.0°C

Mass milk = 4000g°C / 58.0°C

Mass milk =

68.97g of milk must be added

Answer:

The correct answer is - A. ammonia blue.

Explanation:

It is given that there is a mixture of ammonium chloride and sodium hydroxide and if we heat the  Mixture of NH4Cl and NaOH then, the following reaction will take place-

NH4Cl (aq)+ NaOH (aq)==> NaCl (aq) + NH3 (g)+ H2O (l)

In this reaction, there is a release of NH3 gas which is absorbed on litmus paper. We know that NH3 gas is basic in nature and so it will turn litmus paper blue.

Thus, the correct answer would be - NH3 (ammonia ) and Blue

Answer:

Explanation:

Both extraction and wash are technical materials separation processes. Both seek to carry out a separation of liquid-liquid, through the relationship between solvent and solute present in this solution. Both use a separation funnel and are very similar processes where the main difference is the purpose that each one seeks to exercise. The extraction causes the dissolution of the material that the researcher wants to leave behind. this process leaves impurities in the initial solution layer and moves the researcher's desired compound to the second layer of solution. A chemical reaction can take place during this process.

A wash, on the other hand, dissolves all impurities from the solution and extracts the compound desired by the researcher. Therefore, we can conclude that during this process the impurities are moved from one layer to another in the solution until it is discarded from the funnel, leaving the desired compound in the initial layer.

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.

Explanation:

Science is the pursuit and application of knowledge and understanding of the natural and social world following a systematic methodology based on evidence

Answer: In order to increase the rate of reaction between hydrochloric acid and sugar increase the concentration of hydrochloric acid to 2 M because greater concentration results in more collision between the reactants.

Explanation:

More is the concentration of reactant molecules more will be the number of collisions between their molecules. As a result, more readily the products will be formed.

Hence, for the given reaction when concentration of HCl is increased then there will be increase in the number of collisions between reactants.

Thus, we can conclude that in order to increase the rate of reaction between hydrochloric acid and sugar increase the concentration of hydrochloric acid to 2 M because greater concentration results in more collision between the reactants.

Answer:

128.4 g

Explanation:

Step 1: Given data

Step 2: Determine the total mass of the sample

The mass percent of element X in X(NO₃)₂ is 52.55%, that is, there are 52.55 g of X every 100 g of X(NO₃)₂. Then, the mass of X(NO₃)₂ that contains 67.50 g of X is:

67.50 g X × 100 g X(NO₃)₂/52.55 g X = 128.4 g X(NO₃)₂

Answer:

solo I

Explanation:

Según esta teoría para que se produzca una reacción deben cumplirse tres condiciones: Las moléculas de los reactivos tienen que chocar entre sí. Estos choques deben de producirse con energía suficiente de forma que se puedan romper y formar enlaces químicos.

Answer:

The density of water is about 1.0 g/mL at room temperature.

Briefly explain how the density of an aqueous solution at room temperature can be significantly less than 1.0 g/mL.

Give an example of such a solution.

Explanation:

That means 1.0mL of water weighs ---- 1.0g

If any other aqueous solution which has mass less than 1.0g will have density less than 1.0g.

For example aqueous solution of sulfuric acid has density 0.98g/mL.

That means 1mL of sulfuric acid has mass 0.98g.

Dissolving gases or liquids in water whose density are less than that of water will result in an aqueous solution whose density is less than that of water, for example, sulfuric acid which has a density 0.98g/mL.

The density of a substance is the ratio of the mass to the volume of that substance.

Density measures compactness.

More compact substances are more dense.

The density of water is about 1.0 g/mL at room temperature.

This means that 1.0mL of water weighs 1.0g

1.0 mL of an aqueous solution whose mass is less than 1.0g will have density less than that of water.

Dissolving gases or liquids in water whose density are less than that of water will result in an aqueous solution whose density is less than that of water.

An example of an aqueous solution whose density is less than that of water is sulfuric acid which has a density 0.98g/mL.

Sulfuric acid is prepared by dissolving sulphur (vi) oxide, a gas, in water.

This means 1mL of sulfuric acid has mass 0.98g.

Learn more about density at: https://brainly.com/question/406690

Answer:

-713 kJ

Explanation:

Step 1: Write the balaned thermochemical equation

4 NH₃(g) + 3 O₂(g) → 2 N₂(g) + 6 H₂O(g)  ΔΗ = -1267 kJ

Step 2: Calculate the moles corresponding to 38.4 g of NH₃

The molar mass of NH₃ is 17.03 g/mol.

38.4 g × 1 mol/17.03 g = 2.25 mol

Step 3: Calculate the  enthalpy change to burn 2.25 mol of ammonia

According to the thermochemical equation, 1267 kJ are released per 4 moles of ammonia that react.

2.25 mol × (-1267 kJ/4 mol) = -713 kJ

Answer:

[tex]C_3H_7O_3[/tex]

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to infer that the empirical formula of mannitol contains carbon, hydrogen and oxygen, so that the first step is to calculate the moles of C and H contained in the CO2 and H2O, respectively, as the only sources of these two elements in the formula:

[tex]n_C=1.993gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2} =0.0453molC\\\\n_H=0.9519gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O} =0.106molH[/tex]

Next, we calculate the grams and moles of O by subtracting the mass of C and H from the mass of the sample:

[tex]m_O=1.375g-0.0453molC*\frac{12gC}{1molC}-0.106molH*\frac{1.01gH}{1molH}=0.724gO\\\\n_O=0.724gO*\frac{1molO}{16.0gO} =0.0453molO[/tex]

Finally, we divide the moles of C, H and O by 0.0453 as the fewest moles of both C and O to find the mole ratios in the formula:

[tex]C:\frac{0.0453mol}{0.0453mol} =1\\\\H:\frac{0.106mol}{0.0453mol} =2.34\\\\O:\frac{0.0453mol}{0.0453mol} =1[/tex]

To get:

[tex]CH_{2.34}O[/tex]

Which must be multiplied by 3 to get whole numbers for all the subscripts, and therefore obtain:

[tex]C_3H_7O_3[/tex]

Regards!

Answer:

A. 1. Strong acid (sa): Hydrobromic acid: HBr (aq)

2. Strong acid (sa); Hydrochloric acid: HCl (aq)

3. Weak acid (wa); Carbonic acid: H₂CO₃ (aq)

B. H+ (aq) + OH- (aq) ----> H₂O (l)

Explanation:

Strong acids are which ionize completely in aqueous solution into hydrogen ions and the corresponding anion. Examples of strong acids include hydrobromic acid, hydrochloric acid, tetraoxosulfate (vi) acid.

The ionization of hdyrobromic and hydrochloric acids in aqueous solution is given below:

1. Hydrobromic acid: HBr (aq) ----> H+ (aq) + Br- (aq)

Hydrobromic acid in aqueous solution ionizes completely into hydrogen ions and bromide ions

2. Hydrochloric acid: HCl (aq) ----> H+ (aq) + Cl- (aq)

Hydrochloric acid in aqueous solution ionizes completely into hydrogen ions and chloride ions

Weak acids are acids which ionizes only partially in aqueous solutions to hydrogen ions and the corresponding anions. Examples of weak acids are carbonic acid and ethanoic acid. The ionization of carbonic acid in aqueous solution is shown below:

3. Carbonic acid: H₂CO₃ (aq) ⇄ 2 H+ (aq) + CO₃²- (aq)

Carbonic acid ionizes partially only to give hydrogen ions and trioxocarbonate (iv) ions. The unionized acid exists in equilibrium with the ions produced by the partial ionization of the acid.

Part B:

The reaction between hydrochloric acid and barium hydroxide is a neutralization reaction producing barium chloride salt and water.

The net ionic equation of the neutralization reaction is given below :

H+ (aq) + OH- (aq) ----> H₂O (l)

Answer:

Explanation:

From the graphical diagram attached below; we can see the relationship between the concentration of [tex]H_2O_2[/tex] which declines exponentially in relation to the time and it obeys the equation: [tex]\mathtt{y = 0.9951 e^{-8\times 10^{-4}x}}[/tex]

This relates to the 1st order reaction rate, whereby:

The integrated rate law[tex]\mathtt{ [A] = [A]_o e^{-kt}}[/tex]

here:

[A] = reactant concentration at time (t)

[A]_o = initial concentration for the reactant

k = rate constant

As such, the order of the reaction is the first order

Rate constant [tex]\mathtt{k = 8\times 10^{-4} {s^{-1}}}[/tex]

Rate law [tex]\mathtt{= k[H_2O_2]}[/tex]

The integrated rate law [tex]\mathtt{[H_2O_2] = [H_2O_2]_oe^{-(8*10^{-4})t}}[/tex]

From the given table:

the initial concentration of [tex]H_2O_2[/tex] = 1.00 M

∴

We can determine the concentration of the reactant at 4000s by using the formula:

[tex]\mathtt{[H_2O_2] = [H_2O_2]_oe^{-8*10^{-4}(t)}}[/tex]

[tex]\mathtt{[H_2O_2] = (1.00\ M)*e^{-8*10^{-4}(4000)\ s}}[/tex]

[tex]\mathtt{[H_2O_2] =0.0407 \ M}[/tex]

Finally, at 4000s: the average rate is:

[tex]\mathtt{= (8*10^{-4} \ s^{-1})(4000 \ s) }\\ \\ \mathtt{ = 3.256 \times 10^{-5} \ M/s}[/tex]