37.4 g of aluminum chloride reacts with bromine gas to yield aluminum bromide and chlorine gas. How many liters of bromine gas is needed to completely react with aluminum chloride? Assume both gases are at STP.

Answer:

B. Solidification

Hope this helped! :)

Answer:

pH = 8.18

Explanation:

The weak base, X, reacts with HCl as follows:

X + HCl → HX⁺ + Cl⁻

Where 1 mole of X with 1 mole of HCl produce 1 mole of HX⁺ (The conjugate acid of the weak base).

Now, using H-H equation for bases:

pOH = pKb + log [XH⁺] / [X]

Where pOH is the pOH of the buffer (pH = 14 -pOH)

pKb is -log Kb = 5.824

And [X] [HX⁺] are the molar concentrations of each specie

Now, at the neutralization of the half of HX⁺, the other half is as X, that means:

[X] = [HX⁺]

And:

pOH = pKb + log [HX⁺] / [X]

pOH = 5.824 + log 1

pOH = 5.824

pH = 14-pOH

Answer:True

Explanation:

PLZ GIVE ME BRAINLIEST, I NEED IT

Answer:

it can be removed

1. by boiling

Ca(HCO3)2 > CaCO3 + H2O +CO2

2. by treating with calcium hydroxide

Answer:

[tex]\frac{1}{\lambda_{6-5}} \approx 752nm[/tex]

Explanation:

From the question we are told that

Light wavelength  [tex]\lambda_l=655nm[/tex]

Light wavelength atom fall [tex]x_L=5th-2nd[/tex]

Photon wavelength  [tex]\lambda_p=435nm[/tex]

Photon wavelength atom fall [tex]x_P=^th-2nd[/tex]

Generally the equation for the reciprocal of wavelength of emitted photon is is mathematically given by

[tex]\frac{1}{\lambda}=R(\frac{1}{ \lambda _f^2}-\frac{1}{\lambda _i^2} )[/tex]

Therefore for initial drop of 5th to 2nd

[tex]\frac{1}{\lambda_{5-2}}=R(\frac{1}{2^2}-\frac{1}{5^2} )[/tex]

Therefore for initial drop of 6th to 2nd

[tex]\frac{1}{\lambda_{6-2}}=R(\frac{1}{2^2}-\frac{1}{6^2} )[/tex]

Generally we subtract (5th to 2nd) from (6th to 2nd)

[tex]\frac{1}{\lambda_{5-2}}- \frac{1}{\lambda_{6-2}}=R(\frac{1}{2^2}-\frac{1}{5^2} )-\frac{1}R(\frac{1}{2^2}-\frac{1}{6^2} )[/tex]

[tex]\frac{1}{\lambda_{5-2}}- \frac{1}{\lambda_{6-2}}=R(\frac{1}{5^2}-\frac{1}{6^2} )[/tex]

[tex]\frac{1}{\lambda_{5-2}}- \frac{1}{\lambda_{6-2}}=\frac{1}{\lambda_{5-6}}[/tex]

[tex]\frac{1}{\lambda_{5-6}}=\frac{1}{4350nm}-\frac{1}{655nm}[/tex]

[tex]\frac{1}{\lambda_{5-6}}=1.33*10^{-3}[/tex]

Therefore for 6th to 5th stage is mathematically given by

[tex]\frac{1}{\lambda_{6-5}}=(1.33*10^{-3})^{-1}[/tex]

[tex]\frac{1}{\lambda_{6-5}}=751.879nm[/tex]

[tex]\frac{1}{\lambda_{6-5}} \approx 752nm[/tex]

Answer:

To ensure stable reading, insert the conductivity meter in the solution for at least 60 seconds.

Explanation:

The False statement is ; To ensure stable reading, insert the conductivity meter in the solution for at least 60 seconds.

All other statement will affect the experiment's redox potential but  inserting the conductivity meter in the solution for at least 60 seconds will not affect the Redox potential  .

Redox potential is used to describe/measure a system's oxidizing capacity

Answer: The molar mass of X is 61.3g/mol

Explanation:

Depression in freezing point is given by:

[tex]\Delta T_f=K_f\times m[/tex]

[tex]\Delta T_f=T_f^0-T_f=(6.55^0C-5.9^0C)=0.65^0C[/tex] = Depression in freezing point

[tex]K_f[/tex] = freezing point constant = [tex]20.2^0C/mol[/tex]

m= molality  = [tex]\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in Kg}}[/tex]

where,  

Now put all the given values in this formula, we get

[tex]0.65^0C=20.2°C/m\times frac{0.148g}}{M\times 0.075kg}[/tex]

[tex]M=61.3g/mol[/tex]

Thus molar mass of X is 61.3g/mol

Answer:

Explanation:

Dear student, this question appears to be incomplete but in the diagram attached below, I have carefully drawn out the required organic structure.

The aim of this question is to show the mechanism of the given structure and to point out the resonance structure.

The first diagram shows the complete part (i.e. the structure of the organic compound) and its major product(answer).

The second diagram shows the mechanism by which we arrived at the major product.

From the second diagram attached below, the -oxo substituent of the methanol attacks the hydrogen atom as shown below, leading to the formation of the enolate ion (a resonance stabilized anion).

However, this is because the alpha carbon attaches to the hydrogens readily leaves as H⁺ to yield an enolate ion as a result of -C- that is doubly bonded to oxygen (an Electron withdrawing group).

Kindly check out the images attached below to see the mechanism.

Answer:

Use the drop-down menus to make your selections.

What is the difference between wave A and wave B?

✔ Wave A has higher amplitude than wave B.

What is the difference between wave C and wave D?

✔ Wave C has a lower frequency than wave D.

Explanation:

The volume of 0.30 M potassium chromate needed to react with 15 ml of 0.10 M ferric nitrate is 5 ml.

Molarity of a solution is the ratio of its number of moles of solute to the volume of solution in liter. It is the common term used to express the concentration of a solution.

Let V1 and M1 be the volume and molarity of solution 1 and V2 and M2 that of solution 2 which are reacting together.

Thus, V1 M1 = V2 M2.

The volume of 0.30 M potassium chromate is calculated as  follows:

volume = (15 ml × 0.10 M) / 0.30 M = 5 ml.

Therefore, the volume of the potassium chromate solution is 5 mL.

To find more on molarity, refer here:

https://brainly.com/question/8732513

#SPJ1

Answer:

1/16

Explanation:

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

Half-life (t½) = 269 years

Time (t) = 1076 years

Fraction remaining =?

Next, we shall determine the number of half-lives that has elapsed. This can be obtained as follow:

Half-life (t½) = 269 years

Time (t) = 1076 years

Number of half-lives (n) =?

n = t / t½

n = 1076 / 269

n = 4

Thus, 4 half-lives has elapsed.

Finally, we shall determine the fraction of the original amount remaining. This can be obtained as follow:

Let N₀ be the original amount.

Let N be the amount remaining.

Number of half-lives (n) = 4

Fraction remaining (N/N₀ ) =?

N = 1/2ⁿ × N₀

N = 1/2⁴ × N₀

N = 1/16 × N₀

Divide both side by N₀

N/N₀ = 1/16

Thus, the fraction of the original amount remaining is 1/16

Answer:

B

Explanation:The poisonous substances, present in the environment can easily get into the trophic level as living organism depends on each other and environment for food and nutrition. These poisonous substances may not be broken down in the body or excreted easily, efficiently and quickly. Instead, they accumulate in the tissues, and as the living organism eats more, the concentration of these substances increases and they pass from one trophic level to the next. The tertiary consumer being at the top of trophic levels receives the maximum pollutant. This phenomenon is known as biological magnification.

So, the correct answer is option B.

Complete question :

Use the equation qliquid = m × c × ΔT to calculate the heat gained by the cold liquid. Use the specific heat for the liquid you selected.

m = 248.9 g ; T = 72.6°C ; C = 3.73 j/g°C

Use the equation qwater = m × c × ΔT to calculate the heat lost by the hot water. Show your work using the problem-solving method shown in previous rubrics.

M = 237 g ; T = 41.2°C ; C = 4.184 j/g°C

Answer:

Quantity of heat gained = 67401.6 Joules

Quantity of heat lost by hot water = 40854.25 Joules

Explanation:

The heat gained by cold liquid :

Q = m × c × ΔT

m = 248.9 g ; T = 72.6°C ; C = 3.73 j/g°C

Q = 248.9 * 3.73 * 72.6

Q = 67401.6 Joules

Quantity of heat gained = 67401.6 Joules

Heat lost by hot water :

m × c × ΔT

M = 237 g ; T = 41.2°C ; C = 4.184 j/g°C

Q = 237 * 41.2 * 4.184

Q = 40854.25 Joules

Quantity of heat lost by hot water = 40854.25 Joules

Answer:

Imine can be isolated from the reaction mixture as water is continuously removed from the reaction chamber

Explanation:

In this reaction, a non -aqueous solvent  is not used (not mentioned in the question). Thus, we can say that there is continuous removal water under suitable reacting conditions and hence the imine formed is left behind.

Answer:

H2SO4 + 2NAOH➡NA2SO4+2H2O

H2SO4 = 1*2 +32+16*3= 98g

NAOH = (11+16+1)=28

Explanation:

To calculate the number of Mole = Mole *Molar Mass

To calculate for gram = Mole/ Molar Mass

20.5g NaoH/1 * 1 mole NaOH/28g NaoH = 0.73M NaoH

0.73M NaoH /1 * 1 M H2SO4 /2M NaoH = 0.37M H2SO4

0.37 M H2SO4 /1 * 98g H2SO4 /1M H2SO4 = 0.00367g H2SO4

:- 0.00367g of Sulphuric Acid will be needed to complete the reaction.

Answer:

CaCl2 + Na2S --> 2NaCl + CaS?​

Explanation:

CaCl2 + Na2S --> 2NaCl + CaS?​

Answer:

A) w/w % of Fe - 2.44%

B) w/w % of Mg - 4.590%

C) w/w % of Ca - 96%

Explanation:

A) w/w % of Fe in limestone as Fe2O3 = (Mass of Fe2O3 /Mass of limestone) x 100

 0.0357/1.4639 X 100

= 2.438 =2.44

B) w/w % of Mg in limestone as Mg2P2O7 = (Mass of Mg2P2O7 /Mass of limestone) x 100

= 0.0672/1.4639 X 100

= 4.590

C) w/w % of Ca in limestone as CaSO4 = (Mass of CaSO4/Mass of limestone) x 100

= 1.4058/1.4639 X 100

= 96

Answer:Fe =1.71%    Ma=0.99%    Ca=28.24%

Explanation:

Answer: A

Explanation:

Answer:

Explanation:

We have the three equations:

[tex]NH_{3(g)} + HCl_{(g)} => NH_4Cl_{(s)} ..... \Delta H = ? (1)\\2HCl_{(g)} => H_{2(g)} + Cl_{2(g)} .... \Delta H = +184.6 kJ (2)\\2H_{2(g)} + 1/2N_{2(g)} + 1/2Cl_{2(g)} => NH_4Cl_{(s)} ..... \Delta H = -314.2 kJ (3)\\ N_{2(g)} + 3H_{2(g)} => 2NH_{3(g)} .... \Delta H = +184.6kJ (4)[/tex]

(can you double check that it is 184.6kJ for both equations 2 and 4 because it seems unlikely). We need to solve for equation 1 by addition and changing equations 2, 3 and 4. After possibly some trial and error, we can find that if we flip equations 4, multiply equation 3 by 2, add the equations together, and then finally divide by 2, we can get equation 1. We will get the answer of -314.2 kJ. However, I am again skeptical about the delta H values for equation 2 and 4 so double check that. This method might be super confusing and it is really hard to explain. So what I would suggest you to watch videos on Hess' law.

Answer:

(we use hess's law) it is so simple but the second reaction is not correct please right it

Answer:

[tex]T_2=484.8K[/tex]

Explanation:

Hello there!

In this case, according to the the variable temperature and pressure and constant volume, it turns out possible for us to calculate the new temperature via the Gay-Lussac's law as shown below:

[tex]\frac{T_2}{P_2} =\frac{T_1}{P_1}[/tex]

Thus, by solving for the final temperature, T2, we obtain:

[tex]T_2 =\frac{T_1P_2}{P_1}[/tex]

So we plug in the given data to obtain:

[tex]T_2 =\frac{303K*200atm}{125atm}\\\\T_2=484.8K[/tex]

Best regards!

Answer:

True

Explanation:

Markovnikov rule states that the negative part of the addendum is added to the carbon atom having the least number of hydrogen atoms.

This means that during acid-catalyzed addition of water to an alkene, the -OH is added to the more substituted carbon atom while the -H is added to the least substituted carbon atom in obedience to Markovnikov regiochemistry.

Answer:

a. True

Explanation:

Acid-catalyzed addition of water to an alkene yields an alcohol with Markovnikov regiochemistry. The electrophilic H+ adds to the sp2 carbon with the most hydrogens to yield the most stable carbocation intermediate, which then adds water to give the product alcohol.

Answer:

38.4%

Explanation:

We'll begin by calculating the molar mass of Al(ClO₃)₃. This can be obtained as follow:

Molar mass of Al(ClO₃)₃

= 27 + 3[35.5 + (16×3)]

= 27 + 3[35.5 + 48]

= 27 + 3[83.5]

= 27 + 250.5

= 277.5 g/mol

Next, we shall determine the mass of Cl in Al(ClO₃)₃. This can be obtained as follow:

Mass of Cl in Al(ClO₃)₃ = 3 × Cl

= 3 × 35.5

= 106.5 g

Finally, we shall determine the percentage of Cl in Al(ClO₃)₃. This can be obtained as follow:

Mass of Cl in Al(ClO₃)₃ = 106.5 g

Mass of Al(ClO₃)₃ = 277.5 g

Percentage of Cl =?

Percentage of Cl = mass of Cl / mass of Al(ClO₃)₃ × 100

Percentage of Cl = 106.5 / 277.5 × 100

Percentage of Cl = 38.4%

Answer:

1)2Na + Cl2 ----> 2NaCl

2)Ca + Br2 ---->CaBr2

3)K + H2O -----> KOH + H2