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2 years ago

6

Four groups of students were each asked to create table listing two elements and two compounds. Only one group’s table was corre

ct. Which group’s table is correct? [Can someone just please answer]

1 answer:

Alex_Xolod [135]2 years ago

5 0

The answer would be group 2

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Hydroxyl radicals react with and eliminate many atmospheric pollutants. However, the hydroxyl radical does not clean up everythi

Nesterboy [21]

Answer:

ΔHreaction = 263.15 kJ/mol

Explanation:

The reaction is as follow:

OH + CF₂Cl₂ → HOF + CFCl₂

You need to calculate the enthalpy of reaction and for this it is necessary to know the standard enthalpies for each of the compounds. These enthalpies are as follows and can be found in your textbook or on the Internet.

ΔHreaction = ∑ΔHproducts - ∑ΔHreactants

$delta(H)_{reaction} =((1*(-97.8)+(1*(-92))-((1*39)+(1*(-491.15))=263.15kJ/mol$

7 0

2 years ago

Both the acetamido (-NHCOCH3) and bromo (-Br) substituents are ortho, para directors. Even though the acetamido group is much la

sweet [91]

Answer:

Explanation has been given below

Explanation:

Attachment of -Cl group at ortho position to acetamido group can be explained in terms of hydrogen bonding formation.
A hydrogen bonding can be formed between proton attached with N atom in acetamido group and Cl atom at ortho position. This leads to a formation of a stable 5-membered ring structure.
Hence ortho substitution by Cl is favorable process.
There is no hydrogen bonding possible between Br and Cl atom in ortho position. Therefore Cl prefers para position to avoid steric hindrance.
Structure of hydrogen bonded structure has been shown below.

8 0

2 years ago

A patient is prescribed 100mg/day of antibiotic for 2 weeks. The antibiotic is available in vials that contain 20mg/vial of the

Nimfa-mama [501]

I have no idea, but i’m pretty sure you gotta divide

4 0

2 years ago

Read about reverse osmosis and multistage flash distillation .Then identify at least three advantages and three

ss7ja [257]

Answer:

Meaning, Advantages & Disadvantages of - Osmosis, Multiflash Distillatiob

Explanation:

Osmosis is movement of solvent (like water) through semi permeable membrane (like living cell) into solution of higher solute concentration.

Advantages - It assists equalising concentration of solute on two sides of membrane. Reverse Osmosis is used for efficient water softening, it is easy to maintain.

Disadvantages - It needs a lot of energy. A lot of pressure is required for deionisation. Water acidity level increases, as minerals get deionised.

Multistage flash distillation refers to desalination water distilling seawater, by flashing water portions in steams in various stages of concurrent heat exchangers.

Advantages - Its Cost efficient , distillation uses waste heat. It has High gain output ratio. Quality of feedwater is less significant, compared to reverse osmosis.

Disadvantages - It has high operating cost in case of waste heat unavailability. High temperature imply high corrosion & scale formation.

7 0

2 years ago

A 20.7167 g sample of impure magnesium carbonate was heated to complete decomposition according to the equation mgco3(s) → mgo(s

marishachu [46]

The mass of impure $MgCO_{3}$ is 20.7167 g. The decomposition reaction is as follows:

$MgCO_{3}(s)\rightarrow MgO(s)+CO_{2}(g)$

Here, 1 mole of $MgCO_{3}$ gives 1 mole of MgO.

Molar mass of $MgCO_{3}$ and MgO is 84.31 g/mol and 40.3044 g/mol respectively.

Converting number of moles in terms of mass,

$n=\frac{m}{M}$

Here, M is molar mass.

Since, $n_{MgCO_{3}}=n_{MgO}$

Thus, $\frac{m_{MgCO_{3}}}{M_{MgCO_{3}}}=\frac{m_{MgO}}{M_{MgO}}$

On putting the values,

$\frac{m_{MgCO_{3}}}{84.31}=\frac{m_{MgO}}{40.3044}$

Rearranging,

$m_{MgO}=\frac{m_{MgCO_{3}}}{84.31}\times 40.3044$

Or,

$m_{MgCO_{3}}=\frac{m_{MgO}}{0.4780}$ ...... (1)

Let the mass of impurity be $M_{I}$ and mass of impure $MgCO_{3}$ is 20.7167 g thus,

$M_{MgCO_{3}}=(20.7167-M_{I})g$ ...... (2)

Also, mass of impure MgO is 16.8817 g thus,

$M_{MgO}=(16.8817-M_{I})g$ ...... (3)

On comparing equations (2) and (3),

$M_{MgO}=M_{MgCO_{3}}-3.835$

Putting the value of $M_{MgO}$ in equation (1),

$m_{MgCO_{3}}=\frac{M_{MgCO_{3}}-3.835}{0.4780}$

Or,

$0.4780 M_{MgCO_{3}}=M_{MgCO_{3}}-3.835$

Or,

$M_{MgCO_{3}}-0.4780M_{MgCO_{3}}=3.835$

Or,

$0.522 M_{MgCO_{3}}=3.835$

Or,

$M_{MgCO_{3}}=\frac{3.835}{0.522}=7.35 g$

Thus, magnesium carbonate present in the original sample is 7.35 g.

5 0

2 years ago