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Bingel [31]
3 years ago
5

3. What are vaccines? Are there different types?​

Chemistry
2 answers:
Dahasolnce [82]3 years ago
5 0

Answer:

A vaccine is a biological preparation that provides active acquired immunity to a particular infectious disease. Yes there several different types of vaccines.

Inga [223]3 years ago
4 0

Answer:Four types of vaccines are currently available: Live virus vaccines use the weakened (attenuated) form of the virus. The measles, mumps, and rubella (MMR) vaccine and the varicella (chickenpox) vaccine are examples

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please help i dont get chemistry as much as I've only just started and I dont know how to separate all these matters ​
dsp73

is this technical chemistry?

6 0
3 years ago
Sulfonation of benzene has the following mechanism: (1) 2 H2SO4 ⇌ H3O+ + HSO4− + SO3 [fast] (2) SO3 + C6H6 → H(C6H5+)SO3− [slow]
ziro4ka [17]

Question is incomplete, complete question is as follows :

Complete Question : .Sulfonation of benzene has the following mechanism:

(1) 2 H2SO4 ⇌ H3O+ + HSO4− + SO3

[fast]

(2) SO3 + C6H6 → H(C6H5+)SO3−

[slow]

(3) H(C6H5+)SO3− + HSO4− → C6H5SO3− + H2SO4

[fast]

(4) C6H5SO3− + H3O+ → C6H5SO3H + H2O

[fast]

write the overall rate law for the initial rate of the reaction as a fraction.

Rate=k(________/_________)

Answer:

The overall rate law for the initial reaction is = k_{overall} [H_{2}SO_{4}]^{2} [C_{6}H_{6}]

Explanation :

Frist of all, all the common terms are cancelled out and written the overall reaction.

As we know that the rate depednant step is the slowest step of the reaction, rate law is :

                        rate = k_{2} [SO_{3}][C_{6}H_{6}]

But the problem is that SO3 cannot be written in the overall rate law because it is an intermediate.

Rate law for synthesis of S03 is as follows :

                       rate = k_{1}[H_{2}SO_{4}]^{2}

Hence when we substitute equation 2 in equation one,

                   Rate comes out to be =  k_{overall} [H_{2}SO_{4}]^{2} [C_{6}H_{6}]

6 0
3 years ago
In one step in the synthesis of the insecticide Sevin, naphthol reacts with phosgene as shown.
Serhud [2]

Answer:

Explanation:

Chemical equation:

C₁₀H₈O + COCl₂  → C₁₁H₇O₂Cl + HCl

A. How many kilograms of C₁₁H₇O₂Cl form from 2.5×10*2 kg of naphthol?

Given data:

Mass of naphthol = 2.5 ×10² kg ( 250×1000 = 250000 g)

Mass of C₁₁H₇O₂Cl = ?

Solution:

Number of moles of naphthol = mass/ molar mass

Number of moles of naphthol = 250000 g/ 144.17 g/mol

Number of moles of naphthol = 1734.1 mol

Now we will compare the moles of naphthol with C₁₁H₇O₂Cl.

                     C₁₀H₈O       :         C₁₁H₇O₂Cl

                        1               :                1

                       1734.1         :             1734.1

Mass of C₁₁H₇O₂Cl:

Mass = number of moles × molar mass

Mass = 1734.1 mol × 206.5 g/mol

Mass = 358091.65 g

Gram to kilogram:

1 kg×358091.65 g/ 1000 g  = 358.1 kg

B. If 100. g of naphthol and 100. g of phosgene react, what is the theoretical yield of C11H7O2Cl?

Given data:

Mass of naphthol = 100 g

Mass of COCl₂ = 100 g

Theoretical yield of C₁₁H₇O₂Cl = ?

Solution:

Number of moles of naphthol:

Number of moles of naphthol = mass/ molar mass

Number of moles of naphthol = 100 g/ 144.17 g/mol

Number of moles of naphthol = 0.694 mol

Number of moles of phosgene:

Number of moles  = mass/ molar mass

Number of moles =  100 g/ 99 g/mol

Number of moles = 1.0 mol

Now we will compare the moles of naphthol and phosgene with C₁₁H₇O₂Cl.

                     C₁₀H₈O        :         C₁₁H₇O₂Cl

                        1                :                1

                       0.694        :              0.694

                    COCl₂          :             C₁₁H₇O₂Cl

                        1                :                1

                       1.0              :              1.0

The number of moles of C₁₁H₇O₂Cl produced by C₁₀H₈O are less so it will limiting reactant and limit the yield of  C₁₁H₇O₂Cl.

Mass of C₁₁H₇O₂Cl:

Mass = number of moles × molar mass

Mass =  0.694 mol × 206.5 g/mol

Mass = 143.3 g

Theoretical yield  =  143.3 g

C. If the actual yield of C11H7O2Cl in part b is 118 g, what is the percent yield?

Given data:

Actual yield of C₁₁H₇O₂Cl = 118 g

Theoretical yield = 143.3 g

Percent yield = ?

Solution:

Formula :

Percent yield = actual yield / theoretical yield × 100

Now we will put the values in formula.

Percent yield = 118 g/ 143.3 g × 100

Percent yield = 0.82 × 100

Percent yield = 82%

5 0
3 years ago
11. The oxyfuel flame was used for fusion welding as early as the first half of the
Firlakuza [10]

The Oxyfuel gas or flame refers to a group of welding processes that use the flame produced by the combination of a fuel gas and oxygen as the source of heat.

<u>Explanation:</u>

  • Oxy-fuel welding is a process that utilizes fuel gases and oxygen to weld metals. Oxyfuel gas or flame refers to a group of welding processes that utilize the flame delivered by the blending of fuel gas and oxygen as the source of heat.
  • This flame is utilized for cutting and welding of two metallic pieces. This is done due to the heat produced by cutting and welding of two metallic pieces together by heating to the melting point.
  • An oxyhydrogen flame is utilized for cutting and welding of two metallic pieces due to the heat produced by the flame, i.e, 2800 ° C. At this temperature, the metal gets softened effectively and thus it can easily separate or welded together.
7 0
3 years ago
PLEASEEEE HELP MEE!
Ugo [173]

Answer:

2nd option

Explanation:

7 0
3 years ago
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