Answer:
1.03 atm
Explanation:
Primero <u>convertimos 21 °C y 37 °C a K</u>:
- 21 °C + 273.16 = 294.16 K
- 37 °C + 273.16 = 310.16 K
Una vez tenemos las temperaturas absolutas, podemos resolver este problema usando la<em> ley de Gay-Lussac</em>:
En este caso:
Colocando los datos:
- 294.16 K * P₂ = 310.16 K * 0.98 atm
Y <u>despejando P₂</u>:
Answer : The correct answer is 1) AlCl₃ - CH₃Cl 2) HNO₃ -H₂SO₄ at room temperature 3) Fuming HNO₃ -H₂SO₄ at 90-100 ⁰ C heat .
I think this reaction is forming 2,4,6- trinitrotoluene from benzene, since the product is not mentioned. Following are the steps to convert Benzene to 2,4,6 trinitrotoluene .
Step 1: Conversion of Benzene to Toluene .
Benzene can be converted to toluene by Friedel Craft Alkylation of benzene . In this reaction reagent AlCl₃ and Ch3Cl is used . Electrophile CH³⁺ is produced which attached on carbon of benzene and formation of Toluene and HCl occur.

Step 2 : Conversion of Toluene to dinitrotoluene.
Dinitritoluene is prepared from toluene by Nitration . This reaction uses Electrophilic substitution mechanism . The reagents used are HNO₃ and H₂SO₄ at room temperature . These reagents produces NO₂⁺ ( nitronium ion ), a electrophile which attacks on C2 and C4 Carbon atoms of Toluene.
Toluene 
Step 3) Conversion of Dinitro toluene to trinitrotoluene.
This reaction is extended nitration of toluene . Further nitration is done in extreme condition . The temperature of reaction is increased to 90- 100 ⁰ C . Due to which there is more production of NO²⁺ ion occurs from HNO₃ -H₂SO₄ and they attack on C6 carbon atom of dinitrotoluene which forms 2,4,6- trinitrotoluene.
Dinitrotoluene 
So over all reaction uses three reagents in order :

Answer:
b molartiary will decrease
Explanation
Explanation:
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