Answer: 581 gmol
0.581 kmol

Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number
of particles.
To calculate the moles, we use the equation:

1. The conversion for mol to gmol
1 mol = 1 gmol
581 mol= 
2. The conversion for mol to kmol
1 mol = 0.001 kmol
581 mol= 
3. The conversion for mol to lbmol
1 mol = 
581 mol= 
Answer:
Over the past few decades, humans have released so many different chemicals into the air that they have changed the mix of gases in the atmosphere. ... In addition, the exhaust from cars, trucks, and buses releases nitrogen oxides and sulfur dioxide into the air.
Answer:
The correct option is: b. pH 6.4-8.0
Explanation:
Phenol red is a weak acid that is used as a pH indicator and exists in the form of stable red crystals.
<u>The color of the phenol red solution changes from yellow to red when the change in pH is observed. The color of phenol red transitions from yellow to red when the pH is 6.8 - 8.2 or 6.4 - 8.0</u>
Above the pH of 8.2, the phenol red solution turns a bright pink in color.
The value of Kc for the equilibrium is 0.150 mole² / litre ²
<u>Explanation:</u>
<u>Given:</u>
An equilibrium mixture in an 1.00 L vessel contains 5.30 moles of
Mg(OH )₂ 0.800 moles of Mg²⁺ and 0.0010 moles OH₋
We have to find the value of Kc
- Step 1: Find the equilibrium Concentration.
- Step 2: Substitute the values in the equation.
- Step 3: Find the value of Kc.
- I have attached the document for the detailed explanation
The value of Kc for the equilibrium is 0.150 mole² / litre ²
Answer:
See below.
Step-by-step explanation:
Ethers react with HI at high temperature to produce an alky halide and an alcohol.
R-OR' + HI ⟶ R-I + H-OR'
<em>Benzylic ethers</em> react by an Sₙ1 mechanism by forming the stable benzyl cation.
- PhCH₂-OR + HI ⟶ PhCH₂-O⁺(H)R + I⁻ Protonation of the ether
- PhCH₂-O⁺(H)R ⟶ PhCH₂⁺ + HOR Sₙ1 ionization of oxonium ion
- PhCH₂⁺ + I⁻ ⟶ PhCH₂-I Nucleophilic attack by I⁻
If there is excess HI, the alcohol formed in Step 2 is also converted to an alkyl iodide:
ROH +HI ⟶ R-I + H-OH
Thus, benzyl ethyl ether reacts to form benzyl iodide (a) and ethanol (b).
The ethanol reacts with excess HI in an Sₙ2 reaction to form ethyl iodide (c).