0.334 g sample of an unknown halogen occupies 109 mL at 398 K and 1.41 atm. What is the identity of the halogen? R = 0.0821 퐿.푎푡푚푚표푙.퐾
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Ответ:
The halogen is Cl2
Explanation:
Step 1: Data given
Mass of the halogen = 0.334 grams
Volume of the halogen = 109 mL = 0.109 L
Temperature = 398 K
Pressure = 1.41 atm
Step 2: Calculate moles of halogen
p*V = n*R*T
⇒ p = the pressure = 1.41 atm
⇒ V = the volume = 0.109 L
⇒n = the number of moles = TO BE DETERMINED
⇒ R = the gas constant = 0.08206 L*atm/mol*K
⇒T = the temperature = 398 K
n = (p*V)/(R*T)
n = (1.41 * 0.109) / (0.08206 * 398)
n = 0.004706 moles
Step 3: Calculate molar mass
Molar mass = mass / moles
Molar mass = 0.334 grams / 0.004706 moles
Molar mass = 70.97 g/mol
The halogen is Cl2
Ответ:
The halogen is the Cl₂ (chlorine)
Explanation:
Let's solve this by the Ideal Gases Law → P . V = n . R . T
We need the molar mass to identify the halogen
1.41 atm . 0.109L = n. 0.082 . 398K
(1.41 atm . 0.109L) / 0.082 . 398K = n
4.71×10⁻³ = mol
Let's find out the molar mass → 0.334 g / 4.71×10⁻³mol = 70.9 g/mol
The halogen is the Cl₂
Ответ:
Step-by-step explanation:
(x + y)^5 = 1·x^5·y^0 + 5·x^4·y^1 + 10·x^3·y^2 + 10·x^2·y^3 + 5·x^1·y^4 + 1·x^0·y^5
Correct is only A