The correct answer is A. Alkali metals.
I hope this helped
Answer:
Qp > Kp, por lo tanto, la presión parcial de BrF₃(g) aumenta hasta alcanzar el equilibrio.
Explanation:
Paso 1: Escribir la ecuación balanceada
BrF₃ (g) ⇌ BrF(g) + F₂(g) Kp(T) = 64,0
Paso 2: Calcular el cociente de reacción (Qp)
Qp = pBrF × pF₂ / pBrF₃
Qp = 1,50 × 2,00 / 0,0150 = 200
Paso 3: Sacar una conclusión
Dado que Qp > Kp, la reacción se desplazará hacia la izquierda para alcanzar el equilibrio, es decir, la presión parcial de BrF₃(g) aumenta hasta alcanzar el equilibrio.
From,
RAM=element×its relative abudance/total abudance
=((107×13)+(12×109))/25
The answer is=107.96
Answer:
Explanation has been given below.
Explanation:
- Chloroform has three polar C-Cl bonds. Methylene chloride has two polar C-Cl bonds. So it is expected that chloroform should be more polar and posses higher dipole moment than methylene chloride.
- Two factors are liable for the opposite trend observed in dipole moments of methylene chloride and chloroform.
- First one is the number of hyperconjugative hydrogen atoms present in a molecule. Hyperconjugation occurs with vacant d-orbital of Cl atom. Hyperconjugation amplifies charge separation in a molecule resulting higher dipole moment.
- Methylene chloride has two hyperconjugative hydrogen atoms and chloroform has one hyperconjugative hydrogen atom.Therefore methylene chloride should have higher charge separation as compared to chloroform.
- Second one is induction of opposite polarity in a C-Cl bond by another C-Cl bond in a molecule. Higher the opposite induction of polarity, lower the charge separation in a molecule and hence lower the dipole moment of a molecule.
- Chloroform has three C-Cl bonds and methylene chloride has two C-Cl bonds. Therefore opposite induction is higher for chloroform resulting it's lower dipole moment.
