Answer : The molecule 
is a polar molecule.
Explanation :
Polar molecule : When the arrangement of the molecule is asymmetrical then the molecule is polar.
Non-polar molecule : When the arrangement of the molecule is symmetrical then the molecule is non-polar.
The given molecule is,
The electronegativities of oxygen and fluorine are different. The molecular geometry of is bent. As, Fluorine is more elctronegative than the oxygen. So, the arrows putting towards the more electronegative element i.e, fluorine. These arrows do not balance each other. Due to this, the asymmetrical arrangement of these bonds makes the molecule polar.
Hence, the given molecule is polar.
Answer:
The final pressure is 90.1 atm.
Explanation:
Assuming constant temperature, we can solve this problem by using <em>Boyle's Law</em>, which states:
Where in this case:
We <u>input the given data</u>:
- 159 atm * 463 L = P₂ * 817 L
And <u>solve for P₂</u>:
The final pressure is 90.1 atm.
Answer:
No
Explanation:
given that, enthalpy is a state function, that means it depends only on the initial and final states, there is no difference between the enthalpy of a phase transition versus the enthalpy of a heating or cooling process, when the cooling or heating process finish in a change of phase.
It does not matter which way we take to cool or heat the substances the Enthalpy of this process will be the same.
Explanation:
The number of moles of solute present in liter of solution is defined as molarity.
Mathematically, Molarity = 
Also, when number of moles are equal in a solution then the formula will be as follows.
It is given that 
is 8.00 M, 
is 7.00 mL, and 
is 0.80 M.
Hence, calculate the value of 
using above formula as follows.
= 70 ml
Thus, we can conclude that the volume after dilution is 70 ml.
Answer:
pH → 7.46
Explanation:
We begin with the autoionization of water. This equilibrium reaction is:
2H₂O ⇄ H₃O⁺ + OH⁻ Kw = 1×10⁻¹⁴ at 25°C
Kw = [H₃O⁺] . [OH⁻]
We do not consider [H₂O] in the expression for the constant.
[H₃O⁺] = [OH⁻] = √1×10⁻¹⁴ → 1×10⁻⁷ M
Kw depends on the temperature
0.12×10⁻¹⁴ = [H₃O⁺] . [OH⁻] → [H₃O⁺] = [OH⁻] at 0°C
√0.12×10⁻¹⁴ = [H₃O⁺] → 3.46×10⁻⁸ M
- log [H₃O⁺] = pH
pH = - log 3.46×10⁻⁸ → 7.46
