Let's use Boyle's Law here. P1*V1 = P2*V2
Given: (assuming that there are decimals at the end for Sig Figs)
P1 = 900.mmHg
P2 = 1140.mmHg
V1 = ???
V2 = 250.mL
900.mmHg* ??? = 1140.mmHg * 250.mL
??? = 1.27*250.mL
??? = 318.mL
Therefore, the original volume is 318mL.
Answer:
high electronegativity difference between oxygen and hydrogen
Explanation:
The electronegativity value of O and H are 3.5 and 2.1 respectively. The electronegativity difference of the O-H bond in H₂O is 1.4.
- When the elctronegativity difference between two atoms in bond is below 0.5 the bond is nonpolar covalent.
- When the elctronegativity difference between two atoms in bond is between 0.5 and 2.0 the bond is polar covalent.
- When the elctronegativity difference between two atoms in bond is above 0.5 the bond is ionic.
Thus the O-H is a polar covalent bond. This explains the polarity in water molecules.
Based on Pauling Scale, electro negativity of Cl = 3.2, Na = 0.9 and H = 2.1
Thus, Electronegativity difference in

= 3.2 -3.2 = 0
Electronegativity difference in NaCl = 3.2-0.9 = 2.3
Similarly, Electronegativity difference in HCl = 3.2 - 2.1 = 1.1
Thus, among the listed molecules following is the decreasing order of electronegativity difference: NaCl> HCl >
Answer:
Kc = 2.145 × 10⁻⁸¹
Explanation:
Let's consider the following reaction:
O₂(g) ⇄ 2O(g)
The standard Gibbs free energy for the reaction (ΔG°) can be calculated using the following expression:
ΔG° = Σnp. ΔG°f(p) - Σnp. ΔG°f(p)
where,
ni are the moles of products and reactants
ΔG°f(p) are the standard Gibbs free energy of formation of products and reactants
In this case,
ΔG° = 2 × ΔG°f(O) - 1 × ΔG°f(O₂)
ΔG° = 2 × 230.1 kJ/mol - 1 × 0 kJ/mol
ΔG° = 460.2 kJ/mol
With this information, we can calculate the equilibrium constant (Kc) using the following expression:

Answer:
b
Explanation:
because the others are compound and element
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