Answer:
1.22 mL
Explanation:
Let's consider the following balanced reaction.
2 AgNO₃ + BaCl₂ ⇄ Ba(NO₃)₂ + 2 AgCl
The molar mass of silver chloride is 143.32 g/mol. The moles corresponding to 0.525 g are:
0.525 g × (1 mol/143.32 g) = 3.66 × 10⁻³ mol
The molar ratio of AgCl to BaCl₂ is 2:1. The moles of BaCl₂ are 1/2 × 3.66 × 10⁻³ mol = 1.83 × 10⁻³ mol.
The volume of 1.50 M barium chloride containing 1.83 × 10⁻³ moles is:
1.83 × 10⁻³ mol × (1 L/1.50 mol) = 1.22 × 10⁻³ L = 1.22 mL
Coulomb's law mathematically is:
F = kQ₁Q₂/r²
we integrate this with respect to distance to obtain the expression for energy:
E = kQ₁Q₂/r; where k is the Coulomb's constant = 9 x 10⁹; Q are the charges, r is the seperation
Charge on proton = charge on electron = 1.6 x 10⁻¹⁹ C
E = (9 x 10⁹ x 1.6 x 10⁻¹⁹ x 1.6 x 10⁻¹⁹) / (185 x 10⁻¹²)
E = 1.24 x 10⁻¹⁸ Joules per proton/electron pair
Number of pairs in one mole = 6.02 x 10²³
Energy = 6.02 x 10²³ x 1.24 x 10⁻¹⁸
= 746.5 kJ
Answer: Option (5) is the correct answer.
Explanation:
An ionic bond is formed by transfer of electrons between the two chemically combining atoms. Whereas a covalent bond is defined as the bond formed by sharing of electrons between the two chemically combining atoms.
When electronegativity difference is from 0.0 to 0.4 then bond formed between the two atoms is non-polar covalent in nature.
When electronegativity difference is greater than 0.4 and less than 1.7 then bond between the two atoms is a polar covalent bond.
When electronegativity difference is 1.7 or greater than the bond formed is ionic in nature.
Therefore, electronegativity difference of the given species is as follows.
Si-P = 2.1 - 1.8 = 0.3
Si-Cl = 3.0 - 1.8 = 1.2
Si-S = 2.5 - 1.8 = 0.7
Thus, we can conclude that given bonds are placed in order of increasing ionic character as follows.
Si-P < Si-S < Si-Cl
Answer: The standard enthalpy of formation of liquid octane is -250.2 kJ/mol
Explanation:
The given balanced chemical reaction is,
First we have to calculate the enthalpy of reaction 
.
![\Delta H^o=[n_{O_2}\times \Delta H_f^0_{(O_2)}+n_{H_2O}\times \Delta H_f^0_{(H_2O)}]-[n_{C_8H_{18}}\times \Delta H_f^0_{(C_8H_{18})+n_{O_2}\times \Delta H_f^0_{(O_2)}]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo%3D%5Bn_%7BO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28O_2%29%7D%2Bn_%7BH_2O%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28H_2O%29%7D%5D-%5Bn_%7BC_8H_%7B18%7D%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28C_8H_%7B18%7D%29%2Bn_%7BO_2%7D%5Ctimes%20%5CDelta%20H_f%5E0_%7B%28O_2%29%7D%5D)
where,
We are given:
Putting values in above equation, we get:
![-1.0940\times 10^4=[(16\times -393.5)+(18\times -285.8)]-[(25\times 0)+(2\times \Delat H_f{C_8H_{18}(l)}]](https://tex.z-dn.net/?f=-1.0940%5Ctimes%2010%5E4%3D%5B%2816%5Ctimes%20-393.5%29%2B%2818%5Ctimes%20-285.8%29%5D-%5B%2825%5Ctimes%200%29%2B%282%5Ctimes%20%5CDelat%20H_f%7BC_8H_%7B18%7D%28l%29%7D%5D)
Thus the standard enthalpy of formation of liquid octane is -250.2 kJ/mol
Dilution<span> is when you decrease the concentration of a </span>solution<span> by adding a solvent. As a result, if you want to </span>dilute<span> salt water, just add water. ... Add more solute until it quits dissolving. That point at which a solute quits dissolving is the point at which it's </span>saturated<span>.</span>
