In nature, boron is monoatomic. Therefore, its formula is B.
On the other hand, fluorine is diatomic. Therefore, its formula is F2
Now, the basic unbalanced equation is:
B + F2 .........> BF3
Now, we need to balance this equation. As you can see, we have two fluorine moles entering the reaction and 3 formed in the products.
Balancing the equation, we will reach the following balanced reaction:
2B + 3F2 .......> 2BF3
We are asked to calculate the number of moles in a given mass of a substance. To be able to calculate it, we need the molar mass of the substance. For (NH4)2Cr2O7, the molar mass would be <span>252.07 g/mol. We calculate as follows:
0.025 g </span>(NH4)2Cr2O7 ( 1 mol / 252.07 g ) = 0.0001 mol <span>(NH4)2Cr2O7
Hope this answers the question. Have a nice day.</span>
Answer:
0.3023 M
Explanation:
Let Picric acid = 
So,
+
⇄
+ 
The ICE table can be given as:
+
⇄
+ 
Initial: 0.52 0 0
Change: - x + x + x
Equilibrium: 0.52 - x + x + x
Given that;
acid dissociation constant (
) = 0.42
![K_a = \frac{[H_3O^+][Picric^-]}{H_{picric}}](https://tex.z-dn.net/?f=K_a%20%3D%20%5Cfrac%7B%5BH_3O%5E%2B%5D%5BPicric%5E-%5D%7D%7BH_%7Bpicric%7D%7D)
![0.42 = \frac{[x][x]}{0.52-x}}](https://tex.z-dn.net/?f=0.42%20%3D%20%5Cfrac%7B%5Bx%5D%5Bx%5D%7D%7B0.52-x%7D%7D)
![0.42 = \frac{[x]^2}{0.52-x}}](https://tex.z-dn.net/?f=0.42%20%3D%20%5Cfrac%7B%5Bx%5D%5E2%7D%7B0.52-x%7D%7D)
0.42(0.52-x) = x²
0.2184 - 0.42x = x²
x² + 0.42x - 0.2184 = 0 -------------------- (quadratic equation)
Using the quadratic formula;
; ( where +/- represent ± )
= 
= 
=
OR 
=
OR 
=
OR 
= 0.30225 OR - 0.72225
So, we go by the +ve integer that says:
x = 0.30225
x = [
] = [
] = 0.3023 M
∴ the value of [H3O+] for an 0.52 M solution of picric acid = 0.3023 M (to 4 decimal places).
Answer:
-800 kJ/mol
Explanation:
To solve the problem, we have to express the enthalpy of combustion (ΔHc) in kJ per mole (kJ/mol).
First, we have to calculate the moles of methane (CH₄) there are in 2.50 g of substance. For this, we divide the mass into the molecular weight Mw) of CH₄:
Mw(CH₄) = 12 g/mol C + (1 g/mol H x 4) = 16 g/mol
moles CH₄ = mass CH₄/Mw(CH₄)= 2.50 g/(16 g/mol) = 0.15625 mol CH₄
Now, we divide the heat released into the moles of CH₄ to obtain the enthalpy per mole of CH₄:
ΔHc = heat/mol CH₄ = 125 kJ/(0.15625 mol) = 800 kJ/mol
Therefore, the enthalpy of combustion of methane is -800 kJ/mol (the minus sign indicated that the heat is released).