Hello!
To determine [H₃O⁺], we need to apply the Henderson-Hasselback equation, since this is a case of an acid and its conjugate base:
![pH=pKa+log( \frac{[A^{-}] }{[HA]} )](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%20%5Cfrac%7B%5BA%5E%7B-%7D%5D%20%7D%7B%5BHA%5D%7D%20%29)
Now, we use the definition of pH and clear [H₃O⁺] from there:
![pH=-log[H_3O^{+}]](https://tex.z-dn.net/?f=pH%3D-log%5BH_3O%5E%7B%2B%7D%5D%20)
![[H_3O^{+}] = 10^{-pH} =10^{-3,84}=0,00014 M](https://tex.z-dn.net/?f=%20%5BH_3O%5E%7B%2B%7D%5D%20%3D%2010%5E%7B-pH%7D%20%3D10%5E%7B-3%2C84%7D%3D0%2C00014%20M)
So, the [H₃O⁺] concentration is
0,00014 M
Have a nice day!
Answer:
Each molecule contains one atom of A and one atom of B. The reaction does not use all of the atoms to form compounds.
A + B ⟶ Product
Particles: 6 8 6
If six A atoms form six product molecules, each molecule can contain only one A atom.
The formula of the product is ABₙ.
If n = 1, we need six atoms of B.
If n = 2, we need 12 atoms of B. However, we have only eight atoms of B, so the formula of the product must be AB.
Thus, 6A + 6B ⟶ 6AB, with two B atoms left over.
Explanation:
Credit goes to @znk
Hope it helps you :))
Answer:
1.6 L
Explanation:
Using Charle's law
Given ,
V₁ = 1.5 L
V₂ = ?
T₁ = 12 °C
T₂ = 32 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (12 + 273.15) K = 285.15 K
T₂ = (32 + 273.15) K = 305.15 K
Using above equation as:
New volume = 1.6 L
Answer:
C₂H₇F₂P
Explanation:
Given parameters:
Composition by mass:
C = 24%
H = 7%
F = 38%
P = 31%
Unknown:
Empirical formula of compound;
Solution :
The empirical formula is the simplest formula of a compound. To solve for this, follow the process below;
C H F P
% composition
by mass 24 7 38 31
Molar mass 12 1 19 31
Number of
moles 24/12 7/1 38/19 31/31
2 7 2 1
Dividing
by the
smallest 2/1 7/1 2/1 1/1
2 7 2 1
Empirical formula C₂H₇F₂P
