<span>Hydroelectric energy uses the movement of water to spin a turbine and produce electricity. Coal is a fossil fuel that is burned as an energy source, which results in emissions such as carbon dioxide, nitrogen, and sulfur. Coal use is associated with waste products both from the mining process to acquire the coal and also from the actual use of coal itself. Hydroelectric energy does not involve waste, but does pose potential problems to waterways. Coal use is more common than the use of hydroelectric energy.</span>
Answer:
"0.60 g" is the appropriate solution.
Explanation:
The given values are:
Volume of base,
= 30 ml
Molarity of base,
= 0.05 m
Molar mass of acid,
= 400 g/mol
As we know,
⇒ 
On substituting the values, we get
⇒ 
⇒ 
⇒
hence,
⇒ 
On substituting the values, we get
⇒ 
⇒ 
⇒ 
Answer:
11.31g NaClO₂
Explanation:
<em> Is given 250mL of a 1.60M chlorous acid HClO2 solution. Ka is 1.110x10⁻². What mass of NaClO₂ should the student dissolve in the HClO2 solution to turn it into a buffer with pH =1.45? </em>
It is possible to answer this question using Henderson-Hasselbalch equation:
pH = pKa + log₁₀ [A⁻] / [HA]
<em>Where pKa is -log Ka = 1.9547; [A⁻] is the concentration of the conjugate base (NaClO₂), [HA] the concentration of the weak acid</em>
You can change the concentration of the substance if you write the moles of the substances:
[Moles HClO₂] = 250mL = 0.25L×(1.60mol /L) = <em>0.40 moles HClO₂</em>
Replacing in H-H expression, as the pH you want is 1.45:
1.45 = 1.9547 + log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]
-0.5047 = log₁₀ [Moles NaClO₂] / [0.40 moles HClO₂]
<em>0.3128 = </em>[Moles NaClO₂] / [0.40 moles HClO₂]
0.1251 = Moles NaClO₂
As molar mass of NaClO₂ is 90.44g/mol, mass of 0.1251 moles of NaClO₂ is:
0.1251 moles NaClO₂ ₓ (90.44g / mol) =
<h3>11.31g NaClO₂</h3>
Answer:
Explanation:
The reaction is given as:

The reaction quotient is:
![Q_C = \dfrac{[NH_3]^2}{[N_2][H_2]^3}](https://tex.z-dn.net/?f=Q_C%20%3D%20%5Cdfrac%7B%5BNH_3%5D%5E2%7D%7B%5BN_2%5D%5BH_2%5D%5E3%7D)
From the given information:
TO find each entity in the reaction quotient, we have:
![[NH_3] = \dfrac{6.42 \times 10^{-4}}{3.5}\\ \\ NH_3 = 1.834 \times 10^{-4}](https://tex.z-dn.net/?f=%5BNH_3%5D%20%3D%20%5Cdfrac%7B6.42%20%5Ctimes%2010%5E%7B-4%7D%7D%7B3.5%7D%5C%5C%20%5C%5C%20NH_3%20%3D%201.834%20%5Ctimes%2010%5E%7B-4%7D)
![[N_2] = \dfrac{0.024 }{3.5}](https://tex.z-dn.net/?f=%5BN_2%5D%20%3D%20%5Cdfrac%7B0.024%20%7D%7B3.5%7D)
![[N_2] = 0.006857](https://tex.z-dn.net/?f=%5BN_2%5D%20%3D%200.006857)
![[H_2] =\dfrac{3.21 \times 10^{-2}}{3.5}](https://tex.z-dn.net/?f=%5BH_2%5D%20%3D%5Cdfrac%7B3.21%20%5Ctimes%2010%5E%7B-2%7D%7D%7B3.5%7D)
![[H_2] = 9.17 \times 10^{-3}](https://tex.z-dn.net/?f=%5BH_2%5D%20%3D%209.17%20%5Ctimes%2010%5E%7B-3%7D)
∴

However; given that:

By relating
, we will realize that 
The reaction is said that it is not at equilibrium and for it to be at equilibrium, then the reaction needs to proceed in the forward direction.
I hope this help crystal lattic