Answer: materials and design Techniques that reduce the negative environmental impact of a structure
Explanation:
Answer:
4.26 %
Explanation:
There is some info missing. I think this is the original question.
<em>Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is 4.50 × 10
⁻⁴.</em>
<em />
Step 1: Given data
Initial concentration of the acid (Ca): 0.249 M
Acid dissociation constant (Ka): 4.50 × 10
⁻⁴
Step 2: Write the ionization reaction for nitrous acid
HNO₂(aq) ⇒ H⁺(aq) + NO₂⁻(aq)
Step 3: Calculate the concentration of nitrite in the equilibrium ([A⁻])
We will use the following expression.
![[A^{-} ] = \sqrt{Ca \times Ka } = \sqrt{0.249 \times 4.50 \times 10^{-4} } = 0.0106 M](https://tex.z-dn.net/?f=%5BA%5E%7B-%7D%20%5D%20%3D%20%5Csqrt%7BCa%20%5Ctimes%20Ka%20%7D%20%3D%20%5Csqrt%7B0.249%20%5Ctimes%204.50%20%5Ctimes%2010%5E%7B-4%7D%20%20%7D%20%3D%200.0106%20M)
Step 4: Calculate the percent ionization of nitrous acid
We will use the following expression.
![\alpha = \frac{[A^{-} ]}{[HA]} \times 100\% = \frac{0.0106M}{0.249} \times 100\% = 4.26\%](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B%5BA%5E%7B-%7D%20%5D%7D%7B%5BHA%5D%7D%20%5Ctimes%20100%5C%25%20%3D%20%5Cfrac%7B0.0106M%7D%7B0.249%7D%20%5Ctimes%20100%5C%25%20%3D%204.26%5C%25)
1. Common knowledge : Go to the right of periodic table, the atomic radius is decreasing
2. Flurine has 9 protons and lithium has 3 protons. you know that the electron is attracted with the centre of the atom, that's why more proton, more 'energy' that attract to the centre and that's why it make the shape of the atom is smaller
<span>These values also increase from the
bottom to the top of a group because the size of the atom decreases, resulting
in a smaller distance between the nucleus and the valence electron shell, which
increases the attraction between the protons and electrons.</span>
Answer:
25.157 cm³
Explanation:
Data Given:
Mass of Sugar (m) = 40g
Density of sugar given in literature = 1.59 g/cm³
Volume of Sugar = ?
The formula will be used is
d = m/v ........................................... (1)
where
D is density
m is the mass
v is the volume
So
Rearrange the Equation (1)
d x v = m
v = m/ d ................................................ (2)
put the given values in Equation (2)
v = 40g / 1.59 g/cm³
v = 25.157 cm³
volume of 40 g of sugar = 25.157 cm³
