The balanced chemical reaction is written as:
<span>4C(s) + S8(s) → 4CS2(l)
We are given the amount of carbon and sulfur to be used in the reaction. We need to determine first the limiting reactant to be able to solve this correctly.
</span>7.70 g C ( 1 mol / 12.01 g) =0.64 mol C
19.7 g S8 ( 1 mol / 256.48 g) = 0.08 mol S8
The limiting reactant would be S8. We use this amount to calculate.
0.08 mol S8 ( 4 mol CS2 / 1 mol S8 ) ( 256.48 g / 1 mol ) = 78.8 g CS2
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)
B. slows down is your answer, obviously as it approaches carrying capacity, there would be less available space to find in the place of inhabitance, so less and less population units would be able to find the place of inhabitance suitable for living, or can't find enough space to live in.
Hello!
When finding the chemical formula of a compound, we will need to find the charges of each element/bond.
Looking at our period table, sodium has a +1 charge, written as Na 1+, and sulfate has a charge of -2, and it is written as SO4 2-.
Now, we need to make the charges equivalent. To do this, we need to "criss-cross" the charges. This means that sodium will need to additional atoms to make the charges equal, and sulfate will need one.
Therefore, the chemical formula for sodium sulfate is: Na2SO4.
