Response: The moles in 369 grams of calcium hydroxide are 4.98 moles
Reasoning: Given,
Mass of calcium hydroxide = 369 g
Molar mass of calcium hydroxide = 74.093 g/mole
Formula used:

Now substituting the provided values into this formula, you will find the moles of calcium hydroxide.

Thus, the number of moles in 369 grams of calcium hydroxide is, 4.98 moles
Monomers combine through electron sharing during the polymerization process. This leads to the formation of a polymer, which consists of repeating units. The resulting substance has various applications.
<span>128 g/mol
Applying Graham's law of effusion, we can utilize the formula:
r1/r2 = sqrt(m2/m1)
where
r1 = effusion rate of gas 1
r2 = effusion rate of gas 2
m1 = molar mass of gas 1
m2 = molar mass of gas 2
Given that the atomic weight of oxygen is 15.999, the molar mass of O2 = 2 * 15.999 = 31.998.
We can now insert the known values into Graham's equation to find m2.
r1/r2 = sqrt(m2/m1)
2/1 = sqrt(m2/31.998)
4/1 = m2/31.998
Thus, we find m2 to be 127.992.
Rounding to three significant figures yields 128 g/mol</span>
Answer:
0.31%
Explanation:
For the chemical reaction:
I₂ + 2 S₂O₃²⁻ → 2 I⁻ + S₄O₆²⁻
0.043 L multiplied by 0.117 M of sodium thiosulfate gives 5.031x10⁻³ moles of S₂O₃²⁻
5.031x10⁻³ moles of S₂O₃²⁻ produces:
ClO⁻⁻ + 2 H⁺ + 2 I⁻ → I₂ + Cl⁻ + H2O
2.5156x10⁻³ moles of I₂ equates to moles of NaClO
2.5156x10⁻³ moles of NaClO times
yields 0.187 g of NaClO
Thus, the mass percentage composition is:
= 0.31%
I hope this helps!
Answer:NH₃/NH₄Cl
Explanation:
The pH of a buffer can be determined using Henderson-Hasselbalch's equation.
![pH=pKa+log\frac{[base]}{[acid]}](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D)
When the concentration of acid equals that of the base, the pH aligns with the pKa of the buffer. The ideal pH range is pKa ± 1.
Below are the buffers and their corresponding pKa values:
- CH₃COONa/CH3COOH (pKa = 4.74)
Thus, the ideal buffer is NH₃/NH₄Cl.