1.50 mol C3H8 X (3 mol CO2 / 1 mol C3H8) X (44.0 g CO2 / 1 mol CO2) = 198 g CO2
Answer:
14.53ML
Explanation:
V1=218
V2=?
P2=15p1
USING BOYLE'S LAW
P1V1=P2V2
V2=P1V1/P2=P1(218ML)/15P1
=14.53ML
The volume (in mL) of calcium hydroxide, Ca(OH)₂ needed for the reaction is 19.8 mL
<h3>Balanced equation </h3>
2HCl + Ca(OH)₂ —> CaCl₂ + 2H₂O
From the balanced equation above,
- The mole ratio of the acid, HCl (nA) = 2
- The mole ratio of the base, Ca(OH)₂ (nB) = 1
<h3>How to determine the volume of Ca(OH)₂ </h3>
- Molarity of base, Ca(OH)₂ (Mb) = 1.48 M
- Volume of acid, HCl (Va) = 36 mL
- Molarity of acid, HCl (Ma) = 1.63 M
- Volume of base, Ca(OH)₂ (Vb) =?
MaVa / MbVb = nA / nB
(1.63 × 36) / (1.48 × Vb) = 2
58.68 / (1.48 × Vb) = 2
Cross multiply
2 × 1.48 × Vb = 58.68
2.96 × Vb = 58.68
Divide both side by 2.96
Vb = 58.68 / 2.96
Vb = 19.8 mL
Learn more about titration:
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Answer:
3.6
Explanation:
Step 1: Given data
- Concentration of formic acid: 0.03 M
- Concentration of formate ion: 0.02 M
- Acid dissociation constant (Ka): 1.8 × 10⁻⁴
Step 2: Calculate the pH
We have a buffer system formed by a weak acid (HCOOH) and its conjugate base (HCOO⁻). We can calculate the pH using the <em>Henderson-Hasselbach equation</em>.
![pH = pKa +log\frac{[base]}{[acid]} = -log 1.8 \times 10^{-4} + log \frac{0.02}{0.03} = 3.6](https://tex.z-dn.net/?f=pH%20%3D%20pKa%20%2Blog%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%3D%20-log%201.8%20%5Ctimes%2010%5E%7B-4%7D%20%2B%20log%20%5Cfrac%7B0.02%7D%7B0.03%7D%20%3D%203.6)
Explanation:
Standardization is the process of creating protocols to guide the creation of a good or service based on the consensus of all the relevant parties in the industry. ... Standardization also helps in ensuring the safety, interoperability, and compatibility of goods produced.