The volume (in mL) of 0.242 M NaOH solution needed for the titration reaction is 39.44 mL
<h3>Balanced equation </h3>
CH₃CH₂COOH + NaOH —> CH₃CH₂COONa + H₂O
From the balanced equation above,
- The mole ratio of the acid, CH₃CH₂COOH (nA) = 1
- The mole ratio of the base, NaOH (nB) = 1
<h3>How to determine the volume of NaOH</h3>
- Volume of acid, CH₃CH₂COOH (Va) = 46.79 mL
- Molarity of acid, CH₃CH₂COOH (Ma) = 0.204 M
- Molarity of base, NaOH (Mb) = 0.242 M
- Volume of base, KOH (Vb) =?
MaVa / MbVb = nA / nB
(0.204 × 46.79) / (0.242 × Vb) = 1
Cross multiply
0.242 × Vb = 0.204 × 46.79
Divide both side by 0.242
Vb = (0.204 × 46.79) / 0.242
Vb = 39.44 mL
Thus, the volume of NaOH needed for the reaction is 39.44 mL
Learn more about titration:
brainly.com/question/14356286
Answer:
IF the fruit juice contains a high level of vitamin C, THEN the preventative effectiveness against common cold increases.
Explanation:
The hypothesis is a testable explanation of a scientific investigation. It aims at predicting the outcome of the experiment. One feature of the hypothesis is that it must be testable. The hypothesis is usually written in an "IF, THEN" format.
This question is regarding an experiment to test the amount of vitamin C in fruit juice. The vitamin C is thought to be an effective preventative against common cold. Hence, the hypothesis connects the effect on common cold (dependent variable) with the amount of vitamin C (independent variable). The hypothesis can be written as:
IF the fruit juice contains a high level of vitamin C, THEN the preventative effectiveness against common cold increases.
1. Dry Ice (solid carbon dioxide)
2. Iodine
3. Arsenic
4. Naphthalene
Answer:
group 16 period 2 of the periodic table
note: that is not the electronic configuration, that is the Bohr model.
Answer:
The empirical formula is ZnO2
Explanation:
What is the empirical formula for a compound which contains 67.1% zinc and the rest is oxygen?
Step 1: Data given
Suppose the compound has a mass of 100.0 grams
A compound contains:
67.1 % Zinc = 67.1 grams
100 - 67.1 = 32.9 % oxygen = 32.9 grams
Molar mass of Zinc = 65.38 g/mol
Molar mass of O = 16 g/mol
Step 2: Calculate moles of Zinc
Suppose the compound is 100 grams
Moles Zn = 67. 10 grams / 65.38 g/mol
Moles Zn = 1.026 moles
Step 3: Calculate moles of O
Moles O = 32.90 grams / 16.00 g/mol
Moles O = 2.056 moles
Step 4: Calculate mol ratio
We divide by the smallest amount of moles
Zn: 1.026/1.026 = 1
O: 2.056/1.026 = 2
The empirical formula is ZnO2
To control this we can calculate the % Zinc for 1 mol
65.38 / (65.38+2*16) = 0.67.1 = 67.2 %
