The answer you are looking for is A: Adding more batteries will increase the current
Since you have a circuit in series, that means that all of the bulbs would be connected to one wire. If you were add more batteries to that <em>one </em>wire, you would end up with a greater flow of energy throughout the bulbs/wire
Based on the information provided and the assumption that the mole ratio of the acid to the base is 1:1, the molarity of the acid would be 0.0425 M
<h3>Acid/base Titration</h3>
Using the equation:
CaVa = CbVb
This is based on the assumption that the stoichiometric mole ratio of the acid to the base is 1:1.
Ca = CbVb/Va
0.1 x 25.5/60 = 0.0425 M
More on acid/base titration can be found here: brainly.com/question/2728613
Answer:
0.0658 M
Explanation:
The reaction that takes place is:
Na₂SO₄(aq) + BaCl₂(aq) ↔ BaSO₄(s) + 2NaCl(aq)
First we <u>calculate the number of Na₂SO₄ moles</u>:
- 0.4000 g sample * 96.4/100 = 0.3856 g Na₂SO₄
- 0.3856 g Na₂SO₄ ÷ 142.04 g/mol = 2.715x10⁻³ mol Na₂SO₄
Now we <u>convert to moles of BaCl₂</u>:
- 2.715x10⁻³ mol Na₂SO₄ * 1 molBaCl₂/1 molNa₂SO₄ = 2.715x10⁻³ mol BaCl₂
Finally we divide by the volume to <u>calculate the molarity</u>:
- 41.25 mL ⇒ 41.25 / 1000 = 0.04125 L
- 2.715x10⁻³ mol BaCl₂ / 0.04125 L = 0.0658 M
