<h3><u>Answer;</u></h3>
a) It allows electrons to flow from the anode to the cathode.
<h3><u>Explanation</u>;</h3>
- <em><u>Voltaic cell is an electrochemical cell in which a spontaneous chemical reaction produces the flow of electrons</u></em>.
- Electrons are produced by the oxidation reaction occurring at the anode. Electrons flow through the conducting wire from the anode to the cathode. At the cathode these electrons are used to reduce copper(II) ions to copper atoms.
- <em><u>A conducting wire or a wire play connects the two electrodes allowing electrons to flow from the anode to the cathode</u></em>.
Volume:
2.00 x 11.0 x 11.0 => 242 cm³
mass : 213 g
D = m / V
D = 213 / 242
D = 0.880 g/cm³
Answer B
hope this helps!
Answer:
1.53 × 10²² atoms Ag
Explanation:
Step 1: Define conversions
3.271 × 10⁻²² g = 1 atom
Step 2: Use Dimensional Analysis
= 1.52858 × 10²² atoms Ag
Step 3: Simplify
We have 3 sig figs.
1.52858 × 10²² atoms Ag ≈ 1.53 × 10²² atoms Ag
The volume of HCl required is 23.89 mL
Calculation of volume:
The reaction:

As HCl and NaOH react in 1 : 1 ratio.
Volume of NaOH= 718 mL
Concentration= 0.183M
Volume of HCl= ?
Concentration= 5.50M
Using the dilution formula:


Therefore,
Volume of HCl required will be 23.89 mL.
Learn more about neutralization reaction here,
brainly.com/question/1822651
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Answer:
it is because of the bacteria and different things that go into waters such as ponds, rivers, lakes, seas, oceans, streams, and other body of waters and I should also add the chemicals that go into them as well
Explanation: