In order to get the sum of 2.7 and 2.47, we will add both numbers. So, 2.7 plus 2.47 would be 5.17. And the correct number of significant digits in this number is still the same, 5.17, so we have three significant numbers. Why? There are three rules in identifying the significant figures: 1. Non zeros are always significant (which this applies in our sum above). 2. Any zeros in between significant numbers are always significant. 3. <span> The final zero or zeros in the decimal portion only are significant.</span>
Answer:
Do you want an explaniton to? btw the answer is
1 in the 13
2 in the 8
3 in the 42
4 in the Mo
5 in the 96
6 in the 1
7 in the 18
8 in the 2
Answer:
91hours
Explanation:
Firstly, we calculate the quantity of electricity passed.
Q = It
Q = 34.5t
Since we are dealing with cu2+
Cu2+ + 2e- ———> Cu
From the second law of electrolysis, 1 mole of copper ion will need 2 moles of electrons and thus would be needing half faraday of electricity.
Now one faraday would contain 96500C of electricity, 0.5F would contain = 48,250C of electricity .
The number of moles of copper thus deposited = 34.5t/48,250
We know that the mass of copper deposited equals no of moles of copper multiplied by its atomic mass
15kg = 34.5t/48250 * 64
15,000 * 48250 = 34.5t * 64
t = 327,785s
Converting this to hours means we divide by 3,600
327,785/3600 = 91hours
Answer and Explanation:
The galvanic cell is:
Cu(s)| Cu²⁺(aq)|| Cu⁺(aq)| Cu(s)
The first two species before the double bar (||) constitute the <em>anodic half reaction (oxidation)</em>:
Cu(s) ⇒ Cu²⁺(aq) + 2 e-
The two species after the || constitute the <em>cathodic half reaction</em><em> </em><em>(reduction</em>):
Cu⁺(aq) + e- ⇒ Cu(s)
If we multiply the reduction half reaction by 2 (to obtain the same number of electrons than oxidation reaction) and then we add the two half reactions, we obtan the balanced equation:
Reduction (cathode) : 2Cu⁺(aq) +2 e- ⇒ 2Cu(s)
Oxidation (anode) : Cu(s) ⇒ Cu²⁺(aq) + 2 e-
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Total equation: 2Cu⁺(aq) + <em>Cu(s)</em> ⇒ <em>2Cu(s)</em> + Cu²⁺(aq)
Cu(s) is in both reactants side and products side, so we cancel that in both opposite sides to obtain:
2Cu⁺(aq) ⇒ Cu(s)<em> </em>+ Cu²⁺(aq)
If we divide the balanced equation into 2, the smallest possible integer coefficient for Cu⁺(aq) is 1:
Cu⁺(aq) ⇒ 1/2 Cu(s)<em> </em>+ 1/2 Cu²⁺(aq)