Answer:

Explanation:
Hello,
In this case, as the copper's heat loss is gained by the water, the following energetic relationship is:

Therefore the equilibrium temperature shows up as:

Thus, by knowing that water's heat capacity is 4.18J/g°C, one obtains:

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Anode- oxidization
Cathode-reduction
Answer:
a) IUPAC Names:
1) (<em>trans</em>)-but-2-ene
2) (<em>cis</em>)-but-2-ene
3) but-1-ene
b) Balance Equation:
C₄H₁₀O + H₃PO₄ → C₄H₈ + H₂O + H₃PO₄
As H₃PO₄ is catalyst and remains unchanged so we can also write as,
C₄H₁₀O → C₄H₈ + H₂O
c) Rule:
When more than one alkene products are possible then the one thermodynamically stable is favored. Thermodynamically more substituted alkenes are stable. Furthermore, trans alkenes are more stable than cis alkenes. Hence, in our case the major product is trans alkene followed by cis. The minor alkene is the 1-butene as it is less substituted.
d) C is not Geometrical Isomer:
For any alkene to demonstrate geometrical isomerism it is important that there must be two different geminal substituents attached to both carbon atoms. In 1-butene one carbon has same geminal substituents (i.e H atoms). Hence, it can not give geometrical isomers.
Answer:
6.78 × 10⁻³ L
Explanation:
Step 1: Write the balanced equation
Mg₃N₂(s) + 3 H₂O(g) ⇒ 3 MgO(s) + 2 NH₃(g)
Step 2: Calculate the moles corresponding to 10.2 mL (0.0102 L) of H₂O(g)
At STP, 1 mole of H₂O(g) has a volume of 22.4 L.
0.0102 L × 1 mol/22.4 L = 4.55 × 10⁻⁴ mol
Step 3: Calculate the moles of NH₃(g) formed from 4.55 × 10⁻⁴ moles of H₂O(g)
The molar ratio of H₂O to NH₃ is 3:2. The moles of NH₃ produced are 2/3 × 4.55 × 10⁻⁴ mol = 3.03 × 10⁻⁴ mol.
Step 4: Calculate the volume corresponding to 3.03 × 10⁻⁴ moles of NH₃
At STP, 1 mole of NH₃(g) has a volume of 22.4 L.
3.03 × 10⁻⁴ mol × 22.4 L/mol = 6.78 × 10⁻³ L
<span>An organism that contains chloroplasts is able to produce food by the process of
"Photosynthesis"
Hope this helps!
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