Answer:
glucose and oxygen
Explanation:
plants use carbondioxide, water and sunlight to produce their food through photosynthesis so in return they produce oxygen into the air as by product and glucose
Answer: 0.8M
Explanation:
Given that,
Amount of moles of NaCl (n) = ?
Mass of NaCl in grams = 1.40 g
For molar mass of NaCl, use the molar masses:
Sodium, Na = 23g;
Chlorine, Cl = 35.5g
NaCl = (23g + 35.5g)
= 58.5g/mol
Since, amount of moles = mass in grams / molar mass
n = 1.40g / 58.5g/mol
n = 0.024 mole
Now, given that:
Amount of moles of NaCl (n) = 0.024
Volume of NaCl solution (v) = 30.0mL
[Convert 30.0mL to liters
If 1000 mL = 1L
30.0mL = 30.0/1000 = 0.03L]
Concentration of NaCl solution (c) = ?
Since concentration (c) is obtained by dividing the amount of solute dissolved by the volume of solvent, hence
c = n / v
c = 0.024 mole / 0.03 L
c = 0.8 M (0.8M means concentration is in moles per litres)
Thus, the concentration of the solution is 0.8M
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Answer : The correct option is, +91 kJ/mole
Solution :
The balanced cell reaction will be,
Here copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.
First we have to calculate the standard electrode potential of the cell.
![E^0_{[Pb^{2+}/Pb]}=-0.13V](https://tex.z-dn.net/?f=E%5E0_%7B%5BPb%5E%7B2%2B%7D%2FPb%5D%7D%3D-0.13V)
![E^0_{[Cu^{2+}/Cu]}=+0.34V](https://tex.z-dn.net/?f=E%5E0_%7B%5BCu%5E%7B2%2B%7D%2FCu%5D%7D%3D%2B0.34V)
![E^0_{cell}=E^0_{[Pb^{2+}/Pb]}-E^0_{[Cu^{2+}/Cu]}](https://tex.z-dn.net/?f=E%5E0_%7Bcell%7D%3DE%5E0_%7B%5BPb%5E%7B2%2B%7D%2FPb%5D%7D-E%5E0_%7B%5BCu%5E%7B2%2B%7D%2FCu%5D%7D)
Now we have to calculate the standard Gibbs free energy.
Formula used :
where,
= standard Gibbs free energy = ?
n = number of electrons = 2
F = Faraday constant = 96500 C/mole
= standard e.m.f of cell = -0.47 V
Now put all the given values in this formula, we get the Gibbs free energy.
Therefore, the standard Gibbs free energy is +91 kJ/mole
