Answer:
62.98 % of the sample of hydrate is water
Explanation:
Step 1: Data given
Mass of the sample of a hydrate of sodium carbonate (Na2CO3) = 2.026 grams
After heating, the mass of the sample is 0.750 g
Molar mass H2O = 18.02 g/mol
Step 2: Calculate mass of water
Mass water = mass of hydrate - mass of sample after heating
Mass water = 2.026 grams - 0.750 grams
Mass water = 1.276 grams
Step 3: Calculate mass % percent of water
Mass % of water = (mass of water / total mass hydrate) * 100 %
Mass % of water = (1.276 grams / 2.026 grams) *100 %
Mass % of water = 62.98 %
62.98 % of the sample of hydrate is water
Answer:
biotechnology
Explanation:
hope this helps have a good day
You can use the equation ΔS(surr)=q(surr)/T or ΔS(surr)=-q(rxn)/T.
the two equations are equal since we know that the energy the system (reactoin) puts out just goes into the surroundings.
(In other words q(surr)=-q(rxn))
Using the equation, <span>ΔS(surr)=-(-283kJ/298K)=0.9497kJ/K or 949.7J/K
This answer makes sense since the reaction is exothermic which means it released energy into the system which usually causes the entropy to increase.
I hope that helps.</span>
Answer : The 'Ag' is produced at the cathode electrode and 'Cu' is produced at anode electrode under standard conditions.
Explanation :
Galvanic cell : It is defined as a device which is used for the conversion of the chemical energy produces in a redox reaction into the electrical energy. It is also known as the voltaic cell or electrochemical cell.
In the galvanic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.
We are taking the value of standard reduction potential form the standard table.
![E^0_{[Ag^{+}/Ag]}=+0.80V](https://tex.z-dn.net/?f=E%5E0_%7B%5BAg%5E%7B%2B%7D%2FAg%5D%7D%3D%2B0.80V)
![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)
In this cell, the component that has lower standard reduction potential gets oxidized and that is added to the anode electrode. The second forms the cathode electrode.
The balanced two-half reactions will be,
Oxidation half reaction (Anode) : 
Reduction half reaction (Cathode) : 
Thus the overall reaction will be,

From this we conclude that, 'Ag' is produced at the cathode electrode and 'Cu' is produced at anode electrode under standard conditions.
Hence, the 'Ag' is produced at the cathode electrode and 'Cu' is produced at anode electrode under standard conditions.