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Answer:</h3>
The total concentration of ions in a 0.75 M solution of HCl is 1.5 M
That is; 0.75 M H⁺ and 0.75 M Cl⁻
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Explanation:</h3>
- Concentration or molarity is the number of moles of a compound or an ion contained in one liter of solution. It is measured in moles per liter (M).
- The concentration of ions making a compound is determined by the ratio of moles of the compound and the constituents ions.
- For instance, HCl dissociates to give H⁺ and Cl⁻
HCl(aq) → H⁺(aq) + Cl⁻(aq)
- Therefore, since the mole ratio between HCl and the constituent ions H⁺ and Cl⁻ is 1:1, then 0.75 M of HCl dissociates to give 0.75 M H⁺ and 0.75 m Cl⁻
- Hence the total concentration of ions in a 0.75 M solution of HCl is 1.5 M (0.75 M H⁺ and 0.75 M Cl⁻)
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.
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.
Answer:
Explanation:
To solve this problem, we need to obtain the number of moles of the solute we desired to prepare;
Number of moles = molarity x volume
Parameters given;
volume of solution = 500mL = 0.5L
molarity of solution = 0.5M
Number of moles = 0.5 x 0.5 = 0.25moles
Now to know the volume stock to take;
Volume of stock =
molarity of stock = 4M
volume = = 0.0625L or 62.5mL
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