The balanced chemical reaction is:
<span>2Na + 2H2O → 2NaOH + H2
</span><span>
We first use the amount of hydrogen gas to be produced and the molar mass of the hydrogen gas to determine the amount in moles to be produced. Then, we use the relation from the reaction to relate H2 to Na.
53.2 g H2 ( 1 mol / 2.02 g ) ( 2 mol Na / 1 mol H2 ) ( 22.99 g / 1 mol ) = 1210.96 g Na
1210.96 g Na ( 1 mL / 0.97 g ) = 1248.41 mL Na needed</span>
<h3>
Answer:</h3>
0.111 J/g°C
<h3>
Explanation:</h3>
We are given;
- Mass of the unknown metal sample as 58.932 g
- Initial temperature of the metal sample as 101°C
- Final temperature of metal is 23.68 °C
- Volume of pure water = 45.2 mL
But, density of pure water = 1 g/mL
- Therefore; mass of pure water is 45.2 g
- Initial temperature of water = 21°C
- Final temperature of water is 23.68 °C
- Specific heat capacity of water = 4.184 J/g°C
We are required to determine the specific heat of the metal;
<h3>Step 1: Calculate the amount of heat gained by pure water</h3>
Q = m × c × ΔT
For water, ΔT = 23.68 °C - 21° C
= 2.68 °C
Thus;
Q = 45.2 g × 4.184 J/g°C × 2.68°C
= 506.833 Joules
<h3>Step 2: Heat released by the unknown metal sample</h3>
We know that, Q = m × c × ΔT
For the unknown metal, ΔT = 101° C - 23.68 °C
= 77.32°C
Assuming the specific heat capacity of the unknown metal is c
Then;
Q = 58.932 g × c × 77.32°C
= 4556.62c Joules
<h3>Step 3: Calculate the specific heat capacity of the unknown metal sample</h3>
- We know that, the heat released by the unknown metal sample is equal to the heat gained by the water.
4556.62c Joules = 506.833 Joules
c = 506.833 ÷4556.62
= 0.111 J/g°C
Thus, the specific heat capacity of the unknown metal is 0.111 J/g°C
Answer:
AM
Explanation:
to go from moles to grams you multiply by the Atomic Mass or Molar Mass (Atomic Mass for an element and Molar Mass for a compound).
The formula is:
Mass = moles * MM
Answer:
d. The gold(III) ion is most easily reduced.
Explanation:
The standard reduction potentials are
Au³⁺ + 3e⁻ ⟶ Au; 1.50 V
Hg²⁺ + 2e⁻ ⟶ Hg; 0.85 V
Zn²⁺ + 2e⁻ ⟶ Zn; -0.76 V
Na⁺ + e⁻ ⟶ Na; -2.71 V
A <em>more positive voltage</em> means that there is a <em>stronger driving force</em> for the reaction.
Thus, Au³⁺ is the best acceptor of electrons.
Reduction Is Gain of electrons and, Au³⁺ is gaining electrons, so
Au³⁺ is most easily reduced.
A.
everything in existence.
