Answer:
142.82 g
Explanation:
The following data were obtained from the question:
Volume of water = 12 mL
Volume of water + gold = 19.4 mL
Density of gol= 19.3 g/cm³
Mass of gold =.?
Next, we shall determine the volume of the gold. This can be obtained as follow:
Volume of water = 12 mL
Volume of water + gold = 19.4 mL
Volume of gold =.?
Volume of gold = (Volume of water + gold) – (Volume of water)
Volume of gold = 19.4 – 12
Volume of gold = 7.4 mL
Finally, we shall determine the mass of the gold as follow:
Note: 1 mL is equivalent to 1 cm³
Volume of gold = 7.4 mL
Density of gol= 19.3 g/cm³ = 19.3 g/mL
Mass of gold =?
Density = mass /volume
19.3 = mass of gold /7.4
Cross multiply
Mass of gold = 19.3 × 7.4
Mass of gold = 142.82 g
Therefore, the mass of the gold pebble is 142.82 g
Answer:
N₂(g) + 4H⁺(aq) + 4e⁻ → N₂H₄(aq)
Explanation:
The half reaction for the reaction for the reduction of gaseous nitrogen to aqueous hydrazine is;
N₂(g) → N₂H₄(aq)
The balancing the atoms in the half reaction. Hydrogen atom is balanced by adding hydrogen ions (H⁺)
We have;
N₂(g) + 4H⁺(aq) → N₂H₄(aq)
Then we balance the charge on both sides by adding electrons where the positive charge is greater;
we have;
N₂(g) + 4H⁺(aq) + 4e⁻ → N₂H₄(aq)
Carbon dioxide in water is gas/liquid
Alpha particles have have 2 protons,<span> 2 </span>neutrons, ans 0 electron.
Answer:red green and blue hope it helps!,!!!!!!!
Explanation:
