Answer:
I answered first, give me brainliest
Explanation:
Answer:
857.2947
Explanation:
You start with what you’re given 82g of H20 and multiply by 1 mole and the actual grams of h20 which is 18.0 then you do the mole. 1 mol of O2 and 2 mol of H2O times 6.02 x 10^23
Use the volume and density of the gold statue to calculate the mass of the statue.
density = mass / volume, or rearranged, mass = density • volume
Convert the volume of the statue from L to mL so volume unit agrees with density unit
mass of statue = 19.3 g/mL • 1000 mL = 19 300 g
Mass of sand must be same as the mass of the statue, 19 300 g
Use the mass and density of the sand to calculate the volume of sand needed
density = mass / volume, or rearranged, volume = mass / density
volume of sand needed = 19 300 g / 23 g/mL = 8391 mL or 8.391 L
1b
Calculate the density of the statue from the measured mass and volume. If the calculated density agrees with the known density of gold, then the statue is made from pure gold.
density = mass / volume
Convert the mass from kg to g as you want the answer in g/mL so you can compare it to the reference value of gold given in the problem. 16.5 kg • 1000 g / 1 kg = 16 500 g
density of the statue = 16 500 g / 954 mL = 17.3 g/mL
Since this density, 17.3 g/mL is significantly different from the known density of gold, 19.3 g/mL, the statue cannot be made of pure gold. The gold was mixed with a less dense metal.
What are the options the your question
Given:
0.060 mol of NiC2O4
Ksp = 4 x 10⁻¹⁰
1.0 L of solution
Kf of Ni(NH3)6 2⁺ = 1.2 x 10⁹
<span>NiC2O4 + 6NH3 ⇋ Ni(NH3)6 2+ + 2O4 2- </span>
<span>NiC2O4 ⇋ Ni 2+ + C2O4 2- ...Ksp </span>
<span>Ni2+ + 6NH3 ⇋ Ni(NH3)6 2+...Kf </span>
Ksp * Kf = (4 x 10⁻¹⁰) * (1.2 x 10⁹) = 0.48
K = 0.48 = [Ni(NH3)6 2+][C2O4 2-] / [NH3]⁶<span>
</span>0.48 = (0.060)² / [NH3]⁶<span> ... (dissolved C2O4 2- = 0.060M)
</span><span>[NH3]</span>⁶<span> = (0.060)</span>²<span> / 0.48 = </span>0.0036 / 0.48 = 0.0075
NH3 = ⁶√0.0075
NH3 = 0.44 M
