The density of the sample is:
Density = mass / volume
Density = 9.85 / 0.675
Density = 14.6 g/cm³
If the sample has 95% gold, and 5% silver, its density should be:
0.95 x 19.3 + 0.05 x 10.5
Theoretical density = 18.9 g/cm³
The difference in theoretical and actual densities is very large, making it likely that the jeweler was not telling the truth.
Compete Question:
What mass of CDP (403 g mol−1) is in 10 mL of the buffered solution at the beginning of Experiment 1?
Passage: "16 mmol of CDP in 1 L of buffer"
Answer:
6.4 × 10-2 g
Explanation:
we are given from the question that 16 mmol of CDP is in 1 L of buffer
this mean that we have 
moles of CDP in 1 liter of buffer.
so the mass of CDP in one liter of buffer will be calculate as,
mass of CDP = × 403g mol−1
= 
= 6.4 g/L
But because the question
asks us about the mass of CDP in 10 mL of solution, we will go further to calculate it like this:
6.4 g/L × 10 mL
6.4 g/L × 0.01 L = 
Answer:
pH= 2- log3
Explanation:
H2SO4 + H2O -> HSO4^(-) + H30^(+)
0.03M ___ ___
___ 0.03M 0.03M
H30^(+) : C = 0.03M
pH= - log( [H3O^(+)] ) => pH= - log {3× 10^(-2)} => pH = 2 - log3
The shape of BeH2 is linear.
Answer:
The two-step mechanism is a slow mechanism and a fast mechanism. When we combine them, the result is
2H₂O₂ (aq) ⇒2H₂O (l) + O₂ (g)
Explanation:
We know that the decomposition of hydrogen peroxide is catalyzed by iodide ion, which means that the iodide ion will react with the hydrogen peroxide. There is a slow mechanism and a fast one:
H₂O₂(aq) + I₋(aq) ⇒ H₂O(l) + IO₋(aq) this is the slow reaction
IO₋(aq) + H₂O₂(aq)⇒ H₂O(l) + O₂(g) + I₋ (aq) this is the fast reaction
If we cancel the same type of molecules and ions, the final result is:
2H₂O₂ (aq) ⇒2H₂O (l) + O₂ (g)
