Answer:
2.05mg Fe/ g sample
Explanation:
In all chemical extractions you lose analyte. Recovery standards are a way to know how many analyte you lose.
In the problem you recover 3.5mg Fe / 1.0101g sample: <em>3.465mg Fe / g sample. </em>As real concentration of the standard is 4.0 mg / g of sample the percent of recovery extraction is:
3.465 / 4×100 = <em>86,6%</em>
As the recovery of your sample was 1.7mg Fe / 0.9582g, the Iron present in your sample is:
1.7mg Fe / 0.9582g sample× (100/86.6) = <em>2.05mg Fe / g sample</em>
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I hope it helps!
Answer:
b)The molar mass of the particles.
Explanation:
The rate of effusion of a gas is inversely proportional to the square root of its molar mass (Graham's law), a relationship that closely approximates the rate of diffusion. As a result, light gases tend to diffuse and effuse much more rapidly than heavier gases.
Answer:
12 mol O2
Explanation:
This is assuming you're using a combustion equation of
2 H2S + 3O2 -> 2H2O + 2SO2
Next time include the equation because that is where you will find your mol to mol ratio
8 mol H2S x (3 mol O2 / 2 mol H2S) = 12 mol O2
Answer:
0.625 mol
27.5 g
Explanation:
Given data
- Molar mass (M): 44.01 g/mol
- Temperature (T): 19.0°C + 273.15 = 292.2 K
We can find the moles of dinitrogen monoxide gas using the ideal gas equation.
P × V = n × R × T
n = P × V/R × T
n = 0.500 atm × 30.0 L/0.0821 atm.L/mol.K × 292.2 K
n = 0.625 mol
Then, we can find the mass corresponding to 0.625 moles.
0.625 mol × 44.01 g/mol = 27.5 g
