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!
By definition, one mole (one gram molecular weight) of any substance, contains Avogadro’s number of particles; atoms if you are discussing an element, or molecules if a compound. Avogadro’s number has been determined by several methods, all of the accepted values lie within a range of +-1% about the value of 6.022045 x 10^23/gm. That is a large number, in this case approximately; 602,204,500,000,000,000,000,000 molecules of glucose.
From the web :v
Answer:
On treatement of 2 -with hydrogen gas and lindlar catalyst, the major product obtained is cis-2butene.
Answer:
Kc = 168.0749
Explanation:
initial mol: 0.822 0 0
equil. mol: 2(0.822 - x) x x
∴ [ HI ]eq = 0.055 mol/L = 2(0.822 - x) / (1.11 L )
⇒ 1.644 - 2x = 0.055 * 1.11
⇒ 1.644 = 2x + 0.06105
⇒ 2x = 1.583
⇒ x = 0.7915 mol equilibrium
⇒ [ H2 ] eq = 0.7915mol / 1.11L = 0.7130 M = [ I2 ] eq
⇒ Kc = ([ H2 ] * [ I2 ]) / [ HI ]²
⇒ Kc = ( 0.7130² ) / ( 0.055² )
⇒ Kc = 168.0749
Answer: The answer is 0.245
Explanation: That is what it said the answer was